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Chemistry Question Pack: Passage 1 1) This question asks about the possible bonds formed by each of the molecules in the answer choices. Additionally, going through our choices, we see four distinct molecules: Methane, glycine, carbon dioxide, methanol. We have to determine which molecules can form extensive networks of intermolecular hydrogen bonds with both participating. What do we know about hydrogen bonds? Hydrogen bonds are formed when a hydrogen atom bonds with a very electronegative atom. In a hydrogen bond, hydrogen atoms gain a partial positive charge, and the electronegative atom gains a partial negative charge. Let’s also pull up and reference Table 1. We can also draw our four options here. Carbon dioxide does not have any hydrogen atoms, while methane has 4 hydrogen atoms, all bonded to the same carbon. Alternatively, glycine and methanol have hydrogens bonded to different atoms, and are able to accept and donate hydrogens, so an extensive network of hydrogen bonds can be formed.
2) HCl is one of the common strong acids, and this makes sense because archaebacteria thrive in extreme pH. We’re given the molarity of a pool and asked to solve for pH. After reading the question, you can tell right away that it is almost like a standalone question. It can be solved without using the passage. The only information needed to solve for pH is the molarity provided, and our general knowledge. pH can be solved as the negative log of hydrogen ion concentration. Because we know our HCl is going to dissociate completely, the hydrogen ion concentration ends up being the same as our molarity. Our molarity is 0.01 molar which can be rewritten using scientific notation. (If you don’t remember which way to move the decimal point, go back to our mnemonic LARS, which is Left add, right subtract). We start with 0.01 x 10^0. We move the decimal point two places to the right to get a whole number, meaning we subtract two from our exponent to give us 1.0 x 10^-2 molar. Let’s plug in our numbers. We know ph is the negative log of our hydrogen ion concentration. Our hydrogen ion concentration, again, is the same as our molarity. So plug in 1x 10^-2 as our hydrogen ion concentration. So when we take our negative log, we find the pH is 2. For the MCAT, pH is measured as the negative log of hydrogen ion concentration. We’re going to see values between 0 and 14. We know acids will have a pH below 7 on the scale, and bases will have a pH above 7 on the scale. So after solving for our pH of 2, we can go back to our answer choices to see which one of them make sense. Remember we said pH below 7 means an acid, a pH above 7 means a base.
3) This is going to be similar to our last question. It’s tangentially related to what we read in the passage, but the passage isn’t completely necessary to answer the question. This is like a standalone question that most likely can be answered without the use of the passage. First and foremost, to be able to identify the compound with the same geometry as methane, we have to know the geometry of methane itself, then compare with the additional answer choices. Methane has a central carbon atom bonded to four hydrogen atoms. The four hydrogens are organized as far apart as possible from one another at around 109-degree bond angles. We use VSEPR theory and know the atoms in the molecules achieve a geometry that minimizes repulsion between the electrons in the valence shell of the atom. We know methane has a tetrahedral molecular geometry, because it has four electron dense areas, and no lone pairs. Keep in mind we used VSEPR theory to determine methane has a tetrahedral molecular geometry. It is imperative to know the concepts on the MCAT content outline, and this is no different. When given the number of electron dense areas and lone pairs, you should be able to determine the molecular geometry of a compound.
4) The question is ultimately asking us “which answer choice has similar chemical properties to oxygen?” Elements with similar chemical properties to oxygen will most likely accept electrons in a manner similar to oxygen. Elements that are found in the same group on the periodic table tend to have similar characteristics and chemical properties. We can consult with the periodic table of elements, and it’s unlikely we have to revisit our passage to answer the question. The best electron acceptors are those which have the highest electron affinity. Electron affinity increases to the right, and up the periodic table. Looking at our periodic table, we see sulfur is found in the same group as oxygen, meaning it will have similar chemical properties, and also have a high electron affinity compared to our other answer choices. Elements that are found in the same group on the periodic table tend to have similar chemical properties. This periodic table is going to be readily available to you on test day, so make sure you get used to using it as necessary while you are practicing.
Chemistry Question Pack: Passage 2 5) This question is asking us to decide whether limonene or positive carvone will distill first. Why did I phrase it this way? Because even though the test-maker doesn’t mention the two components by name, we recall the components of caraway seed oil. Here we have our two structures from the passage. Like I mentione, we’ll look at the structure of both carvone and limonene to compare. Carvone has a hexane ring, just like limonene. The biggest difference is the additional carbonyl group, which I’ve circled in red. How does that affect the properties of the two compounds, and finally the boiling point? We’ll have to rely a bit on our content. Oxygen is more polar than carbon, which makes carvone more polar than limonene. More polar compounds have higher boiling points. Said differently, limonene has a lower boiling point than carvone. The additional double bond is also an indicator that carvone has a higher boiling point. Now what are we doing with this information? Let’s discuss the experiment as well. During distillation, temperature is slowly increased until one of the components begins to vaporize out of the solution. This vapor is then condensed into a different container. The component that vaporizes first will do so at a lower temperature than the second. Therefore, the component that has the lower boiling point (limonene) will vaporize first. So before even looking at our answer choices, we have a predicted answer, with some solid reasoning behind the prediction. We can go back to our answer choices and see if this matches any of our answers.
6) Here we have an excerpt from our passage, as well as the figure. When I pull up these excerpts from the passage, it’s not so you go back and re-read every single detail, but rather so I can demonstrate my thought process. I show you exactly what I’m referencing from the passage so you can follow along at the same time. When you’re actually practicing, and when you’re taking the exam, I don’t recommend going back to the passage unless you’re looking for details in the form of data or figures. The passage says an ebulliator was lowered into the distillation flask to introduce small air bubbles into the system. The author gives us the explicit function of the ebulliator. That ebulliator creates an area of imperfections for gases to collect and start boiling. So essentially it’s a boiling chip. In perfectly clean, smooth glassware there’s nowhere for gas bubbles to start forming at boiling temperatures. Without the ebulliator, you might super-heat your liquid, or there may be bumping or splashing. The ebulliator is also helping the solution begin boiling at the correct temperature.
7) We know limonene and positive carvone are separated based on their boiling points, but we can revisit the passage to see what the author says specifically about the purpose of the experiment. Here we have an excerpt from the passage. It says Because the two compounds have very different boiling points, the chemist decided to separate them by vacuum fractional distillation. The entire experiment revolves around this boiling point difference, and separating the two components of caraway seed oil by vacuum fractional distillation. We’re looking for the answer choice that would most likely improve the separation between the limonene and positive carvone. Let’s go through our answer choices and see which would most improve the experiment.
8) To answer this question, we’re going to look at our answer choices and see which carbons we want to look at specifically. We know right away by glancing at the answer choices that our efforts should be focused on carbons 2, 5 and 7 because those are the only carbons listed in the answers. We have to remember the distinction between our positive and negative carvone enantiomers. The difference is the positive carvone is the enantiomer that rotates plane-polarized light in the positive direction, and the negative carvone is the enantiomer that rotates plane-polarized light in the counterclockwise, or negative direction. Here we have carvone with carbons 2, 5, and 7 labeled. We’re told Carvone exists as two different enantiomers in the passage. Enantiomers are non-superimposable mirror images of each other that differ in configuration. Enantiomers have opposite absolute configurations at every chiral carbon. Chiral carbons have 4 distinct substituents. So, with that in mind, let’s dig into the structure of our carvone, so we can find the ways positive and negative carvone differ. The passage mentions the two are enantiomers, meaning they will have opposite configurations at chiral carbons. Let’s first look at carbons 2, 5 and 7 to see which have a stereocenter and four distinct substituents. – Carbon 2 has a double bond, so not a stereocenter Ultimately, we’re going to look for an answer choice that has carbon 5.
9) The test-maker is asking the effect of a malfunction in the vacuum distillation in which the vacuum leaks. If the vacuum is not working correctly, it’s not properly lowering the pressure inside our distillation apparatus. We want to know what happens to the boiling points of carvone and limonene at higher pressures. The question is simply asking the correlation between pressure and boiling point. Vacuum distillation is distillation performed under reduced pressure, which ultimately decreases the boiling point of compounds. Why would we do this? Because sometimes compounds can have very high boiling points that are hard to reach. These high temperatures could adversely affect the compound itself-we can see denaturing, or the experiment just won’t work. A leak in the apparatus will undo this vacuum effect. That means a higher pressure, and we should see an increase in the boiling point of the compounds.
Chemistry Question Pack: Passage 3 10) The values to calculate the number of sodium ions will come from the passage, but we’ll be solving the problem with general knowledge. One of the most important things to keep in mind for this question or for any math problems in general, is we have to keep our units correct as we work through our question. We can pull up part of the passage here as reference. The passage mentions there are 7.15 g of Na2CO3 in the colorless solution, and also gives the molar mass. We can find the number of moles of sodium carbonate by dividing the 7.15g in the solution by the molar mass to get 0.0250 moles sodium carbonate. Let’s also take a closer look at the formula for sodium carbonate. For every 1 mole of Na2CO3, there are 2 moles of sodium ions (Na+), so the solution contains 0.050 mol Na+. Avogadro’s number is known as the number of particles (in this case it is ions, but it can be any unit) per mole of a substance. (6.022 x 1023 ions per mole)(0.05 mol Na+) = 3.00 x 1022 This corresponds to answer choice B. I’ve also written everything out below so it’s easier to follow the math: 11) This question is simply testing our ability to know about the acidity or basicity of sodium carbonate. Litmus paper is used to test substances of at various pHs. Red litmus paper turns blue in base, while blue litmus paper turns red in acid. If red litmus paper is added to an acid, it remains red, and if blue litmus paper is added to a base, it remains blue. The first sentence from our passage mentioned the class was experimenting with solubility equilibrium. We expect that in an experiment dealing with solubility equilibrium, the solids are dissolved in water, and the base particles that make up the solids will be form. Sodium carbonate is a water-soluble salt, so it’s neutral and wouldn’t change the color of the litmus paper. However, when sodium carbonate is in water, it’s broken down into sodium ions, and we have our carbonate ions. When carbonate ions are in water, they can accept protons, therefore leaving bicarbonate and hydroxide ion products. We know bicarbonate from pH buffering in the human body, but we’re focused on the hydroxide ions for now. Hydroxide ions are strongly basic, so we’re expecting a basic pH.
12) This is a basic math problem. The precipitation reaction forms nickel carbonate as a product-we just need to know exactly how much (maximum number of moles). After we balance the equation discussed in the passage, we find the number of moles of nickel carbonate. Find the number of moles nickel. Divide grams of sample by molar mass to get moles of NiSO4. There is an equal number of moles of Ni: 6.57 g NiSO4/(262.84 g/mol) = 0.025 mol. We can find the number of moles of Na2CO3 by dividing the 7.15g by the molar mass to get 0.0250 moles Na2CO3. For every 1 mole of Na2CO3, there is 1 mole of CO3 2-. There is no limiting reactant between the two as there is an equal number of moles. I’ve worked out the math with the essential information from the passage below: We can pick our correct answer, answer choice A: 0.025 mol. 13) From the passage, we learned that during experimentation a gas formed, and we need to identify it. We’ll rely on the passage for details about the experiment, but we will use general knowledge to balance chemical equations and quantify the number of moles. Second paragraph of our passage is above. Two of the initial solutions are mixed, and then “To one portion is added an excess of 6 M HCl, which results in the disappearance of the precipitate and a rapid evolution of a gas.” We can write down the essential information like we did in our previous problems. Normally, I have no problem typing everything out, but for math problems and chemical equations this tends to be easier to read: Two possible precipitates form: Na2SO4 or NiCO3. What kind of metals form green precipitates? Normally chromium, iron, and nickel form green precipitates. This hints at the fact that our precipitate is likely nickel carbonate NiCO3. (Na2SO4 would likely be white) We add hydrochloric acid to our green precipitate (our nickel carbonate), the precipitate dissolves and a mystery gas is evolved. Wow, so what just happened? Where’s did our precipitate just go? And why is there gas everywhere? Nobody freak out! Let’s break this down by writing out the double replacement reaction Solid nickel chloride doesn’t help answer this specific question, but we might recognize carbonic acid from respiratory physiology. What we know about carbonic acid is it exists in chemical equilibrium with carbon dioxide gas and water. In fact, we expect there are only going to be trace amounts of aqueous carbonic acid, and instead predominantly carbon dioxide and water. We mentioned earlier in our breakdown that our precipitate is likely nickel carbonate NiCO3, so we can automatically eliminate answer choices A and B. Looking at C and D, only C matches what we expect to see in the experiment (carbon dioxide). 14) Different colors were formed during the experiment because of one specific ion. We’re just asked to identify this ion, so we’ll reference the passage to find out the components in the colored solutions and details of the solution, but the answer may ultimately come from general knowledge and our knowledge of the colors that different metals ions form. Let’s glance at our answer choices quickly. We have sulfate and nickel as the two recurring ions that may cause the color of the solution. I mentioned something about this in our previous question. What kind of metals form green precipitates? Normally chromium, iron, and nickel form green precipitates. What do they have in common? Complex ions containing transition metals are usually colored, and then similar ions from non-transition metals (like sulfur) are not. What does that mean for our question? That suggests that the partly filled d-orbitals (which are characteristic of transition metals) should be involved in generating the color in some way. Answer choice C is going to be our best choice here. Nickel is a transition metal, and we said the reason for the color of the solutions is the unfilled d-orbitals. It does not have to do with charge. Chemistry Question Pack: Questions 15-19 15) This is a standalone question, but we’re going to rely on something presented to us in the question stem to answer, rather than our general knowledge alone. We’ll have to compare the structure of these two compounds and their anions. Alternatively, we can say Compound 1 would be a stronger acid than Compound 2 because Compound 1 has a more stable conjugate base when it loses a hydrogen atom. Phrasing our question differently, why is the conjugate base of Compound 1 more stable than the conjugate base of Compound 2? What do we know about about anions? Anions are negatively charged ions. The carboxyl group loses its H and becomes negatively charged carboxylate ion. Why does this happen? Carboxyl groups will ionize often, meaning carboxyl groups in general are acidic functional groups. Ionizing means the carboxyl can break its double bond, which allows it to go to a lower energy state and increase its stability. Then after the hydrogen atom leaves, the remaining 2 oxygen atoms share the remaining negative charge in the carboxylate.
16) We have 4 different substituents around the central carbon. The shaded-in triangles mean a substituent is coming out of the screen. The dotted lines mean the substituents are away from view, into the screen. We can say the grouped substituent orientations should stay the same. What does this mean? The carboxylic acid (COOH) and the hydroxyl group (OH) should be in the same orientation, while the hydrogen atom and the methanol should be in the same orientation. Next thing we want to make sure, is whether the compounds are in an R or S configuration. To do that, go from the highest priority substituent and draw a curved arrow toward the lowest priority substituent. In our question, we would do that from carboxylic acid to hydrogen>the arrow is curving left. Our compound is S-configuration (S stands for sinister, or “left” in latin). We’re ensuring the compounds that had identical orientations in compound A continue to have identical orientations in the correct answer choice. We also want the correct answer choice to have S-configuration a. Answer choice A is an enantiomer of compound A, it has R-configuration. b. Answer choice B is just answer choice A rotated 90 degrees and is also is an enantiomer of compound A, it has R-configuration c. Answer choice C is also an enantiomer of compound A. It has R-configuration. It’s the same as answer choice B, but again rotated 90 degrees. d. Answer choice D is the exact same structure as compound A, but it is rotated 90 degrees on screen. It’s identical to compound A and has S-configuration. That means we can pick answer choice D as our best option here. 17) We’re breaking a triglyceride up into its base components, and we want to pay close attention to our R groups and our R’, or R prime, groups. What are triglycerides made of? They’re glycerol molecules combined with three fatty acids. There’s a dehydration reaction and the glycerol forms ester bonds to link with the three fatty acids. Here we’re undoing our triglyceride. Our glycerol is hydrolyzed, meaning our ester bonds are broken, to give 2 fatty acids R, 1 fatty acid R’, and 1 mol glycerol a. Answer choice A has three R groups, which contradicts what our initial question said. We’re expecting 2 R groups and 1 R’ group b. Answer choice B has three R’ groups. This again contradicts our question stem. c. Answer choice C has two R groups and one R prime group. We like this answer choice of the ones we’ve gone through. d. Answer choice D has one R group and two R prime groups which contradicts our breakdown of the question. We stick with our best answer, answer choice C. 18) We’re given compound I and compound II and at first glance it looks like the only difference is a nitrogen atom in compound I and one fewer hydrogen atom, while compound II has a carbon atom and an additional hydrogen where compound I has a nitrogen atom. This question is about solubility. When we look at different solutes and solvents, remember the mantra that’s always drilled in our heads during chemistry classes “like dissolves like.” A polar substance will dissolve best in a polar solvent; a nonpolar substance will dissolve best in a nonpolar solvent. This question is asking which of the two Compounds (I or II) is more similar to water. Nitrogen and oxygen can form hydrogen bonds, while carbon does not form hydrogen bonds. Hydrogen bonding happens between a hydrogen atom and an electronegative atom. For the sake of the MCAT especially, we think of fluorine, oxygen, and nitrogen as the electronegative atoms in hydrogen bonding (FON atoms). We’re expecting the compound with nitrogen (Compound I) is more soluble in water.
19) Most likely, the student is going to be separating the starting material from the ending material. The initial material has carbonyl group; the ending product has a hydroxyl group. The only additional information we’re given about the reaction is that the initial material reacts with sodium borohydride. Visualize this thin-layer chromatography experiment. Rf is the distance travelled by the materials, so materials that interact more strongly with the silica plates will have lower Rf (less distance travelled), materials that interact less strongly will have higher Rf values. In thin-layer chromatography, glass is coated with a thin layer of adsorbent, often silica gel or aluminum oxide, and the mixture is spotted on this layer, similar to paper chromatography. Different substances are going to ascend the plate at different rates and be separated based on their polarity when using thin-layer chromatography. Both the hydroxyl group and the carbonyl group are polar functional groups, but the hydroxyl group can form hydrogen bonds-and that’s what makes it more polar than carbonyl groups. This means the product with the hydroxyl group binds with the silica gel more strongly and will have a lower Rf
Chemistry Question Pack: Passage 4 20) CDP is the monomer discussed in Passage 4. In this question, we’re given the molar mass and asked about Experiment 1 from the passage specifically, so we’ll likely need to go back for specific details about the experiment. We have to make sure we pay particularly close attention to our units and prefixes. First bit of information: from our question we’re told the Molar mass of CDP is 403 g/mol and we need to find the mass CDP (in grams) in 10 mL of the solution in experiment 1. We’re told in Experiment 1, the chemist dissolved 16 mmol CDP in 1L solution. This gives us a molarity of 16mM, which can be converted into 0.016 M. There are 1000 millimoles in a single mole. Next, to find the number of grams in a single liter of solution, multiply: 0.016 molar x 403 g/ 1 mol = 6.4 grams CDP per liter of solution Remember, we need to know the mass of CDP in 10 mL of solution. Let’s adjust our units once again. We need to know how many liters are in 10 mL. There are 1000 mL in 1 L, so this means a volume of 0.01 L. See how important it is to keep track of units? Multiply: 6.4 grams CDP/L x the volume 0.01L = We’re left with 0.064 g CDP OR 6.4 x 10-2 g Remember our mnemonic LARS for converting to scientific notation, which is Left add, and in this case, right subtract. I want you to note how important it is to keep track of your units during every math problem. Just knowing the proper units can help you get to the right answer, even if you’re unsure exactly how to solve a problem. Glancing at our choices, the initial digits stay the same in all 4 answers, but the exponent next to the 10 varies. This should really hammer the point home, you have to be careful with your units and prefixes to get the correct answer. This question was a math problem, so we can eliminate any incorrect values that don’t match our calculated value. We’re left with our correct answer, answer choice C: the mass of CDP in 10 mL of solution is 6.4 x 10-2 grams. I’ve also handwritten the key points here so it’s easier to follow my math: 21) Recall from the passage and the previous question that Experiment 2 improved the yield of the polymer that was obtained in Experiment 1. In this new scenario, the chemist is going to increase the amount of a component in the experiment to increase the yield of polymer. We said the polymer was a product in Equation 1, so we’re trying to increase the yield of products.
22) We’re solving for the equilibrium constant of Equation 1, which we write out here. We had to go back to the passage to find Equation 1, but do we need any additional information to find the equilibrium constant? We should know the equilibrium constant for a general reaction, so we should be able to apply that to Equation 1. That makes this a standalone question. What do we know about equilibrium constants for reactions? The equilibrium constant (Keq) for any general reaction is the following: aA+bB⇌cD+dD The products of the reaction (C and D) are placed in the numerator, and their concentrations are raised to the power of the coefficients from the balanced equation. The reactants (A and B) are placed in the denominator, with their concentrations raised to the power of their coefficients. The concentrations of both aqueous solutions and gases change during the progress of a reaction. For reactions involving a solid or a liquid: the amounts of the solid or liquid will change during a reaction, but their concentrations won’t change during the reaction. Instead, the values for solids and liquids will remain constant. Because they are constant, their values are not included in the equilibrium constant expression. Given this, we can put Equation 1 in the proper format: Let’s look at our four choices and decide: Answer choice A correctly lists the concentration of the polymer in the numerator, and the concentration of hydrogen phosphate raised to the power of its coefficient (n) in the balanced equation. The denominator also correctly lists the concentration of the reactants (which is our CDP) raised to the power of its coefficient (n) as well. This is going to be our best answer choice. Answer choice D is similar, but note we’re only getting the concentration of CP, not the polymer. Answer choice B does not raise the concentration of hydrogen phosphate to the power of its coefficient in the numerator. Answer choice C does not raise the concentration of CDP to the power of its coefficient in the denominator. Answer choice A is our correct answer. 23) We’re comparing the concentration of our two products. We can write out our Equation 1 that we need from our passage. In the balanced equation, we can add a “1” in front of a molecule that doesn’t have a coefficient already. Every 1 molecule of (CP)n that is formed, there are n HPO4 2- that are formed. So in the solution, the ratio of (CP)n to HPO4 is 1/n. Compared to the concentration of HPO4, concentration of CPn is 1/n[ HPO4] For every 1 molecule of polymer that’s formed, there’s N hydrogen phosphates formed. So in the solution, the ratio of polymer to hydrogen phosphate is 1/n. And compared to the concentration of hydrogen phosphate, the concentration of the polymer is 1/n times the concentration of hydrogen phosphate. So that’s one way of going through this. We can also look and think about this problem a bit differently as well. We can also say every time you make one polymer particle, another hydrogen phosphate is formed as well. The actual number of polymers formed can be large, meaning “n” can be any number for the sake of this question. The polymer, and the hydrogen phosphate are produced at a 1:n ratio. To find the concentration of the polymer in terms of the concentration of the hydrogen phosphate, you have to divide the concentration of the hydrogen phosphate by N. Let’s go through our 4 answer choices:
24) In other words, we’re trying to compare the concentration of hydrogen phosphate anion to the concentration of dihydrogen phosphate in buffer solution. There are two big details we want to note: We are given pKa here in the question stem, and the question references the buffer solution from Experiment 1. Let’s look back at Experiment 1 in the passage. 16 millimol CDP dissolved in 1 L aqueous solution with PNP and magnesium ion. Then we’re told the solution is buffered at a pH of 8.7. Remember, we’re answering a question about acid/base equilibria. We were given the pKa for dissociation of hydrogen phosphate anion to the concentration of dihydrogen phosphate. What is the concept being tested here? Looking closely, we’re comparing an acid and its conjugate base. The conjugate base is the hydrogen phosphate anion, and the acid is dihydrogen phosphate. The chemist buffers the solution at pH 8.7 A little background here, and some key points you want to note:
pH=pKa + log[A–]/[HA] [A–] corresponds with the conjugate base (hydrogen phosphate anion), while [HA] corresponds with the acid (dihydrogen phosphate). Plug in the values we’ve been given. The passage tells us pH=8.7 and question tells us pKa is 6.7 2.0= log [A–]/[HA] Anti log on both sides means: 10^2 = [A–]/[HA] and ratio is 100 : 1
Chemistry Question Pack: Passage 5 25) The question explicitly says “according to the passage.” You may have noticed this doesn’t always mean you have to flip back to the passage to answer the question. For example, we should remember from the passage that the cleavage is catalyzed by a base. We’ll keep that in mind for the time being, but what I want you to realize is, ideally, you only go back to the passage rarely for specific details. However, as I go through these solutions you’ll find I pull up excerpts from the passage so you can follow my thought process, and so I’m thorough. Our passage says large-scale synthesis of Olestra starts with a base-catalyzed cleavage. So this confirms what we said previously about it being base catalyzed. The figure in the passage shows us base-catalyzed cleavage of a triglyceride. So, our correct answer should be a base. That’s all the information we needed from the passage, but we still need to be able to identify a base. Strong acids and strong bases are listed on AAMC’s content outline, so make sure you know these most common ones. And we also want to know their properties. Strong bases will dissociate completely in an aqueous solution while yielding hydroxide ions. We expect the compound the worker uses to catalyze the cleavage of the triglyceride could be one of the 8 bases listed here.
26) In other words, we’re given stats for the combustion of 10 peanuts and asked about the stats of a single peanut. We need the specific heat of water, which should be general knowledge: 1 cal/g*oC. That should be the only information we may need from the passage, so this is almost like a standalone question. We’re going to have to use dimensional analysis, so we’ll write our numbers down and keep our units consistent. Our final answer is in kilocalories. We’re dealing with 1 kg of water in a calorimeter. So to use the specific heat for water, we can convert the units on the 1 kilogram of water, to grams. Dimensional analysis, multiply by a factor (1000 grams/1kg) which gives us 1000 grams water. Our temperature is already in Celsius. So last unit, 1 kilocalorie is 1000 calories, and we raised temperature 50 degrees Celsius with 10 peanuts. So solving for the energy content of 10 peanuts, we get 50,000 calories, or 50 kilocalories because our answers are in kilocalories. Lastly, we want the kilocalories for a single peanut, so we can divide the 50 kilocalories in 10 peanuts by 10. We get an energy content of 5 kilocalories per peanut. I’ve handwritten my work below so it’s easier for you to follow along: Our calculation corresponds to answer choice C: 5 kcal. 27) Macronutrients provide a certain number of dietary calories per gram. We have to figure out how many are in a one gram sample of olestra. We know the calories per gram of other major macronutrients (that includes fat), but not of Olestra.
However, the passage said the Olestra is not metabolized. What does that mean? There are no dietary calories contributed to human consumers. We know that any value that isn’t 0 is incorrect. We can pick answer choice A as our correct answer. 28) The structure of glycerine was provided in the passage, but we should know the structure. These structures come in AAMC’s content outline categories 5B and 5D. If you need to review the content outline, we have it on our website with the information we think is most important to know about each topic. We can compare glycerine’s structure with the structure of isopropyl alcohol: Glycerine has three carbons, each also bonded to a hydroxyl group. Isopropyl alcohol is our simplest secondary alcohol, and has one hydroxyl group. What are our big differences here? We look at our glycerine, there are three hydroxyl groups (labeled in red). We look at our isopropyl alcohol, we see a single hydroxyl group. The hydroxyl groups in each molecule, they’re able to hydrogen bond. We know that hydrogen bonding is strong. It takes lots of energy to break them, so an increase in the number of hydrogen bonds increases the boiling point of our molecule. Just looking at the number of potential hydrogen bonds in each molecule, the boiling point of glycerine should be much higher.
Chemistry Question Pack: Passage 6 This passage and the related questions relied heavily on knowing periodic table trends. Make sure to review if you’re not comfortable with any of the topics: https://jackwestin.com/resources/mcat-content/the-periodic-table-variations-of-chemical-properties-with-group-and-row 29) This is almost like a standalone question. It’s tangentially related to the passage we just read, but we don’t need the passage to answer this question. Our passage was more of a history lesson, and didn’t provide any specific details about the elements in the periodic table. Instead, we need to pull up the periodic table: We can locate the four answer choices on the table. Our atomic radius is highest all the way to the left, and at the bottom of the table, so we can immediately pick the leftmost and downmost element as our answer choice, which is sodium. We were able to compare all 4 answer choices at once and pick the best answer: Answer choice A. The MCAT knows you most likely won’t know every specific value for elements on test day. The radius of two different elements could be nearly identical, but the testmaker knows you can’t possibly memorize such minute details before taking the exam. That’s why it’s important to know periodic trends. 30) Similar to our last question, we don’t need the passage to answer this question. We’re given 4 atoms, and we’re picking the answer choice that has the largest first ionization energy. Our passage didn’t actually provide specific details about the elements in the periodic table. Where is ionization energy the highest? To the top and to the right of the periodic table. The first ionization energy is the energy that is needed to knock off the first valence electron of an atom. In general, ionization energy increases to the right across a period. And it also increases up a group because of decreasing radii. So, a noble gas like krypton will have the highest nuclear charge and is also the least willing to give up a valence electron. It already has a noble gas valence electron configuration, so it’s not going to want to change that. An alkali metal like potassium has a much lower ionization energy. Potassium can give up a valence electron and achieve noble gas valence electron configuration, so it’s willing to part with the valence electron. We’re left with our best answer, answer choice D: Krypton. 31) We have a strontium isotope, and the mass number we’re given is 90. We’re solving for the combined number of protons, neutrons, and electrons in the isotope. This is our third straight question where we’re going to rely more on the periodic table than we are our passage to answer the question. Find Strontium on the periodic table (element 38). We know an uncharged element has an equal number of positively charged protons and negatively charged electrons. We also know the mass number is the sum of all nucleons. I mentioned strontium is element 38 on the periodic table. So, the number of protons and electrons both should be 38. We’re given a mass of 90. So, solving for neutrons is the total mass of 90 minus the atomic number 38, or the number of protons. This gives us 52 neutrons. We can add up 38 protons, 52 neutrons, and 38 electrons giving us a total sum of 128. Quick recap: We know an isotope’s mass number is going to be the sum of the number of protons and neutrons, so those two combined is 90. We just have to add the number of electrons, and we have our combined sum of 128 in a matter of seconds.
32) In other words, we’re given 4 sets of elements and asked to identify which pairs of atoms is most likely to form an ionic bond. The question stem explicitly says we’re focused on electronegativity. Electronegativity trends allow us to determine which types of bonds are formed between elements. It’s the highest to the right, and toward the top of the periodic table. Electronegativity differences are measure on something called the Pauling Scale, but they are very precise. For the sake of our trends, large electronegativity differences mean an ionic bond is formed. A small electronegativity difference means a nonpolar covalent bond. A bond in between is a polar covalent bond. We’re trying to find the pair that is most likely to form an ionic bond. We said electronegativity increases from left to right across a period, and also increases from the bottom to the top of the periodic table. Generally, when a metal and nonmetal from opposite ends of the periodic table bond, an ionic bond is formed because of the huge electronegativity difference. In this case, we can pick the pair that matches our description-Calcium and iodine. Quick recap: when elements from opposite ends of the periodic table bond, an ionic bond is formed. A bond is considered an ionic bond because there is a complete transfer of electrons. And that’s exactly what happened in answer choice C. Chemistry Question Pack: Questions 33-37 33) We’re give the planar structure and the chair confirmation of cholic acid. There’s a circled hydroxyl group in the figure; it’s in the planar structure. We’re trying to identify the same hydroxyl in the identical molecule, but in its chair confirmation. This is an organic chemistry question from AAMC’s content category 5B which talks about intermolecular interactions, and in this question, we’re trying to visualize a molecule in two different views. The chair confirmation in particular minimizes strain, so it’s standard practice to switch from the planar structure to the chair confirmation. The circled hydroxyl group is found on a secondary carbon. What does a secondary carbon mean? It means the hydroxyl group is bonded to a carbon, that’s bound to two other carbon atoms. What else can we recognize about the hydroxyl group? It looks like it’s two carbons removed from the fusion of the third hexose ring and the pentose ring. The hydroxyl group is also five carbons removed from each of the other hydroxyl groups.
34) We’re covering a nucleophilic substitution. Alcohols have hydroxyl groups, and we know hydroxide ions are poor leaving groups. Weak bases are the best leaving groups, but hydroxide ion is strong base. But this question specifically focuses on the mechanism by which the acid catalyst enhances the reaction. Adding an acid allows for the conversion of our alcohol to its protonated state (its oxonium ion). This allows water to be the leaving group instead of the more difficult hydroxide ion. We’re assuming our answer will at least be tangentially related to this process.
35) What exactly can we use the IR spectrum for? To tell us which functional groups are found in our starting material and products. First and foremost, we want to know about what occurred in the reaction in the question stem before we make observations about the infrared spectrum of the reaction mixture. This is as simple as knowing what occurred in this reaction. The aldehyde (our CHO) is replaced with a primary alcohol (CH2OH). The single OH group is present in both the reactants and products. Our OCH3 is also present in both the reactants and products. Our answer should include something about the aldehyde group being replaced with a primary alcohol.
36) The name Proton Sponge could imply the compound is basic if it is a good proton acceptor. What makes a good nucleophile, and what do we know about the structure of the compound? The compound is aromatic and has tertiary amino groups that are located on the same side of the compound. We already hypothesized that the compound was basic based on its name and its structure-it’s a good proton acceptor. Nucleophilicity depends on how readily a species will donate electrons. So basic compounds are typically good nucleophiles, but a big factor in determining nucleophilicity is the structure of the compound itself. Substitution reactions don’t proceed when there is steric hindrance like we’re seeing in our compound here. We mentioned both tertiary amino groups are on the same side of the compound.
37) The compound shown contains 3 amino acids. We’re trying to use these 3 same amino acids to form different tripeptides. A tripeptide means a peptide derived from three amino acids, the same as the compound in the question stem. Do we have to know the identity of the amino acids in the compound? In general, we do, but for this question? Not at all. We have to know the possible number of peptides given we have one of each of our three amino acids. The formula for the number of peptides if we have one of each of three amino acids is 3 factorial, or 3*2*1, or 6. What’s the general formula? N! (N factorial) if we’re trying to find the possible number of different peptides that contain each of N amino acids. So, if n=4, then 4! Is 24.
Chemistry Question Pack: Passage 7 38) We were told different properties of elements and bonds in the passage. This question focuses sulfur hexafluoride, and we have to describe its bonding and electrical conductivity. That means we can use the information given in the passage, likely about covalent and ionic compounds. The passage says nonmetallic elements make up covalent compounds. Ionic compounds are a metal and nonmetal. Sulfur and fluoride are both nonmetals, they’re found on the right side of the periodic table. We can start forming our prediction-the bonding should be covalent, and we’ll want to kill any ionic answer choices. The passage says many covalent compounds are liquids or gases (so they are aqueous) and they will not conduct electricity, meaning our compound will be a nonconductor. We’re looking for an answer that mentions covalent bonding and a nonconductor in terms of electrical conductivity.
39) The periodic table is based on trends, so all this is asking is about a specific trend, where nonmetals are typically found. Metals are generally found on the left side and middle of the periodic table, the nonmetals are found on the right side, Metalloids separate the metals and nonmetals and share characteristics of metals and nonmetals. We’re looking for an answer that mentions the right side of the periodic table.
40) We’re told trends in the passage, and now we’re asked to pick examples of both ionic and covalent compounds. Correct answer will have an ionic compound first, and a covalent bond 2nd. The definition of ionic and covalent will come from the passage, but we’ll need to use the periodic table as well. The passage said that covalent compounds are comprised of nonmetallic elements. Ionic compounds are made up of a metal and nonmetal. So, let’s find our compounds on the periodic table, and see which pair lists an ionic compound first, and a covalent bond 2nd. We can start going through our answer choices and compare.
41) We’re given four solid compounds in our answer choices. We have to decide which compound has the most ionic character. The passage tells us what makes compounds ionic and covalent. That’s what we’re going to use to pick the correct answer here. We’ll also need the periodic table again to see where each element is found. Once again, the passage said that covalent compounds are comprised of nonmetallic elements. Ionic compounds are made up of a metal and nonmetal. So, let’s find the elements that make up our compounds on the periodic table. We’re looking for a metal and nonmetal. And we want the compound with the most ionic character. That usually means an element near the bottom left of the periodic table where electronegativity is the least. And the top right of the periodic table where electronegativity is greatest.
42) We have to explain why HCl is not an ionic compound. The passage went into details about ionic compounds, and we have to explain why HCl matches these criteria, even though it is not an ionic compound. Ionic compounds are solid at room temperature, made of a metal and nonmetal, and aqueous solutions conduct electricity. Covalent compounds are liquids, gases, or solids, mostly nonmetals, and aqueous solutions do not conduct electricity. HCl is a covalent compound. But one of the answers lists a property of an ionic compound.
Chemistry Question Pack: Passage 8 43) Diamond is formed from graphite at extreme pressures and it’s formed from carbon. This question is asking about the structure and bonding of diamond and which elements its similar to. Elements found in the same group will often have similar chemical and physical properties, which is why we can group them together and give the columns names. So, we likely want to pick elements that are in the same group as carbon. We’ll be using the periodic table to find the elements in the same period as carbon. The passage says that elements subjected to high pressures will assume the structure and bonding characteristics of a heavier element in the same column of the periodic table. So which pair of elements is found in the same group as carbon? Choice A is aluminum and gallium. Choice B is silicon and germanium. Choice C is phosphorus and arsenic. Choice D is sulfur and selenium. Only silicon and germanium are both in the same column as carbon. The correct answer is answer choice B. 44) This is a multi-level question. First, we have to know the type of bonding that’s present in solid nitrogen. Next, we have to know about the bonding in methanol. We’re ultimately trying to explain the increase in energy density in the solid nitrogen that was discussed in the passage. Energy density is how much energy the nitrogen has, compared to its volume. The only information we’re given in the passage related to this question is nitrogen having an energy density that’s equal to or greater than hydrocarbon fuels. The actual reasoning is going to come from general knowledge. We’re comparing solid nitrogen with methanol, which is a liquid. We can visualize here-a solid is much more compact than a liquid. Next, we know diatomic nitrogen contains triple bonds. For homonuclear, diatomic molecules, there’s no electronegativity difference. Normally we have to use our periodic trends, but in this case both nitrogens have equal electronegativity values. The bonding is going to be covalent. Methanol has the -OH group, so there is hydrogen bonding taking place. We classify hydrogen bonding as intermolecular forces. Chemical bonds are much stronger than inter-molecular forces. Last thing we want to mention, the nitrogen molecules are well packed in small volume, so their energy density is so much higher than that of the methanol liquid. Solid molecules are much closer to each other, compared to liquids.
45) The passage tells us about the general background of solid nitrogen, but doesn’t go into specific characteristics. We’re using general knowledge to explain the entropy change that accompanies a phase change to a solid. Entropy decreases following a phase change from a gas to a liquid to a solid. Solids have the fewest microstates. They’re by far the most ordered state. Glancing at the answer choices as well, we’ll have to explain delta S as well. ΔS represents the change in entropy of the system and it is expected to be negative to correspond to a decrease in entropy. Entropy will increase if there are additional microstates in a system (solid>liquid>gas). That’s when delta S would be positive. We’re looking for an answer along the lines of: delta S is negative and entropy will decrease.
46) The passage mentions solid form of nitrogen is stable at above 65 GigaPascals, but we’re not given much additional information. We’re going to have to know how gases behave under various conditions. This is general knowledge, and we’re not just using the passage to answer our questions. As we mentioned, the passage mentions solid form of nitrogen is stable at above 65 GPa. 65 Gigapascals is 65 x 10^9 pascals, or a BIG number. Continuing with this, at 180 GPa, we can’t assume the ideal gas law is in play, just because the solid is stable. Gases deviate from the ideal gas law at low temperatures and at high pressures. 180 gigapascals is an extremely high pressure. So, we would expect deviations due to the pressure difference, but not necessarily because of the room temperature.
Chemistry Question Pack: Passage 9 47) We’re given four different mixtures here and told specific details about the components in each set of mixtures. Each component is present at a concentration of 0.1 molar. To answer this question, we want to know which mixture has a pH closest to 7. We’re going to need the passage to get specific details, but we need to know how to analyze the details about the pH and Kas using general knowledge. Quick overview of our answer choices. First component of each mixture listed is one of the four acids from the passage. Second component of each mixture is the conjugate base that reacted to sodium ion from sodium hydroxide. We’re using the details from the passage. We have a table that gives us Ka values. What general knowledge do we need to know? When concentrations are equal like we have in this question. pH of our buffer is equal to pKa of our acid, so in this case we want a pKa of close to 7. We’ll have to convert our Ka to pKa. pKa can be found by taking the negative logarithm of Ka. We can approximate this using the following formula: -log (y-10^-x) =x-0.y formula. The only option here that will give us a value of close to 7 is hypochlorous acid, which gives an approximate value of 7.7. We can eliminate the other answer choices for being incorrect values-we already compared all of the answers. Keep our correct answer choice, answer choice A: HClO(aq) and NaClO(aq) 48) What is dissolution? Another word for dissolving. So we’re given the same reactant in every answer choice, our ammonia nitrate, and we have to show the equation for what exactly is taking place. There wasn’t anything in the passage that would hint at the reactants in this dissolution, so we’ll use general knowledge to pick the correct equation. We’re going to need to use solubility rules. First and foremost, we know ammonium ion (NH4+) and nitrates (NO3-) are soluble in water. We have an ionic salt being dissolved in water, so we’re predicting there to be ionization that occurs with our two ions being ammonium ion and nitrate ion. Let’s consider our four options:
49) We’re noting a change in the pH of a solution when strong base is added. Specifically, we have 2 mL of 0.1 molar sodium hydroxide being added to our solution. The question wants to know how much of change in pH we will see. We’re told we’re adding the strong acid to a solution containing 0.1 molar hypochlorous acid and 0.1 molar sodium hypochlorite. This was something we needed to know for an earlier question. When an acid and conjugate base are found in equal concentration, then the pH of a buffer equals the pKa of the acid. Said differently, we’re at the half-equivalence point of the titration curve. The half-equivalence point is the point at which pH=pKa. And the concentration of acid in solution is equal to the concentration of the conjugate base in solution. At this point when a strong base like sodium hydroxide is added, pH would still become more basic, but change very slowly.
50) This question is going to rely on knowing our solubility rules, and properties of sodium nitrite. Sodium nitrite is a salt. Salts containing group I elements (such as Na+) are soluble. They’re usually listed as solubility rule #1 because they are completely soluble. The nitrite ion is basic, it can pull out a H+ from water, leaving hydroxide ion OH-. What do we know about hydroxide ions? They increase pH.
Chemistry Question Pack: Questions 51-55 51) We’re given an equilibrium equation above and we have to give a name for the transition from the compound on the left to the compound on the right. First thing we want to do is break down what’s going on in the figure. A hydrogen atom from this CH2 moves to the oxygen atom on this carbonyl group. The formula on both sides of the equation is the same, the only difference is the connectivity. We essentially are dealing with constitutional isomers. Again, the connectivity is different, but if we count the atoms on both sides, they are the same.
52) Glycine is an amino acid, and polarity is the result of uneven distribution of electron density. It has a specific property that makes it a dipolar ion that we’re looking to identify. We’re working with amino acids, and dipolar ions. How can we either link these using our content outline? Or what do we know about each individual topic? Amino acids can exist as zwitterions, or dipolar ions, at neutral pH. So first things first, what is a zwitterion? Zwitterions are molecules with a positive charge and a negative charge in different functional groups. Amino acids are amphoteric-they have both an acidic carboxyl group, and a basic amino group. So why are amino acids so polar? Because of the uneven distribution of electron density.
53) This is a standalone question, so we’re using the figure in the question stem, and using external knowledge to answer this question. First things, we want to number our carbons. We have several substituents and functional groups attached to our carbons around the ring. However, only one carbon is attached to multiple substituents-that’s going to be carbon 1 (I’ve labeled it on the right side of the figure below). If we look at the straight-chain structure, this is the carbon that was part of the carbonyl group, so in this case, we can also call it the anomeric carbon. The difference between our alpha and beta-D-glucoronides is going to be the side on which the -OR group is found. In the compound shown here in our question stem, the C-1 -OR group is on the same side of the pyranose ring as the carboxyl group. In alpha-D-glucuronide, the C-1 -OR is on the opposite side. So, our difference comes at the C-1 carbon. Comparing answer choices here is going to be easy. We already broke down the structure and compared all of the carbons. We know the C-1 carbon is going to be where the configuration difference comes in. Answer choice A is the correct answer. 54) To answer this question, we’re going to be using stoichiometry and dimensional analysis. That means it’s going to be crucial to keep our units straight. Note our final answer is given in milliliters, and all answers are given in whole numbers. Stoichiometry’s going to help us balance the equation given. And we’ll more clearly see the relationship between the substances in our reactants and products. If we rewrite our equation, we need an additional H on the left side, so we add a 2 coefficient in front on HCl. We now need another chlorine on the right side and an extra sodium. We add a 2 coefficient in front of our sodium chloride as well. We’re also told conditions are STP. That means standard temperature and pressure. 273 Kelvin and a pressure equal to 1 atm. We’re dealing with carbon dioxide, which is a gas. Gases occupy a volume of 22.4 L for 1 mole of ideal gas. I’ve handwritten my work here, and I’ll break down the process right below. We need to know the mL of a 2 molar solution of sodium carbonate to produce 11.2 L of CO2. We can quickly convert to moles of CO2 using our STP information. If 1 mole occupies 22.4 L of volume, ½ a mole will occupy 11.2 L. For every ½ mole of CO2, there are also ½ mole Na2CO3. Both have the same coefficient. Next, divide the ½ moles of sodium carbonate by molarity of 2.0 moles/liter. We get 0.25 liters. But we said our final answer should be in mL. So we use dimensional analysis once again. Multiply by a factor (1000 mL/1 L) and get 250 mL of the 2 molar sodium chloride will produce 11.2 L of carbon dioxide at STP. This is a math problem, and we did minimal rounding, so the calculated value can be used as our predicted value. We said our prediction is 250 mL. Comparing our answer choices. Answer choice B is the correct answer. What are the key takeaways here? Always keep your units and conversions straight. If that means writing them out like I did here, then please make sure you do so for every dimensional analysis question. We want to build these good habits. The next thing, we want to make sure we start making the connections to our content outline as you work through these problems. We were told the reaction was at STP, and we knew right away what that meant and how to use that information. The better you know the content outline, and the more you practice using it, the better off you’ll be on exam day. 55) We’re shown a reaction that at first glance looks balanced. We’re going to analyze the reaction to see which species is oxidized, and which is reduced. That’s going to help us find the oxidizing and reducing agent in the reaction. What we want to be careful about the most, is the verbiage. This can get tricky. In the reaction, a molecule, atom, or ion in the oxidizing agent is reduced-meaning it gains electrons, and we’ll see a more negative oxidation number. A molecule, atom, or ion in the reducing agent is oxidized-meaning it loses electrons and we see a more positive oxidation number. Let’s look at our specific reaction: We’ll go one by one through the reactants first, and use our rules for oxidation numbers.
Products:
What was reduced? Manganese was reduced from +4 to +2. What was oxidized? Oxygen from -1 to 0. The manganese dioxide is the oxidizing agent. It is an electron acceptor-it gains electrons. Hydrogen peroxide is the reducing agent. Reducing agent will lose electrons.
Chemistry Question Pack: Passage 10 56) Table 1 is essentially a summary of all of the various conditions in the experiment, and how they affected the rate of Reaction 1. Which of the options can be added in excess without influencing the rate? We can pull up Table 1 and see how Solution A and B are affected. We can determine the role of each of our four answer choices to answer correctly. As I mentioned, we need to pull up Table 1 to answer our question. Answer choices are near the top right of the figure, and the breakdown of the two solutions is right below the table. When we have an experimental passage, it’s not uncommon to be asked about specific components of an experiment. The test-maker wants to see if you understand the experiment and the various components that go into the experiment. The four answer choices listed correspond to the 4 compounds in Solution A and Solution B. All of the aqueous compounds are directly related to Reaction 1 and Reaction 2 in the passage. Starch on the other hand, was not found in the reactants or products. It was used as an indicator. Adding excess starch doesn’t affect the rate of the reactions, because starch is simply an indicator. Answer choice D is going to be our best answer choice because starch doesn’t participate in the reactions. 57) In the passage we learned that “The students added 1 drop of 0.1 M CuSO4(aq) to Tube 6.” To answer this question, we simply have to explain the function of this drop. We can go back to the passage to check the effect of CuSO4 on the rate of Reaction 1. Copper sulfate is added to Tube 6. It was one of the lowest times in the rightmost column. This is despite all of the other conditions being identical to Tube 1. Looking at Tube 6, we see a time that is ~33% faster than Tube 1, with the other conditions being identical. Seemingly, the copper sulfate increases the rate of the reaction. That sounds like an enzyme or catalyst of sorts. The AAMC outline emphasizes that we need to be able to demonstrate the understanding of important components of scientific research, meaning you have to be able to read and understand experiments. In this specific case, we have to know what a catalyst is, and how it is used in experiments. Enzymes are catalysts that are able to lower activation energy without being used up or being destroyed in the reaction. Catalysts work to increase the rate of a reaction by lowering activation energy.
58) We’re comparing Tube 1 and Tube 4. We’re also explaining why the rate of the reaction decreases in Tube 4. We’ll go back to Table 1 once again. We should be able to see a decreased rate of reaction in Tube 4. We want to justify why that is. Starch is used as an indicator that turns solution dark blue. We recall that from the passage. The time column of the table is measuring the time it takes for the solution to turn dark blue. Once again, we’re explaining why Tube 1 shows less time in that column that Tube 4. Why did Tube 1 turn more rapidly than did Tube 4? Tubes 1 and 4 have the same volume conditions, but the only difference we see is with temperature. Temperature in Tube 1 is 10 degrees higher. The lower temperature in Tube 4 correlates to a decreased rate of reaction.
59) We’re using the variables given in the passage and looking at them in a different format: a line graph. We can go back to Table 1 once again. We want a graph that shows moles of S4O62–(aq) in Tube 6. First thing we want to note, is the reaction turns dark blue in 19 seconds. Dark blue is the result of the starch indicator telling us thiosulfate was used up in Reaction 2, and the maximum amount of tetrathionate has been formed at this point. What do we expect the graph to look like? We expect the product to steadily increase from 0-19 seconds (19 seconds is the max number of moles), then we should see the graph plateau at that point. a. Graph A implies the moles continue rising, even though the reaction is complete in 19 seconds. We expect the line will continue to rise until 19 seconds, then plateau b. Graph B does exactly what we expect, rises for 19 second, then plateaus. This is now our best option. c. Graph C says the moles increase after 19 seconds, but the reaction is complete at that point, not starting d. Graph D says the number of moles increases, but then subsequently decreases. Not the case. The moles do not disappear. We’re going to stick with our best option, answer choice B. Chemistry Question Pack: Passage 11 60) We had a reaction take place at 300 Kelvin, and the question is asking us to write out the net reaction. To answer this question, we need to see the figure in the passage to see what reaction actually took place. We have our figure above. Look at the figure to see that for every X in reactants (or before) side of Figure 1, two Ys bind to form the ultimate Y2X product. So, the ratio of X reactants to Y reactants in the net reaction will be 1:2.
61) This question is very straightforward if you’ve taken the time to read the passage. To answer this question, we can use the results of the experiment to predict the reaction rate at 400 K. What was the biggest thing we should’ve picked up from reading this experimental passage? As temperature decreases, reaction rate decreases. As temperature increases, reaction rate increases. This ties the whole experimental passage together. The purpose of the experiment was to explain the relationship between temperature and the amount of product formed, and reaction rate. So, what do we expect to happen here? Answer choice A: Reaction rate should increase. 62) We already know the reaction in the passage is exothermic, so there’s no real reason to go back to the passage. Because it’s exothermic, the potential energy for our product will be lower than for our reactants. Energy (that’s shown on our y-axis in our answer choices) is going to be lower at the end of our reaction. Next thing we want to note, do we have to overcome an activation energy barrier? We absolutely do, and we can do that simply by increasing temperature. a. Answer choice A shows a reaction where we overcome the activation energy threshold. But energy ends up in the same location on the y-axis. We’ll still hold on to this answer choice for now to compare with our additional answers b. Answer choice B shows an endothermic reaction. That contradicts what we’re told in the question stem. That’s no better than choice A c. Answer choice C shows the activation energy as a threshold of sorts. But we know it’s not a dip in energy as shown. In the experiment we were able to overcome the activation energy barrier with an increase in temperature. The final energy is lower than our initial energy here though, so that part matches our breakdown at least d. Answer choice D shows an exothermic reaction, and the proper activation energy barrier that matches our breakdown. Answer choice D is going to be our best answer. We can eliminate the remaining answer choices that we already invalidated 63) To answer this question, we want to visualize the product molecule, our Y2X, and determine the molecular shape. We can go back to the passage, but the issue is, the passage doesn’t tell us about the identity of our reactants. There’s no way to know how many electrons are found in our product molecules. So even though we know our central X atom is attached to two electron dense areas, our Y atoms, we can have any number of valence electrons on our X atom. There are two possible options: The molecular geometry can either be linear or bent, depending on the identity of the individual reactants, and the number of electron dense areas in our final molecule.
Chemistry Question Pack: Passage 12 64) Triacylglyerols are formed through the joining of three fatty acids to a glycerol backbone by a dehydration reaction. We want to know specifically about the triacylglycerols and its number of stereogenic centers. We have Reaction 1 in the passage that showed us the structure of triacylglycerol. We’re going to be using general knowledge to identify the number of stereogenic centers. We can draw out the reaction of our glycerol with the three fatty acids to form triglyceride. We know the structure of glycerol. We’re reacting with three fatty acids. Ultimately, we form this triacylglycerol. Any carbons attached to four different groups will be a stereogenic center. The top carbon and the bottom carbon along the glycerol backbone are bonded to two hydrogens each. Carbon 2 is attached to four different groups, so we have a single stereogenic center. Looking at our answer choices, we’re given numbers from zero to three. We already predicted there is only a single sterogenic center on carbon 2. Our correct answer is going to be B: there is 1 stereogenic center in the product triacylglycerol. 65) In the passage we were shown reaction 1 that produced fatty acid salts. This question wants us to pick the formula that represents the general structure of the fatty acid salts produced in the reaction. We have to know the general formula of a fatty acid, and then the general formula for fatty acid salts that were produced in Reaction 1. We can use the reaction in the passage as a reference point, but the answer here is going to come from external knowledge. We have reaction 1 listed here. Again, first step, let’s draw out the general formula for a fatty acid. General formula here is going to be our R group attached to a carboxyl group. Our fatty acid salt was formed through saponification with sodium hydroxide, so we’re expecting our fatty acid salt to have a corresponding sodium ion attached. The general formula for the fatty acid salt is going to be our R group. At the end, we have CO2 minus and a positive sodium. This is going to be the general formula for the sodium salt of the fatty acids.
66) We want an explanation for the four different fatty acid salts that were the result of the saponification of the triacylglycerol. Normally we’d expect three, but in this case, there were four, and why is that? The passage actually mentioned there were four fatty acid salts produced, but we didn’t get any specific reasons. Saponification is the formation of a salt following the mixture of a long-chain carboxylic acid with a strong base. In this case, the strong base was sodium hydroxide, and there were four fatty acid salts instead of three. The two most likely outcomes would be if the original molecule was broken down into 4 acyl groups that turned into fatty acid salts, or one of the fatty acid salts was counted twice for some reason. Is it possible there were 4 acyl groups? Not likely. We broke down a triacylglycerol in basic conditions. There’re only going to be 3 fatty acid units. In fact, the passage explicitly says it was pure triacylglycerol. Now why might one of the fatty acid salts get counted twice? One of the fatty acids must have reacted. Unsaturated fatty acids can undergo addition and elimination under basic conditions, meaning cis/trans isomers were produced. This is a possibility that would explain why there were four different fatty acid salts.
67) The example reaction in the passage used sodium hydroxide to saponify a triacylglycerol and formed glycerol and fatty acid salts. We want to know how much. And a quick glance at the answers shows this is not going to be a simple quantitative answer. To answer this question, we’re going to need to know about triacylglycerols and the breaking of the ester bonds during saponification. Triacylglycerols consist of glycerol attached to three fatty acid chains. The strong base that is typically used in the saponification process is an alkali hydroxide. Just like in our passage, that alkali hydroxide is commonly sodium hydroxide. In the presence of the strong base, there’s cleavage of the ester bond which releases glycerol and fatty acid salts known as soap. Additionally, the triacylglycerols contain three ester linkages. So, our strong base, our sodium hydroxide, has to be present to cleave these three linkages.
68) We want to know about the fatty acid salts that were formed in Reaction 1 in the passage. Are the solubility properties going to come from the passage, or external knowledge? Likely from external knowledge. We weren’t given much information about the fatty acid salts that we shouldn’t have known coming into the passage. We still need to break down the structure of the fatty acid salts, then we can explore the solubility properties. I’ve handwritten below, and explained exactly how to break this question down. We have our triacylglycerol and the 3 equivalents of strong base. Our products are our glycerol and the sodium salt of our fatty acid. We’re focused on the fatty acid salt to answer our question. We have a hydrocarbon chain, and we also have this charged CO2- and Na+. What’s the golden rule for solubility? Like dissolves like. So first, Lipids are hydrophobic and nonpolar and are not very soluble in water. While structures may vary based on the type of lipid, all lipids have low solubility in water, and higher solubility in nonpolar organic solvents. But, we have the charged group at the end of our salt. That’s going to be soluble in polar solvents. For example, table salt can dissolve in water because both are polar. Again, like dissolves like. So, we’re expecting our fatty acid salts to be able to dissolve partly in polar solvents and partly in nonpolar solvents.
Chemistry Question Pack: Questions 69-73 69) We’re given the mass of each individual component in a compound, so we have to find the corresponding empirical formula. We’re going to need to know the molar mass of each individual component of the compound, so we actually need our periodic table. Empirical formula is the smallest ratio of numbers to represent the proportions of the atoms. That means we’re going to need to divide the grams of each individual component in the compound to find the number of moles of each. For example, 12.0 grams of carbon in the compound divided by the mass on the periodic of 12.0 gives 1.0. After dividing all three components, we have a ratio of 1:2:1. Carbon:hydrogen:oxygen. Can this ratio be simplified any further? No, it cannot. Before we go through each answer choice individually, let’s be careful here. We mentioned empirical formulas can’t be simplified. If we look at answer choices C and D, can those be simplified? Yes, they can, so we’re skeptical about those answer choices right away since they aren’t proper empirical formulas.
70) The way we’re going to approach this question is to pay attention to subscripts and the equivalents of different components in the prepared solution. First thing we want to note, is we’re focused on Na+ in each of our initial solutions. By looking at the subscripts, we can see there is 1 equivalent of Na+ in the NaHCO3 solution (no subscript), and 2 equivalents in the Na+ Na2CO3 solution (subscript 2 on the sodium). Given that information and also noting the molarity of each solution, there is also 0.010 M Na+ in the NaHCO3 solution and 0.020 M Na+ in the Na2CO3 solution. However, we’re not done yet. We’re mixing these two solutions to form a third solution. The test-maker makes this easy for us and tells us there are equal volumes of the two solutions. What does that mean? The final molar concentration of Na+ is going to be the average of the two solutions being mixed: (0.010 M + 0.020 M) / 2 = 0.015 M Na+(aq). You might be looking at the question stem and this calculated answer and wondering about units. To find the molar concentration, we divide the number of moles of a substance by the volume of the solution. We are already given molar concentrations, so we simply find the relative amount contributed by both. We have the proper units.
71) This question is asking which of the two Compounds (1 or 2) can form hydrogen bonds water. This boils down to a difference in electronegativity between elements determining the type of bond that will form between the elements. Hydrogen bonds are formed when hydrogen bonds with a very electronegative atom. The hydrogen gains a partial positive charge, and the atom gains a partial negative charge. Nitrogen is an electronegative element relative to hydrogen, but carbon is not. Like I mentioned, hydrogen bonding happens between a hydrogen atom and an electronegative atom. For the sake of the MCAT especially, we think of fluorine, oxygen, and nitrogen as the electronegative atoms in hydrogen bonding (FON atoms). We’re expecting the compound with nitrogen (Compound 2) is more soluble in water. We answered an objective question here, so we can actually compare all of our answers at once here. Answer choice B is the only answer choice that contains Compound 2, and does not contain Compound 1. 72) To answer this question, we need to know the structure of the molecules in the four answer choices. Our answers have a central carbon surrounded by one to four bromines. We’re more focused on the carbon here. Look at the figure above and you see it’s tetrahedral and bond angles are all roughly 109o when all substituents are identical. Electrons are all shared equally between carbon and the 4 bromines, meaning the vector sum of the bond polarities is zero. Where do we see this happening? In answer choice A. Carbon is sp3 hybridized in all of our four answer choices, but there’s only one answer choice in which the vector sum of the bond polarities is zero. 4 of the same substituents, all contributing the same bond polarity. We can stick with our best answer, answer choice A. 73) To answer this question, we have to know the difference between chlorine’s normal electron configuration, and its electron configuration in NaCl. An ionic bond is formed, meaning the sodium atoms are cations while chlorines are anions. Chlorine has an extra electron, giving it a full octet in this case with 18 electrons, or the same electron configuration as argon. A neutral chlorine atom will have 7 valence electrons and a total of 17 electrons, with two being found in the first shell, eight being found in the second shell, and seven remaining electrons that are found in the outermost shell as valence electrons. Chlorine is very reactive and instead is typically found as a diatomic molecule or part of a compound (like in NaCl). With 18 electrons, chlorine has a full third shell with electrons found in the s-subshell and p-subshell.
Chemistry Question Pack: Passage 13 74) During the passage we were shown different fertilizers and corresponding equations. This question is asking specifically about the fertilizer shown in Equation 2. We want to know the effect of moist soil conditions (additional water in the reactants). We’re going to need the equation from the passage primarily. And we need to know the effect of adding more reactant to a chemical equation. Glancing at our answer choices, we want to answer in terms of ionization, and the effect on hydroxide ion. We have Equation 2, and we want to know the effect of additional moisture, or more water reactant. A quick look at our equation shows that there is ionization taking place. There’s the release of hydroxide ions. What happens when we add reactants to a chemical equation? This is Le Chatelier’s principle. The reaction equilibrium should shift toward the products. We’re expecting more hydroxide ion to be released. We know hydroxide is charged (hence the ion). What do we expect in terms of our degree of ionization? We’re going from neutral water and ammonium monohydrogen phosphate reactants, to ionized products like the hydroxide ion. That means our degree of ionization increases
75) We were told common fertilizers contain ionic salts of nitrogen. So why can’t plants use the nitrogen in the atmosphere instead as a sort of fertilizer? To answer this question, we have to know properties of nitrogen to explain why plants can’t simply use nitrogen from the atmosphere as fertilizer. Nitrogen from the atmosphere is going to be diatomic nitrogen. The two nitrogen atoms are held together by a triple bond, and it’s going to be found in a gaseous state. If need be, we can have our periodic table and periodic trends ready.
76) We’re given a component of equation 1, one of the products. We have to explain whether it is a conjugate acid or base of different compounds. We only need the definition of conjugate acids and bases to answer. Even though the question references Equation 1, that doesn’t mean we need to go back. Oftentimes the testmaker will reference something in the passage, but you can use external knowledge to answer. That’s what we’re doing here. We save so much time here just by knowing the definition of conjugate acids and bases. Step 3: A conjugate base is the compound that remains after an acid has donated a proton during a chemical reaction. A conjugate base is formed when an acid loses a proton. A conjugate acid is formed when a base gains a proton. Hydrogen phosphate is the conjugate base of dihydrogen phosphate ion. The dihydrogen phosphate ion loses a proton to form the hydrogen phosphate. Hydrogen phosphate is the conjugate acid of phosphate ion. Phosphate ion gains a proton to form hydrogen phosphate.
77) To answer this question, we’re referencing Equation 2 from the passage and we’re asked about the equilibrium constant. We need to explain two things here: why the concentration of H2O is omitted, and how it relates to the salt in Equation 2. We’ll need the passage to reference equation 2, and we’re going to need to know about acid-base equilibria. We have equation 3 given above. First and foremost, our water is a liquid. For reactions involving a solid or a liquid. The amounts of the solid or liquid will change during a reaction, their concentrations and densities won’t change. Those values remain virtually constant. Because they’re constant, the values are not included in the equilibrium constant expression. Next order of business, the fertilizer is our salt here. The ammonium hydrogen phosphate. When we read through the passage, we should’ve noticed the formation of hydroxide ions in the product. Does that make the salt acidic or basic? It’s basic.
Chemistry Question Pack: Passage 14 78) Experiment 2 involved electrolysis cells, and we want to know what is taking place at the cathode. To answer this question, we need to know how electrolytic cells work. Electrical energy is used to drive a nonspontaneous redox reaction. The oxidation half reaction occurs at the anode, and the reduction half reaction takes place at the cathode. Electrons flow from the positive anode to negative cathode.
79) To answer this question, we can figure out the electron configuration without using the passage. We just need to know about the reaction of calcium and water. Then, we can find the electron configuration of the ion. Another thing we want to note is we’re dealing with ions, so there’s a strong chance we’ll be using the periodic table as well. What do we know about the reaction? We know the resulting solution is basic, meaning hydroxide in the product. Let’s write out the equation: Ca(s) + 2H2O(g) → Ca(OH)2(aq) + H2(g) We have solid calcium reacting with liquid water. That yields calcium ion, hydroxide ion like we just mentioned, and hydrogen gas. We can balance the equation by adding a 2 as the coefficient in front of water, and in front of our hydroxide ions. The charge on our calcium ion is positive 2. Quickly reference the periodic table to see how many electrons calcium normally has. Elemental calcium is element number 20, meaning it has 20 protons and 20 total electrons typically. In this case, we’re dealing with calcium ion. The calcium loses 2 electrons to get a +2 charge and giving it a full octet, or the same electron configuration as argon, a noble gas. Argon’s electron configuration is 1s22s22p63s23p6. First energy level holds 2 electrons, 2nd energy level holds 8 electrons, and 3rd energy level will hold an additional 8 electrons.
80) Experiment 1 gave us brief observations about the reaction between the metals and water. This question is asking to pick a trend to explain the reactivities of the different metals. We’re going to need to use Table 1 from the passage to find the observations made about the different metal ions, then we’re going to use the periodic table and periodic trends to help answer this question. I mentioned we’re going to need the passage to answer our question, so reference Table 1 above. Magnesium is an outlier. There’s no obvious reaction and the resulting solution is neutral. This could mean no magnesium ion even formed. Next thing we want to note, is the sodium and potassium reacted the strongest. Those ions formed fairly readily it looks like. Let’s locate all five of our elements on the periodic table next to see if we can see any trends. Our five elements were magnesium, calcium, lithium, sodium, and potassium. The magnesium was an outlier because there was no obvious reaction, meaning there likely wasn’t an ion formed. The sodium and potassium reacted the strongest, meaning they ionized more readily. Lithium, sodium and potassium are all group 1 elements. Calcium and magnesium are group 2 elements. So, it seems like the further right and up the periodic table we go, the less likely we are to form an ion. That corresponds to what are referred to as E trends-electronegativity, electron affinity, and ionization energy.
81) In this case, we’re redoing Experiment 1 with calcium and trying to identify the gas that evolved from the experiment and its volume. Note, we’re given pressure and temperature, and we shouldn’t need the passage to answer this question. We’re going to use the ideal gas law to solve for the volume of the gas, and the volume is given in mL. That means we’ll have to solve for moles and we’ll have all the information necessary to find volume. Next thing we’re going to do is write out the reaction taking place. We actually used this reaction two questions ago: Ca(s) + 2H2O(g) → Ca(OH)2(aq) + H2(g) We have solid calcium reacting with liquid water. That yields calcium ion, hydroxide ion, and hydrogen gas. Make sure to balance the equation. We needed to identify the gas produced to answer this question, and we can see it’s hydrogen gas. Let’s move to the rest of our question: We have 0.4 grams calcium, let’s solve for moles of calcium so we can subsequently solve for moles of hydrogen gas. Molar mass of calcium is 40.1 g/mol, so we can use dimensional analysis and solve for moles. We have 0.01 mol calcium. Looking at our balanced equation above, for every mole of solid calcium, we have 1 mole of hydrogen gas. That means we also have 0.01 moles hydrogen gas. Now is when our ideal gas law comes in. PV=nRT. We can plug in our values and solve for our unknown volume. Isolating volume, we get 0.25 mL. We said our answers are in mL, so we can use dimensional analysis to get 250 mL. I’ve handwritten everything here to allow you to follow more closely:
Chemistry Question Pack: Passage 15 82) To answer this question, we’re going to be comparing formulas for compounds 1 and 2. Compound 1 is B2P2H2N2R4. Compound 2 is B3P3H3N3R6. The molecular formula here is not the same for the 2 compounds, but we can check the empirical formula. The empirical formula is the smallest ratio of numbers to represent the proportions of the atoms. Empirical formula for Compound 1 is B1P1H1N1R2. Empirical formula for Compound 2 is B1P1H1N1R2. So, in this case, even though the molecular formula of Compound 2 represented a larger compound, its empirical formula contained a smaller ratio of whole numbers, which was the same as that of Compound 1. Let’s jump into our answers and remember that both compounds have the same empirical formula, but the molecular formula for Compound 2 represents a larger compound.
83) We’re given specific quantitative values and conditions here. We have to solve for the maximum volume of PH3 gas (one of the products in Equation 1a) we can get given the conditions. We’ll need to pull up Equation 1a, and we’ll need the molar ratios of the reactants and products. We can use these molar ratios to find the volume of phosphine gas. We’ll use external knowledge to manipulate our ratios, and to determine the volume of gas per mole. Answers are given in mL and liters. We have Equation 1a above, and the conditions we were given right below. Our equation is already balanced. Equation 1a shows that there is a ratio of 2 moles R2NBCl2 to 5 moles of LiPH2 to produce 3 moles phosphine gas. Conveniently, the volumes given are in the same ratios as our molar ratios. That means we have 0.003 moles phosphine according to our molar ratio. Look at pressure and temperature. This reaction takes place at STP, meaning 1 mole of a gas will occupy a volume of 22.4L. Pay attention to every aspect of the question and what each number could potentially mean. Using the volume 22.4 liters/1 mole, multiply by the 0.003 moles phosphine to find the volume to be 0.0672L or we can convert using dimensional analysis to 67.2 mL. I’ve handwritten the math to solve this question below:
84) We can draw out the Lewis structure, but we should check the periodic table of the elements first. We need to know the number of valence electrons on phosphorus, and the phosphorus is bonded to three hydrogen atoms. So, we expect three of the valence electrons will be part of the bonds with hydrogens. That should leave two as a lone pair. That gives 8 electrons around phosphorus, and each hydrogen is accounted for. The central phosphorous in this molecule has five valence electrons and only binds with three hydrogen atoms. The two remaining valence electrons will be unbonded, so that lone pair would be the only one in the molecule. Our final predicted answer is a central phosphorus with 5 valence electrons. The one lone pair, and three will be a part of bonds with hydrogens. Total of 8 electrons around phosphorus. a. Answer choice A looks exactly like our drawn-out Lewis structure above. Let’s still compare to our additional answer choices b. Answer choice B only has 6 total electrons around the central phosphorus. There should be an additional lone pair. This is incorrect. c. Answer choice C has too many electrons around the central phosphorus. The phosphorus should only have 8 electrons around it. This answer choice is incorrect as well. d. Answer choice D is similar to answer B. We only have 6 electrons around the central phosphorus. This is also incorrect. The central atom here should have 8 electrons around it. The correct answer is answer choice A. 85) In this question, we’re given a line graph showing how Compound 1’s volume varies with temperature. We want to compare this with what the graph of Compound 2 would look like. We’re going to need to know the details about Compound 1 and Compound 2 from the passage. We’ll also use external knowledge to properly compare the two compounds. Let’s take a look at both of our compounds. Then we’ll analyze our variables; we want to know if our variables are directly related or inversely related. We’re told both compounds are in the gaseous state and at constant pressure. Volume is the dependent variable on the line graph. When temperature increases, volume will also increase. Next, we write out our ideal gas law: Pv=nrt where p and r will be the same for both compounds. Pressure is constant. The number of moles in Compound 2 is fewer than Compound 1 because Compound 2 has a greater molecular weight. So, we’re expecting that at the same temperature, volume will be less in Compound 2. I’ve demonstrated below:
Chemistry Question Pack: Passage 16 86) Silver sulfide was formed following the reaction of hydrogen sulfide and silver. We want to say what type of reaction forms the silver sulfide. To answer this question, we’re going to use the passage to identify the reaction in the question stem, then we’ll be using external knowledge to explain the reaction type. Let’s take a look at Equation 2 from the passage. The silver from the coins replaces the hydrogen in the hydrogen sulfide compound. This looks like a single replacement reaction. General formula for a single replacement, or substitution reaction is A + BY > AY + B. We can match up Equation 2 with this general formula. The element silver replaced the hydrogen in the compound in the reactants. The hydrogen is then found alone in the products.
87) In other words, if 2 moles of sulfate ion are consumed, how much hydrogen sulfide is produced (in grams)? We’re going to need Equation 1 from the passage, then we’re going to use dimensional analysis to convert between moles and molar masses. Notice our answers are given in grams, and are rounded to whole numbers. We have our balanced equation from the passage. Look at the coefficients on the sulfate ion and the hydrogen sulfide. Both are the same. So, for every mole of sulfate ion consumed, there is one mole of hydrogen sulfide produced. We’re told 2 moles of sulfate ion are consumed in the question stem, so we also have 2 moles of hydrogen sulfide produced. Now we can use dimensional analysis. Molar mass of hydrogen is 1 gram/mol. Molar mass of sulfur is 32 grams/mol. So two hydrogen atoms and one sulfur atom together in hydrogen sulfide have a molar mass of 34 grams/mole. Multiply this molar mass by the number of moles-we have 2 moles. This gives us a maximum number of grams produced of 68 grams. I’ve handwritten the solution below so it’s easier to follow:
88) What is the ionization constant? It’s essentially telling us the strength of an acid. Remember strong acids dissociate well, but weak acids do not. So, in essence, we’re asked if the hydrogen sulfide is a weak or strong acid. While the passage told us about hydrogen sulfide in particular, we’re going to have to know about the ionization constant coming into the exam. The passage explicitly told us hydrogen sulfide is a weak acid. It was a small detail that the author added to the end of a sentence, but luckily we read slowly and we included that bit in our mental summary. Some quick background here related to the ionization constant for an acid. The acid dissociation constant, Ka, is a quantitative measure of the strength of an acid. A strong acid will dissociate completely in solution. So a high value of Ka will correspond with a stronger acid that dissociates well. A low value of Ka will correspond with a weaker acid that dissociates only slightly. So, we expect our hydrogen sulfide to have a small ionization constant.
89) This topic is always tricky for students because of the verbiage, so make sure to pay close attention as you’re working through it. We need to know the reducing agent in equation 2. We’re not asked the species that is reduced. We want to know the reducing agent from the equation in the passage. We’re going to need to analyze Equation 2 and we’re going to use external knowledge to actually identify the reducing agent. We have Equation 2 from the passage and we need to identify the reducing agent. The reducing agent is the species being oxidized. The oxidizing agent is the species that is reduced. A species that’s oxidized loses electrons. A species that’s reduced, gains electrons. In Equation 2, silver reacts with hydrogen sulfide to produce silver(I) sulfide and hydrogen. Silver goes from 0 charge in the reactants, to +1 in the products. It donates electrons and is the reducing agent/the species being oxidized.
Chemistry Question Pack: Questions 90-94 90) In the question stem we’re shown the phase diagram for water and asked the phase shift between two points. This is a standalone question and we’re going to have to read the phase diagram given. We’re going to answer the question using the diagram itself and our external knowledge about the phase diagram of water. Our phase diagram of water has two variables on its axis, pressure and temperature. So as these conditions vary, so do the phases of water. The phases vary from solid, to liquid, or gas based on the variation in pressure and temperature. The phase all the way to the left is the solid phase. The rightmost phase is the gaseous phase. The topmost and middle phase is the liquid phase. This should inherently make sense. Ice is present at the lowest temperatures, steam is present at the highest temperatures, and liquid is present in between these two phases. Point A is liquid, and Point B is gas. Evaporation, or vaporization, is when molecules are heated and have enough energy to transition from the liquid to the gaseous phase. That means Point A to Point B is vaporization.
91) To answer this question, we’re going to use external knowledge to explain the Bohr model. Niels Bohr proposed an update to Ernest Rutherford’s model. The Bohr model focuses more on electrons and their movement around the central nucleus. That’s consistent with what we’re being asked in the question stem. In Bohr’s model, electrons move at a constant speed and in fixed orbits. Bohr’s model explained that the energy of an electron depends on the size of an orbit. Smaller orbits correspond to lower energy. Moving away from the nucleus will mean an increase in energy level with each orbit of electrons. Electrons that move from energy levels further away from the nucleus to an energy level closer to the nucleus will emit energy.
92) In this question, we’re given a hypothetical soluble metal hydroxide and given a corresponding molar solubility. We have to solve for the Ksp value for this hypothetical metal hydroxide. First thing we’re going to do is write out the general formula for the solubility product constant, our Ksp. Remember with Ksp, we raise the product to that coefficient power (and also multiply the concentration by that coefficient. A lot of students will forget this step or think we’re counting it twice. Make sure to not skip this). Write out the Ksp: M(OH)2 = [M][OH-]2. We can plug in our molar solubility, S, and multiply the concentration by 2 for our hydroxide: We have [s]*[2s]^2 When we multiply numbers raised to exponents, we can add the exponents. Multiplying this out, we get 4s^3. I’ve hand written this out below to make it easier to follow:
93) All of the 4 molecules in the question stem look like they have similar formulas, but only one is liquid at room temperature. We have to explain why water is an outlier, while the other three are gases. Water is unusual. It has high boiling and freezing points, it has a high specific heat, high surface tension, and high polarity. We would expect it to be gas, rather than liquid considering its molecular weight, but it’s not. We expect water to behave in a certain way given its molecular weight, but it doesn’t. Why is that? It’s because water molecules can form hydrogen bonds. Hydrogen bonding is a type of dipole-dipole interaction in which a hydrogen is covalently bound to a very electronegative element. The big three are fluorine, oxygen, and nitrogen. Hydrogen (2.1) is a relatively electronegative element. Oxygen (3.5) is more electronegative than sulfur (2.5), selenium (2.6) and Telluriium (2.1). Water is a polar covalent molecule. The rest are closer to nonpolar covalent.
94) To answer this question, we’re going to use external knowledge to compare partial pressures of two components of an equation. First thing we’re going to do before we jump into our partial pressures is to write out balanced equation to visualize better: CH4(g) + 2O2 (g) > CO2 (g) + 2H2O First thing we want to notice, is there are 2 moles water to 1 mole of carbon dioxide, meaning the effect on total pressure is a ratio of 2:1. We know the total pressure is 1.2 torr, so let’s use our ratios. For every 3 total parts of our gases, 2 parts are water, and 1 part is carbon dioxide. 2/3 of our total pressure is 0.8 torr water, while 1/3 of the total pressure is 0.4 torr carbon dioxide.
Chemistry Question Pack: Passage 17 95) We’re asked specifically about solution A and given the cation portion. We want to know the corresponding anion component, or negatively charged component of the solution. We’re going to need to use the passage. The two tables in our passage provided good summaries of the experimental results. At the very least, we’re going to be the results of mixing silver with the 4 anions. We have Table 1 here, and the information about Solution A. We were told Solution A contains silver ion. And the passage also says “Solution A was colorless.” What does that mean? It’s completely soluble in water. Look at the table. We’re looking at the anion component of the solution. Anions are negatively charged ions that’ve gained one or more electrons. So, we’re looking at our Ag+ column. Going down the column, there’s only one anion that does not have a reaction with silver ion. That’s our fluorine ion.
96) We’re being asked why we had the results we did during experimentation. We got results in Table 1. We mixed anions and cations, but there were only certain situations in which precipitates formed. This question is asking why precipitates formed between anions and cations in the situations when they did. We want to identify why precipitates form in general. There are two ways to do this, we can look at solubility tables, and we can determine the concentration of the ions to see if their reaction quotient is greater than the solubility product constant: the Ksp value. We get this value when a solution is saturated. So how can we read Ksp? A lower Ksp equals lower solubility of compounds in water. Once the Ksp is exceeded, precipitates would form.
97) The four cations that were listed around Table 1 are shown as possible answer choices for this question. We want to know which of the four options had the fewest visible reactions with anions. Looking at Table 1, we either had colored precipitate formed, or we had no reaction, meaning we have a soluble compound. We’re going to need to use the passage to see which mixtures were soluble and which are insoluble. We also have to reference our external information to know the definition of soluble and insoluble. Take a look at Table 1 once more. We have the results of mixing aqueous solutions of cations and anions here. We have to determine which cation allowed for the greatest number of soluble compounds. Cations are positively charged ions, while anions are negatively charged ions. Table 1 shows cations along the top part of the table. Soluble compounds are going to be the mixture results that correspond to no reactions. So, we can go through our cations one by one to compare. Silver ion forms one soluble compound. Calcium ion forms three soluble compounds (only one precipitate). Copper and iron ions both form two soluble compounds each. So which cation allowed for the greatest number of soluble products? Our calcium ion. That means we’ve analyzed all 4 of our answer choices listed here. We compared the number of soluble compounds allowed by each cation. We’re left with our correct answer choice: Answer choice B, calcium ion. 98) In other words, barium cation is toxic, but mixing or dissolving it in water can possibly minimize this toxicity. We want to know which of the compounds listed as answer choices with corresponding Ksp values would be the safest for mammals to consume. Precipitates form following Ksp being exceeded. Lower Ksp means lower solubility of compounds in water. A lower Ksp means the most amount of solid precipitate and least amount of toxic, aqueous barium ion. Do we have to know the specifics about each compound in our answer choices? Not necessarily. All we’re focused on here is finding the most favorable Ksp value. The lowest Ksp value is the safest, and also our correct answer.
99) We have the same cation in both solutions, but iron cation is bonded to chlorine ion in one solution, fluorine ion in another. We have to distinguish between the two in the question stem using one of the answer choices listed. We’re going to need to use the passage to see which mixtures were soluble and which are insoluble. We should be able to answer the question using just Table 1. Looking at our table, what’s the first thing we notice? There is no reaction for both iron chloride and iron fluoride. These compounds are soluble. We need to find a way to differentiate between the two. So, we have to get creative with the rest of our chart. Our question stem said we can add a cation or anion to differentiate between the solutions. So which ions can help us differentiate? Let’s keep looking through our table from right to left. If we add copper to the solutions, would we have able to differentiate between iron chloride and iron fluoride? Well, both fluorine ion and chlorine ion don’t react with copper ion, so we’d have soluble solutions. Next, check the calcium ion. Calcium fluoride forms a precipitate, but calcium chloride is water soluble. So FeCl3 and FeF3 can be differentiated by adding Ca2+. Next, can silver ion differentiate the two? AgF is soluble, and AgCl forms a white precipitate. So FeCl3 and FeF3 can be differentiated by adding either calcium ion or silver ion.
Chemistry Question Pack: Passage 18 100) From the passage we recall that KHP was used while the strong base (sodium hydroxide) was standardized. We want to know specifically the number of moles KHP in student A’s sample. We’re going to need to use information from the passage about student A’s KHP sample; we’ll have to manipulate the numbers we get from the passage to solve for the number of moles. Looking back at some information from our passage above, we have Equation 2 that shows us how KHP is used in the standardization. We’re also told the molar mass of 204.2 here. We have a bit of Table 2 right below, we’re told the mass of KHP in student A’s sample is 0.5500 grams. If we have the grams of KHP, we can use the molar mass and solve for the number of moles. The molar mass of KHP can also be written as 204.2 grams/mole. When we’re using dimensional analysis, we want to cancel out units that aren’t our final units. In this case, we want our final answer in moles. So, we can divide the 0.5500 grams by molar mass. The grams unit cancels, and we can estimate our answer here. 0.5500 grams divided by 2 x 10^2 grams per mole, gives us 0.275 x 10^-2 moles. Or in proper scientific notation, approximately 2.75 x 10^-3 moles. We move the decimal point one place to the right, so we subtract 1 from the exponent. Review the handwritten solution right below as that can be easier to visualize: Keep in mind that we rounded our answer, just in case we have to be more accurate with our calculation. Glancing at the four answers, we have a coefficient in each one, following by 10 raised to different exponents. Which exponent is consistent with our calculation? Answer choice C. We did some rounding, so our answer here isn’t exact, but AAMC doesn’t expect it to be. This question is done this way on purpose. The other answer choices are so far away from one another so we can be confident in answer choice C. 101) We recall students prepared the sodium hydroxide solution in Equation 1, and we have to explain whether any of the three changes in state functions occurred. Recall from the passage the temperature of the solution rose during the mixing process in Equation 1 and that the process is exothermic. We can pull up part of the passage, but we’ll need external knowledge to explain the changes in state functions-we’re dealing with change in Gibbs free energy. ΔG=ΔH–TΔS where ΔG is the change in Gibbs free energy, ΔH is the change in enthalpy, T is temperature (kelvin), and ΔS is the change in entropy. We said heat is given off, so the process is exothermic. That means ΔH is negative. Entropy also increases as the sodium hydroxide is dissolved and moves from solid to aqueous. The more positional probabilities that are available for an atom, the more entropy it has. Aqueous solutions have far more positional options than solids, so they have more entropy. That means ΔS is positive. If we plug in a negative value for change in enthalpy, and a positive value for change in entropy, ΔG, or the change in Gibbs free energy, is always negative because absolute temperature has to be positive also. We have a negative ΔH, we have a positive ΔS, and we have a negative ΔG. That means options I and III are our correct options. Every answer choice ends in “only” so we’re looking for an answer choice that matches our breakdown and predicted answer exactly in this case. Our correct answer is going to be answer choice C, options I and III ONLY. 102) Recall the instructor prepared a solution of sodium hydroxide at the beginning of the experiment. We have to calculate the approximate molarity of the solution. We’re going to use the passage to find the specific information we need about sodium hydroxide. We’re also going to use dimensional analysis to solve for molarity. The excerpt from the passage above says 8 grams of NaOH dissolved in 2L of H2O. Divide 8 grams of NaOH by the given molar mass to find number of moles of sodium hydroxide (40.0 grams/mol) = 0.2 mol NaOH. Molarity is moles of solute per liter of solution, so it can be found through dividing 0.2 mol NaOH by the 2 liters of H2O to get a molarity of 0.1M. See my handwritten solution below which may be clearer:
103) We’re going to use external knowledge about equivalence pH and equivalence points to answer this question. Coming in, we should know that a weak acid and a strong base, like this situation, would yield an equivalence point at a pH of above 7. But how do we know this? The pH at the equivalence point will be the same as the pH of the sodium benzoate and water formed. The question stem mentions that benzoic acid is a weak acid. And we know NaOH is a strong base. How do we know that?
We should have our common strong bases memorized (above), but also group I hydroxides are typically strong bases. But, because benzoic acid is a weak acid, benzoic acid’s conjugate base can react with water to produce hydroxide ion. Hydroxide ions cause pH to increase, or be more basic. We can check our work here, and we should note this general trend. If we only know the equivalence point occurs at a pH above 7, we could conclude the acid is a weak acid, because sodium hydroxide is a strong base. If benzoic acid was a strong acid, the equivalence point would be at pH 7.
104) This is almost like a Standalone question. No need to go back to the passage to answer, except if you didn’t know chlorobenzoic acid’s formula. Chlorobenzoic acid’s formula is actually given in Table 1 as HC7H4ClO2. A conjugate base is the compound that remains after an acid has donated a proton during a chemical reaction. The conjugate base is C7H4ClO2–.
105) To answer this question, we need information from Table 1, Table 2, and information about equivalence weights. Equivalence weight of an acid is the quantity of the acid that yields one mole of hydrogen ions. For succinic acid, find the molar mass using Table 1. We have H2C4H4O4. We can use the periodic table and the corresponding subscripts to solve for molar mass. 2+48+4+64= 118 g/mol Student E’s equivalent wt. for succinic acid is found to be 59.1. This means that 59.1 grams of succinic acid, when dissociated, will yield one mole of hydrogen ions. Think about what that means. Because this is only one half of the molar mass of succinic acid, the titration needs another mole of NaOH per mole of succinic acid. There are two titratable hydrogen atoms per molecule.
Chemistry Question Pack: Passage 19 106) To answer this question, we’re locating and identifying the block of the periodic table where we’ll find aluminum, and we’re picking from the 4 blocks: S, P, D, and F. This is almost like a Standalone question so there’s no need to go back to the passage to answer. Our passage was focused on aluminum, but we can answer this question using just the periodic table: We are dealing with the properties of a specific element, aluminum, so it’s expected we use our periodic table. Identify our element of interest, aluminum. It’s in group 13 and atomic number is 13 as well. How do we classify blocks of the periodic table? Blocks are split and named based on the valence electrons or valence shells of elements. In general, the first two groups are S-block, the transition metals are D-block, the right side of the table is P-block, and the F-block elements are the lanthanides and actinides (shown at the bottom of the periodic table). Where does aluminum land? P-block.
107) We want to know the oxidation number of a specific atom here, the aluminum in sodium aluminate. This is like a Standalone question just like our last question. We can answer this question without using the passage. The passage was about aluminum, but we don’t need to know specifics about aluminum from the passage to answer this. We’ll go one by one through the components of our compound, and use our rules for oxidation numbers. Keep in mind the sum of all oxidation numbers in our compound is going to be zero because our compound is uncharged. Oxidation number of alkali metals is +1 so we can write that above our sodium. Oxidation number of an ion like hydroxide ion is equal to the charge on the ion, so we have -1. We have four hydroxides, so a total of -4. Now we can solve for the oxidation number of aluminum. Remember, we want the compound to be neutral. That means aluminum has an oxidation number of +3. We have a numerical value, so we can quickly go through our answer choices. We’ll stick with our correct answer choice C, the oxidation number of aluminum in the compound is +3. 108) Just like the previous questions in this question set, we won’t need the passage to answer this question. We’re going to use VSEPR theory. We need to break down the compound one step at a time. Aluminum has a charge of +3. Every fluorine has a charge of -1. So, we have a central aluminum, surrounded by 6 ligands. The molecular geometry and electron geometry is going to be octahedral because of the 6 electron-dense areas according to VSEPR theory. So main takeaway here? Know your content! This question can be answered in a matter of seconds because we know VSEPR theory.
109) We’re going to use the information in the passage, and then use dimensional analysis to solve for the aluminum oxide required. Note our units are in kilograms, and the question stem says “approximately.” I’ll write out the process here, then I’ll include handwritten notes that are easier to follow visually. 2 moles of liquid aluminum oxide (Al2O3) forms 4 moles of pure, solid aluminum. Next step, 100 kg of Al is how many moles? Molar mass of aluminum is 27 g/mol, so we’re going to round to 25 g/mol. Next: dimensional analysis: 100 kg is 100,000 g. Dividing 100,000 by 25 gives 4000 moles of solid aluminum. One thing we want to remember and note from our math lecture. This is an overestimation. We rounded our divisor down. So the actual number of moles should be slightly less than 4,000. Next thing we want to do, is multiply by the ratio of moles (2 moles aluminum oxide / 4 moles pure aluminum) = 2,000 moles aluminum oxide. Multiply by the molar mass of aluminum oxide. Molar mass is 102, we can round that down to 100. Notice again, we rounded a number, but this time we rounded our multiplier down. This helps compensate for our last set of rounding where we said we had an overestimation. 2000 moles aluminum oxide x ~100 g/mol molar mass= 200,000 g aluminum oxide, or approximately 200 kg
Rest assured AAMC knows we’re not doing this math with calculators. They’re not going to make the calculations very easy on purpose, but typically we see questions like this. We’re expected to approximate, and the answers differ by large orders of magnitude. The answer choices they usually give us differ in how they’re obtained, rather than the details of the specific calculations. Another thing we can potentially do is look at the answer choices beforehand and get a sense of how close our approximations need to be. In most cases, the most important thing is to keep our units straight, and do our calculations properly. Rounding is expected of us. 110) We’re identifying the role of aluminum hydroxide in Equation 1. We’re going to use the equation in the passage, then we’re going to analyze aluminum hydroxide using what we know about acids and bases. Glancing at the answer choices, we have to decide between Lewis and Bronsted acids and bases. Lewis acids and Lewis bases deal with the movement of electrons. Lewis bases donate electrons, and lewis acids are electron acceptors. Bronsted-Lowry acids donate protons and Bronsted-Lowry bases accept protons. Be aware that every Bronsted-Lowry acid and base will also be a Lewis acid or base. Therefore, it’s unlikely our answer will be Bronsted acid or base. We can only have one correct answer choice. The excerpt from the passage says aluminum hydroxide reacts with sodium hydroxide to produce sodium aluminate. What happens to our aluminum here? Is it donating or accepting electrons? The aluminum in aluminum hydroxide is accepting that electron pair from the hydroxide. Lewis acids are electron acceptors, so our answer is going to be a Lewis acid. We already compared the 4 choices briefly. We said our predicted answer is that the aluminum in aluminum hydroxide is accepting that electron pair from the hydroxide. That makes it a Lewis acid by definition. That’s answer choice A. We also said every Bronsted-Lowry acid and base will also be a Lewis acid or base, but not necessarily the other way around. We can eliminate answer choices C and D because it’s not possible for a question to have two correct answers. That means we’re left with our correct answer choice: Lewis Acid, answer choice A. 111) We’re asked about the final step in producing pure aluminum through electrolysis. We’re going to have to use what we know about galvanic and electrolytic cells. First step, we need to know how electrolytic cells work. Electrical energy is used to drive nonspontaneous reactions. The oxidation half reaction occurs at the anode, and the reduction half reaction takes place at the cathode. Electrons flow from the positive anode to negative cathode. Next, in a simple Galvanic cell, chemical energy is converted to electrical energy. The oxidation half reaction occurs at the anode, while the reduction half reaction takes place at the cathode. So, the definition of both cathode and anode is the same for both. The reduction takes place at the cathode and oxidation occurs at the anode. Aluminum ion going from a +3 charge in aluminum oxide, to a 0 charge in solid aluminum is a decrease in the oxidation state of aluminum. This means aluminum is reduced.
Chemistry Question Pack: Passage 20 112) We want to know the effect on our experiment if a gas besides oxygen enters into our experimental apparatus. The experimental apparatus is used to determine ultrasmall concentrations of oxygen, so we want to know how adding another gas to be reduced by the apparatus will affect our concentration reading. The passage specifically mentions how the apparatus works in detail, so we should be able to identify the effect of adding another gas. We were told coulometry is a means of monitoring gases. Coulometry measures the amount of matter transformed during the reaction, but is not going to specifically find the amount of oxygen. That’s a small distinction, but an important one. So, what do we expect happens to the accuracy of our reading? Accuracy will go down-there’s now an increased amount of reduced gas being measured. The experimental setup is not going to discriminate between oxygen and a new gas.
113) To answer this question, we’ll be using the diagram in the passage, and we’ll be explaining a specific aspect of the diagram: the cadmium electrode. We’ll be using external knowledge to explain the role of the specific electrode. The passage says reduction occurs at the silver electrode and the electrochemical reaction is completed at the cadmium electrode. In a Galvanic cell, the oxidation half reaction occurs at the anode, while the reduction half reaction takes place at the cathode. In this case, the silver electrode is the cathode where the reduction half reaction takes place; the cadmium electrode is the anode where oxidation takes place. In Equation 3, oxidation is taking place, the cadmium is losing two electrons.
114) PPM is parts per million. We’re finding a numerator, that when divided by 1,000,000, will give us 1%. So the “part” in “parts per million” to make it equal to 1%. We can solve this using the question stem alone; we just need to use dimensional analysis. I’ve handwritten the process above so it’s easier to follow along. First, write out the ratios. 1 part per 100 is equal to 1%. An unknown part per 1,000,000 is also equal to 1%. We’re simply solving for X here. Isolating X gives us 10,000 parts per million. Dimensional analysis to the rescue! It’s not uncommon to get an extra two or three questions right per section just by keeping your units straight and using dimensional analysis. We did a simple math conversion without any rounding. Answer choice C matches our calculated 10,000. 115) In other words, can methane gas replace oxygen in the experiment to be able to find its concentration? We’re going to need to know specifics about the experiment from the passage, but we’ll have to know about the properties of oxygen. Additionally, we’ll have to know specifically about methane to see if it’s a candidate for determination by coulometry described in the passage. The experiment requires a gas to be reduced at the anode, exactly how oxygen is reduced in the passage. We would have to reduce methane gas at the cathode. Remember, the oxidation half reaction occurs at the anode, while the reduction half reaction takes place at the cathode.
Chemistry Question Pack: Questions 116-120 116) To answer this question, we can quickly go through what we know about catalysts. Catalysts are chemical compounds that increase the rate of a reaction by lowering the activation energy required to reach the transition state. Unlike reactants, a catalyst is not consumed as part of the reaction process. Catalysts do not change the equilibrium constant of the reaction. Biggest example of a catalyst we’ll see on the MCAT is undoubtedly enzymes. When an enzyme binds its substrate, it forms an enzyme-substrate complex. This complex lowers the activation energy of the reaction and promotes its rapid progression by providing certain ions or chemical groups that actually form covalent bonds with molecules as a necessary step of the reaction process. Enzymes also promote chemical reactions by bringing substrates together in an optimal orientation and lining up the atoms and bonds of one molecule with the atoms and bonds of the other molecule. This can contort the substrate molecules and facilitate bond-breaking. The active site of an enzyme also creates an ideal environment, such as a slightly acidic or non-polar environment, for the reaction to occur.
117) Let’s think about the definition of conjugate bases. A conjugate base is the compound that remains after an acid has donated a proton during a chemical reaction. In other words, a conjugate base is formed when an acid loses a proton. A conjugate acid is formed when a base gains a proton. When we’re looking at the bisulfate ion, we can remove a proton from HSO4– which gets us SO42-. This is a straightforward and objective answer. If you accidentally find the conjugate acid instead, you might pick answer choice D. However, we solved for our correct conjugate base: SO42- which is answer choice C. 118) This question just boils down to reading the phase diagram carefully. Our phase diagram of water has two variables on its axes, pressure and temperature. As these conditions vary, so do the phases of water. The phases vary from solid, to liquid, or vapor based on the variation in pressure and temperature. The phase all the way to the left is the solid phase. The rightmost phase is the gaseous (vapor) phase. The topmost and middle phase is the liquid phase. This should inherently make sense. Ice is present at the lowest temperatures, steam is present at the highest temperatures, and liquid is present in between these two phases. Note, in this question specifically, we’re looking at negative 0.1oC, while the x-axis shows -3 oC and 0.01 oC. That means we’re looking at somewhere between the two labeled points along the x-axis. We can start at 1.0 torr along the y-axis which intersects our phase diagram at -3.0 oC. As we move slightly right (toward -0.1 oC) we’re in the vapor phase. We’re told pressure increases from 1.0 torr to 200 atm. 200 atm is roughly 152,000 Torr, so there’s quite a bit of pressure increase. As we trace up from the x-axis from -0.1 oC, we note we go from vapor to solid at around 2 to 3 torr. However, as we keep increasing in pressure, we will once again cross into another phase. This time we cross into the liquid phase. So, we expect to go from vapor to solid to liquid.
119) First thing we want to note is, what do we know about sodium hydroxide? It’s a strong base:
Strong bases will dissociate completely in an aqueous solution while yielding hydroxide ions. pOH can be solved as the negative log of hydroxide ion concentration. The sodium hydroxide will dissociate completely, and hydroxide ion concentration ends up being the same as our molarity. Our molarity is 0.001 molar which can be rewritten using scientific notation. (If you don’t remember which way to move the decimal point, go back to the mnemonic LARS, which is Left Add, Right Subtract). We start with 0.001 x 10^0. We move the decimal point three places to the right to get a whole number, meaning we subtract three from our exponent to give us 1.0 x 10^-3 molar. Let’s plug in our numbers. We know pOH is the negative log of our hydroxide ion concentration. Our hydroxide ion concentration, again, is the same as our molarity. So plug in 1x 10^-3 as our hydroxide ion concentration. So when we take our negative log, we find the pOH is 3. However, we’re looking for pH. To find pH, we can simply subtract pOH from 14: 14-3 = 11.
120) A gas is considered ideal if its particles are so far apart that they do not exert any attractive forces upon one another. In real life, there is no such thing as a truly ideal gas, but at high temperatures and low pressures (conditions in which individual particles will be moving very quickly and be very far apart from one another so that their interaction is almost zero), gases behave close to ideally. For ideal gases, at Standard Temperature (273K or 0 oC) and Pressure (1 atm), 1 mole of any gas will occupy a volume of 22.4 L.
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Which of the following equations shows the net reaction when the temperature was 300 K
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