Some G-Chem questions

[SBK]

---

Members don't see this ad.

I would GREATLY appreciate some help on these questions.

After going through like 200 MCAT General Chem questions, I'm stuck on understanding why these are 'correct' and the internet is not helping me out. Could any goons please help me with any of these. It would be immensely appreciated.


For the hydration of carbon dioxide, which of the following statements is true?
A. Carbon dioxide is a Lewis base, because it accepts a pair of electrons from water.
B. Carbon dioxide is a Brønsted–Lowry base, because it accepts a proton from water.
C. Water is a Brønsted–Lowry base, because it donates a proton to carbon dioxide.
D. Water is a Lewis base, because it donates a pair of electrons to carbon dioxide.
E. Water is a Lewis acid, because it donates a pair of electrons to carbon dioxide.

As reagents, nitrogen gas and oxygen gas react to form nitrogen monoxide. The reaction is endothermic,
and equilibrium is established in a rigid container. Which of the following changes to the system favor the
formation of nitrogen monoxide?
A. Increasing the pressure
B. Increasing the temperature
C. Decreasing the temperature
D. Adding neon gas
E. Removing oxygen gas

An uncatalyzed reaction, A → B, has a rather large negative ∆G of –3000 J/mol. Which of the following
statements must be true?
A. The activation energy of the reverse reaction is greater than that of the forward reaction.
B. The activation energy of the reverse reaction is lower than that of the forward reaction.
C. If a catalyst is added, the value of ∆G for the catalyzed reaction will be smaller in magnitude than the
value of ∆G for the uncatalyzed reaction.
D. If a catalyst is added, the value of ∆G for the catalyzed reaction will be larger in magnitude than the
value of ∆G for the uncatalyzed reaction.
E. Since the magnitude of ∆G is so large, the reaction occurs very rapidly.

Which of the following is a redox reaction?
A. TiCl
4(g) + 2 Mg(l) ∆ Ti(s) + 2 MgCl2(l)
B. NH
3(g) + CO2(g) + H2O(l) NH4HCO3(aq)
C. NaCl(s) + NaHSO
4(s) ∆ HCl(g) + Na2SO4(s)
D. NaOH(aq) + H
3PO4(aq) NaH2PO4(aq) + H2O(l)
E. None of the above

Many transition elements are strongly attracted to a magnetic field, because:
A. they form only cations.
B. most of them are solids at room temperature.
C. most of them are diamagnetic.
D. most of them are paramagnetic.
E. most of them have completely filled s orbitals.

Of the following, which is the strongest acid?
A. Hydrofluoric acid
B. Sulfurous acid
C. Nitric acid
D. Hydrocyanic acid
E. Phosphoric acid

What is the pH of 1.0 M NH3 (Kb = 2 x 10 –5)?
A. 9.6
B. 10.8
C. 11.7
D. 12.4
E. 13.1

An indicator has Ka equal to 8 x 10 –7
. What is the ratio of the concentration of the acid form of the indicator
to the concentration of the basic form, [HIn] / [In–], in a buffer solution in which [H+] = 10–5 M?
A. 1/0.08
B. 8/1
C. 1/0.8
D. 1/8
E. 0.08/1

The primary reason for the nonexistence of NaCl 2 is that:
A. Cl has a low electron affinity.
B. the Cl2–ion has been observed only as a ligand in coordination complexes of transition metal ions.
C. NaCl2 would have very high lattice energy.
D. the expected repulsion between the Cl atoms is strong.
E. Na has a high second ionization energy.

If hypochlorous acid is titrated with sodium hydroxide, the pH at the equivalence point is:
A. equal to one-half the pH at the half-equivalence point.
B. less than 7.
C. equal to 7.
D. greater than 7.

Sodium chloride readily dissolves in water even though the reaction is endothermic. Which of the following
statements best explains this phenomenon?
A. Heat is released when the sodium and chloride ions are hydrated.
B. Heat is required for the sodium and chloride ions to be hydrated.
C. The negative entropy change of dissolution results in a net decrease in Gibbs free energy.
D. The positive entropy change of dissolution results in a net decrease in Gibbs free energy.

Of the following pairs of atoms, which have the greatest difference in size?
A. Ni and Cu
B. F and I
C. K and Br
D. Cs and F
E. Li and Br

NO2 is soluble in carbon tetrachloride. As it dissolves, it dimerizes into N2O4. If the equilibrium concentration
of NO2 is 0.5 M at 20 C, what is the equilibrium concentration of N 2O4 given that Kc = 28?
A. (0.05)(28) M
B. (0.05)2(28) M
C. (0.05)2/28 M
D. 28/(0.05) M
E. 28/(0.05)2 M


Given the following half-cell potentials,

Zn2+ + 2 e– → Zn (E ° = –0.76 V)
Cr3+ + 3e– → Cr (E ° = –0.74 V)

Determine the cell potential whose overall redox reaction under standard conditions is
3 Zn(s) + 2 Cr3+(aq) → 3 Zn2+(aq) + 2 Cr(s)
A. –0.80 V
B. –0.02 V
C. +0.02 V
D. +0.80 V

 

[loveoforganic]

I'd say 1 is d. A lewis base is defined as an e- donor. The nucleophilic oxygen of water attacks the electrophilic carbon of co2.

I'd say 2 is b. This is LeChatlier's principle.

I'd say 3 is a. The reaction is exothermic, so it has to take more energy to reach the TS going back to the reactants than to the products.

5 is either c or d, don't remember. It has to do with unpaired electrons in the MO diagram.

6 is C. Nitric acid is the only strong acid in that list. Just have to memorize your strong acids.

7 is C. Kb = [OH-][HA] / [A-]; [A-] is about equal to cA-. [OH-] = [HA] cA- = 1M so [OH-]^2 / 1 = 2*10^-5. [OH-] = .00447 14 - pOH = pH = 11.65

11 is D. Entropy (S) increases significantly as the crystal's ions separate. dG = dH - TdS. S is large enough to outweigh dH

That's all I have time for right now, GL!

 

[sleepy425]

I'll take the rest...

Number 4 is A. The surefire sign of a redox reaction is going from the elemental form of an element to having the element in a salt, or vice versa. So you start out with Mg on the left side, with an oxidation state of 0, and you oxidize it to a salt, where Mg has an oxidation state of +2. And Ti starts out as a salt with an oxidation state of +4 and it gets reduced to elemental Ti with an oxidation state of 0. Choice B is just nucleophilic attack of carbon dioxide by water, followed by ammonia pulling a proton off of water to make the ammonium carbonate salt. Choice C is an acid/base reaction where Cl- pulls off a proton from HSO4- to make SO4(2-) and HCl. And the counterion for the sulfate anion is 2 sodium cations. Choice D is a very simple acid/base reaction.

Number 5 is D, because paramagnetism is a strong response to the applied magnetic field, whereas diamagnetism is a weak response to the applied magnetic field.

Number 8 deals with indicators. People have a lot of trouble with indicators, but indicators are no different from buffered solutions. So here's how an acid/base indicator works: You're doing some acid base reaction and you know that, at the end point, which is where you've consumed all of your analyte and added one more drop of titrant, there will be a significant change in pH. You choose an indicator that has a pKa that is in between the pH at the equivalence point and the pH at the end point (so basically, say your analyte is a base and you're titrating with an acid. your pH starts at 11, but when you reach the equivalence point the pH is down to 9. Since there is no more base to buffer the solution, when you add one more drop of your acid, the pH drops down to 4 or so. So you choose an indicator with a pKa between 4 and 9, so say 7. At pH 9, the indicator is all in its basic form, and at pH 4, it's all in the acidic form. Since the acidic and basic forms have different colors, the indicator tells you when you're done).

Ok, so that was a brief explanation of indicators, just to show you that it follows simple acid/base chemistry. So they give us the Ka of the indicator, and they give us the proton concentration. For me, it is easiest to work with these as pKa and pH, so from simple math, I know that the pH is 5. The pKa isn't as easy to do in your head, but you know that 8*10^-7 is between 10^-6 and 10^-7, so the pKa has to be between 6 and 7, but closer to 6. Ok, now you can do the henderson hasselbach equation, but it's much easier to use this trick:

if pH=pKa, [HA]=[A-]
if pH is one more than pKa, you have 10 times more of the basic form than the acidic form
if pH is 2 more than pKa, you have 100 times more of the basic form than of the acidic form
if pH is one less than pKa, you have 10 times less basic form than acidic form
if pH is 2 less than pKa, you have 100 times less basic form than acidic form

So in our case, our pH is a little more than 1 unit less than our pKa, so that means that we have a little more than 10 times less of the basic form than of the acidic form. So that means that [HIn]/[In-] must be a little more than 10. C is close to 1, B is less than 10, and A is a little more than 10, so the answer is A.

Number 9 is E because you can't get Na(2+) because pulling off the second electron from sodium is very hard because Na(+) has 8 electrons in its valence shell

Number 10 is a weak acid titrated against a strong base, so the equivalence point is going to be greater than 7, so it's choice D.

Number 12 is B. You have to go down 3 rows of the periodic table to get from F to I, and atomic radius increases rapidly as you go down rows. The others are fewer rows down from each other (there is an effect as you go down the columns-radius gets smaller but it is a less significant effect and doesn't really make a difference here).

Number 13 is B. The reaction is 2NO2---->N2O4. Kc=[N2O4]/[NO2]^2. So plugging in the values gives us 28*.05^2.

Number 14 is D. They give you the balanced equation, so basically you have to figure out what combination of the two half reactions adds up to that balanced equation. If you flip the zinc reaction and multiply by 3, and add that to 2 times the chromium reaction, you get the balanced equation. When you flip the zinc reaction, the sign of the potential is flipped. So now we have 3*.76+2*(-.74) so that's just .80 V.

 

[free09]

SBK, can you post the correct answers for verification.Thanks.

 

[SteinUmStein]

Given the following half-cell potentials,

Zn2+ + 2 e– → Zn (E ° = –0.76 V)
Cr3+ + 3e– → Cr (E ° = –0.74 V)

Determine the cell potential whose overall redox reaction under standard conditions is
3 Zn(s) + 2 Cr3+(aq) → 3 Zn2+(aq) + 2 Cr(s)
A. –0.80 V
B. –0.02 V
C. +0.02 V
D. +0.80 V

Number 14 is D. They give you the balanced equation, so basically you have to figure out what combination of the two half reactions adds up to that balanced equation. If you flip the zinc reaction and multiply by 3, and add that to 2 times the chromium reaction, you get the balanced equation. When you flip the zinc reaction, the sign of the potential is flipped. So now we have 3*.76+2*(-.74) so that's just .80 V.

Please correct me if I'm wrong, but at some point I was told that the reactions are not multiplied by the reaction coefficients when calculating cell potentials. If that's true, then flipping the zinc reaction and adding the chromium reaction would yield: .76+(-.74)= .02, or answer C.

 

[sleepy425]

Please correct me if I'm wrong, but at some point I was told that the reactions are not multiplied by the reaction coefficients when calculating cell potentials. If that's true, then flipping the zinc reaction and adding the chromium reaction would yield: .76+(-.74)= .02, or answer C.

yes, sorry, I've been really frikin ******ed lately. yes, you're right, it should be C

 

[JohnnytheNomad]

I'd say 1 is d. A lewis base is defined as an e- donor. The nucleophilic oxygen of water attacks the electrophilic carbon of co2.

I'd say 2 is b. This is LeChatlier's principle.

I'd say 3 is a. The reaction is exothermic, so it has to take more energy to reach the TS going back to the reactants than to the products.

5 is either c or d, don't remember. It has to do with unpaired electrons in the MO diagram.

6 is C. Nitric acid is the only strong acid in that list. Just have to memorize your strong acids.

7 is C. Kb = [OH-][HA] / [A-]; [A-] is about equal to cA-. [OH-] = [HA] cA- = 1M so [OH-]^2 / 1 = 2*10^-5. [OH-] = .00447 14 - pOH = pH = 11.65

11 is D. Entropy (S) increases significantly as the crystal's ions separate. dG = dH - TdS. S is large enough to outweigh dH

That's all I have time for right now, GL!

2 is A. The question says the reaction is already at equilibrium, adding heat just enables the reaction to get to equilibrium faster, where as if you add pressure, this is LeChat, look a the moles of gas. Adding pressure favors the least amount of moles of gas, thus pushing the reaction to the right.

 

[JohnnytheNomad]

I'd say 1 is d. A lewis base is defined as an e- donor. The nucleophilic oxygen of water attacks the electrophilic carbon of co2.

I'd say 2 is b. This is LeChatlier's principle.

I'd say 3 is a. The reaction is exothermic, so it has to take more energy to reach the TS going back to the reactants than to the products.

5 is either c or d, don't remember. It has to do with unpaired electrons in the MO diagram.

6 is C. Nitric acid is the only strong acid in that list. Just have to memorize your strong acids.

7 is C. Kb = [OH-][HA] / [A-]; [A-] is about equal to cA-. [OH-] = [HA] cA- = 1M so [OH-]^2 / 1 = 2*10^-5. [OH-] = .00447 14 - pOH = pH = 11.65

11 is D. Entropy (S) increases significantly as the crystal's ions separate. dG = dH - TdS. S is large enough to outweigh dH

That's all I have time for right now, GL!

I think i have to disagree with you on 7. the kb is 2 E -5 , i thought you could think of pkb = pOH, if this is true, just using logic this would mean a pkb between 4 and 5, so the pH = 14 - pOH, so the pH would have to be between 9 and 10, making the answer A. Also, just look at the base you are using to make the solution alkaline. This is a weak base. Please let me know if I am wrong. [email protected]

 

[JohnnytheNomad]

I'll take the rest...

Number 4 is A. The surefire sign of a redox reaction is going from the elemental form of an element to having the element in a salt, or vice versa. So you start out with Mg on the left side, with an oxidation state of 0, and you oxidize it to a salt, where Mg has an oxidation state of +2. And Ti starts out as a salt with an oxidation state of +4 and it gets reduced to elemental Ti with an oxidation state of 0. Choice B is just nucleophilic attack of carbon dioxide by water, followed by ammonia pulling a proton off of water to make the ammonium carbonate salt. Choice C is an acid/base reaction where Cl- pulls off a proton from HSO4- to make SO4(2-) and HCl. And the counterion for the sulfate anion is 2 sodium cations. Choice D is a very simple acid/base reaction.

Number 5 is D, because paramagnetism is a strong response to the applied magnetic field, whereas diamagnetism is a weak response to the applied magnetic field.

Number 8 deals with indicators. People have a lot of trouble with indicators, but indicators are no different from buffered solutions. So here's how an acid/base indicator works: You're doing some acid base reaction and you know that, at the end point, which is where you've consumed all of your analyte and added one more drop of titrant, there will be a significant change in pH. You choose an indicator that has a pKa that is in between the pH at the equivalence point and the pH at the end point (so basically, say your analyte is a base and you're titrating with an acid. your pH starts at 11, but when you reach the equivalence point the pH is down to 9. Since there is no more base to buffer the solution, when you add one more drop of your acid, the pH drops down to 4 or so. So you choose an indicator with a pKa between 4 and 9, so say 7. At pH 9, the indicator is all in its basic form, and at pH 4, it's all in the acidic form. Since the acidic and basic forms have different colors, the indicator tells you when you're done).

Ok, so that was a brief explanation of indicators, just to show you that it follows simple acid/base chemistry. So they give us the Ka of the indicator, and they give us the proton concentration. For me, it is easiest to work with these as pKa and pH, so from simple math, I know that the pH is 5. The pKa isn't as easy to do in your head, but you know that 8*10^-7 is between 10^-6 and 10^-7, so the pKa has to be between 6 and 7, but closer to 6. Ok, now you can do the henderson hasselbach equation, but it's much easier to use this trick:

if pH=pKa, [HA]=[A-]
if pH is one more than pKa, you have 10 times more of the basic form than the acidic form
if pH is 2 more than pKa, you have 100 times more of the basic form than of the acidic form
if pH is one less than pKa, you have 10 times less basic form than acidic form
if pH is 2 less than pKa, you have 100 times less basic form than acidic form

So in our case, our pH is a little more than 1 unit less than our pKa, so that means that we have a little more than 10 times less of the basic form than of the acidic form. So that means that [HIn]/[In-] must be a little more than 10. C is close to 1, B is less than 10, and A is a little more than 10, so the answer is A.

Number 9 is E because you can't get Na(2+) because pulling off the second electron from sodium is very hard because Na(+) has 8 electrons in its valence shell

Number 10 is a weak acid titrated against a strong base, so the equivalence point is going to be greater than 7, so it's choice D.

Number 12 is B. You have to go down 3 rows of the periodic table to get from F to I, and atomic radius increases rapidly as you go down rows. The others are fewer rows down from each other (there is an effect as you go down the columns-radius gets smaller but it is a less significant effect and doesn't really make a difference here).

Number 13 is B. The reaction is 2NO2---->N2O4. Kc=[N2O4]/[NO2]^2. So plugging in the values gives us 28*.05^2.

Number 14 is D. They give you the balanced equation, so basically you have to figure out what combination of the two half reactions adds up to that balanced equation. If you flip the zinc reaction and multiply by 3, and add that to 2 times the chromium reaction, you get the balanced equation. When you flip the zinc reaction, the sign of the potential is flipped. So now we have 3*.76+2*(-.74) so that's just .80 V.

I don't agree with number 8 either. You are great with it up to being between one or two less. This being equal to 1/10 or 1/100. so the answer should be between .1 and .01, making the answer .08/1 = .08. Please correct me if I am wrong on these.

 

[sleepy425]

I don't agree with number 8 either. You are great with it up to being between one or two less. This being equal to 1/10 or 1/100. so the answer should be between .1 and .01, making the answer .08/1 = .08. Please correct me if I am wrong on these.

I think you must have misread the question. It's asking for the ratio of [HIn]/[In-]. Since the pH is less than the pKa of the indicator, you have to have more in the acidic form than in the basic form. So the answer has to be greater than 1. My answer is correct, it's 1/.08.

 

[sleepy425]

2 is A. The question says the reaction is already at equilibrium, adding heat just enables the reaction to get to equilibrium faster, where as if you add pressure, this is LeChat, look a the moles of gas. Adding pressure favors the least amount of moles of gas, thus pushing the reaction to the right.

No, loveoforganic was correct, 2 is B. The number of moles of gas is the same on both sides. It's N2+O2<----->2NO

So it's 2 molecules of gas on each side, so pressure does not shift equilibrium.

Adding heat does favor NO formation. The trick they commonly show is that, if it's endothermic, you can include "heat" as a reactant, so it's like adding more reactant. This is a crude explanation, the real reason is derived from statistical mechanics, but it holds for all cases.

 

[sleepy425]

I think i have to disagree with you on 7. the kb is 2 E -5 , i thought you could think of pkb = pOH, if this is true, just using logic this would mean a pkb between 4 and 5, so the pH = 14 - pOH, so the pH would have to be between 9 and 10, making the answer A. Also, just look at the base you are using to make the solution alkaline. This is a weak base. Please let me know if I am wrong. [email protected]

pKb does not equal pOH. Kb is the equilibrium constant for the expression:

NH3<---->NH4+ + OH-

So at equilibrium the expression is x^2/(1-x)=2*10^-5

Solving for x and taking the negative log gives you pOH, and then 14 minus that is your pH, which is 11.65

 

[robrobbberts]

I'll take the rest...

Number 4 is A. The surefire sign of a redox reaction is going from the elemental form of an element to having the element in a salt, or vice versa. So you start out with Mg on the left side, with an oxidation state of 0, and you oxidize it to a salt, where Mg has an oxidation state of +2. And Ti starts out as a salt with an oxidation state of +4 and it gets reduced to elemental Ti with an oxidation state of 0. Choice B is just nucleophilic attack of carbon dioxide by water, followed by ammonia pulling a proton off of water to make the ammonium carbonate salt. Choice C is an acid/base reaction where Cl- pulls off a proton from HSO4- to make SO4(2-) and HCl. And the counterion for the sulfate anion is 2 sodium cations. Choice D is a very simple acid/base reaction.

Number 5 is D, because paramagnetism is a strong response to the applied magnetic field, whereas diamagnetism is a weak response to the applied magnetic field.

Number 8 deals with indicators. People have a lot of trouble with indicators, but indicators are no different from buffered solutions. So here's how an acid/base indicator works: You're doing some acid base reaction and you know that, at the end point, which is where you've consumed all of your analyte and added one more drop of titrant, there will be a significant change in pH. You choose an indicator that has a pKa that is in between the pH at the equivalence point and the pH at the end point (so basically, say your analyte is a base and you're titrating with an acid. your pH starts at 11, but when you reach the equivalence point the pH is down to 9. Since there is no more base to buffer the solution, when you add one more drop of your acid, the pH drops down to 4 or so. So you choose an indicator with a pKa between 4 and 9, so say 7. At pH 9, the indicator is all in its basic form, and at pH 4, it's all in the acidic form. Since the acidic and basic forms have different colors, the indicator tells you when you're done).

Ok, so that was a brief explanation of indicators, just to show you that it follows simple acid/base chemistry. So they give us the Ka of the indicator, and they give us the proton concentration. For me, it is easiest to work with these as pKa and pH, so from simple math, I know that the pH is 5. The pKa isn't as easy to do in your head, but you know that 8*10^-7 is between 10^-6 and 10^-7, so the pKa has to be between 6 and 7, but closer to 6. Ok, now you can do the henderson hasselbach equation, but it's much easier to use this trick:

if pH=pKa, [HA]=[A-]
if pH is one more than pKa, you have 10 times more of the basic form than the acidic form
if pH is 2 more than pKa, you have 100 times more of the basic form than of the acidic form
if pH is one less than pKa, you have 10 times less basic form than acidic form
if pH is 2 less than pKa, you have 100 times less basic form than acidic form

So in our case, our pH is a little more than 1 unit less than our pKa, so that means that we have a little more than 10 times less of the basic form than of the acidic form. So that means that [HIn]/[In-] must be a little more than 10. C is close to 1, B is less than 10, and A is a little more than 10, so the answer is A.

Number 9 is E because you can't get Na(2+) because pulling off the second electron from sodium is very hard because Na(+) has 8 electrons in its valence shell

Number 10 is a weak acid titrated against a strong base, so the equivalence point is going to be greater than 7, so it's choice D.

Number 12 is B. You have to go down 3 rows of the periodic table to get from F to I, and atomic radius increases rapidly as you go down rows. The others are fewer rows down from each other (there is an effect as you go down the columns-radius gets smaller but it is a less significant effect and doesn't really make a difference here).

Number 13 is B. The reaction is 2NO2---->N2O4. Kc=[N2O4]/[NO2]^2. So plugging in the values gives us 28*.05^2.

Number 14 is D. They give you the balanced equation, so basically you have to figure out what combination of the two half reactions adds up to that balanced equation. If you flip the zinc reaction and multiply by 3, and add that to 2 times the chromium reaction, you get the balanced equation. When you flip the zinc reaction, the sign of the potential is flipped. So now we have 3*.76+2*(-.74) so that's just .80 V.

#12 is definitely D. Cs is one of the largest atoms there is. its all the way on the bottom left, F is all the way on the top right.. making it one of the smallest atoms.



and i think you over complicated number 8.

it gives the the Ka, and states that in the solution [H+] = 10–5 M.

so, set up Ka = [H+][A-]/[HA] where A- = [In-] and HA = [HIn] .. and just solve for [HIn]/[In-] .. its much simpler.

 

[sleepy425]

#12 is definitely D. Cs is one of the largest atoms there is. its all the way on the bottom left, F is all the way on the top right.. making it one of the smallest atoms.



and i think you over complicated number 8.

it gives the the Ka, and states that in the solution [H+] = 10–5 M.

so, set up Ka = [H+][A-]/[HA] where A- = [In-] and HA = [HIn] .. and just solve for [HIn]/[In-] .. its much simpler.

yeah, you're right about 12, I think I didn't see the Cs or something, but yeah, definitely, D. As for number 8, I didn't overcomplicate it at all. I went through the explanation because a lot of people get freaked out by indicators because they don't know how indicators work. So I just explained how they work. I used pKa and pH because I like to, it doesn't overcomplicate it, it's exactly the same thing as the Ka equation. But yeah, Cs is definitely bigger than I, so it'd be D for that one.

 

[robrobbberts]

yeah, you're right about 12, I think I didn't see the Cs or something, but yeah, definitely, D. As for number 8, I didn't overcomplicate it at all. I went through the explanation because a lot of people get freaked out by indicators because they don't know how indicators work. So I just explained how they work. I used pKa and pH because I like to, it doesn't overcomplicate it, it's exactly the same thing as the Ka equation. But yeah, Cs is definitely bigger than I, so it'd be D for that one.

I must confess, i didn't actually read everything you wrote.. I just saw the huge block of text and was like whoa! 🙂

 

[sleepy425]

I must confess, i didn't actually read everything you wrote.. I just saw the huge block of text and was like whoa! 🙂

lol, it's cool. Thanks for catching the other mistake!

 

[Mikey_Med]

I would GREATLY appreciate some help on these questions.

After going through like 200 MCAT General Chem questions, I'm stuck on understanding why these are 'correct' and the internet is not helping me out. Could any goons please help me with any of these. It would be immensely appreciated.


For the hydration of carbon dioxide, which of the following statements is true?
A. Carbon dioxide is a Lewis base, because it accepts a pair of electrons from water.
B. Carbon dioxide is a Brønsted–Lowry base, because it accepts a proton from water.
C. Water is a Brønsted–Lowry base, because it donates a proton to carbon dioxide.
D. Water is a Lewis base, because it donates a pair of electrons to carbon dioxide.
E. Water is a Lewis acid, because it donates a pair of electrons to carbon dioxide.

As reagents, nitrogen gas and oxygen gas react to form nitrogen monoxide. The reaction is endothermic,
and equilibrium is established in a rigid container. Which of the following changes to the system favor the
formation of nitrogen monoxide?
A. Increasing the pressure
B. Increasing the temperature
C. Decreasing the temperature
D. Adding neon gas
E. Removing oxygen gas

An uncatalyzed reaction, A &#8594; B, has a rather large negative &#8710;G of –3000 J/mol. Which of the following
statements must be true?
A. The activation energy of the reverse reaction is greater than that of the forward reaction.
B. The activation energy of the reverse reaction is lower than that of the forward reaction.
C. If a catalyst is added, the value of &#8710;G for the catalyzed reaction will be smaller in magnitude than the
value of &#8710;G for the uncatalyzed reaction.
D. If a catalyst is added, the value of &#8710;G for the catalyzed reaction will be larger in magnitude than the
value of &#8710;G for the uncatalyzed reaction.
E. Since the magnitude of &#8710;G is so large, the reaction occurs very rapidly.

Which of the following is a redox reaction?
A. TiCl
4(g) + 2 Mg(l) &#8710; Ti(s) + 2 MgCl2(l)
B. NH
3(g) + CO2(g) + H2O(l) NH4HCO3(aq)
C. NaCl(s) + NaHSO
4(s) &#8710; HCl(g) + Na2SO4(s)
D. NaOH(aq) + H
3PO4(aq) NaH2PO4(aq) + H2O(l)
E. None of the above

Many transition elements are strongly attracted to a magnetic field, because:
A. they form only cations.
B. most of them are solids at room temperature.
C. most of them are diamagnetic.
D. most of them are paramagnetic.
E. most of them have completely filled s orbitals.

Of the following, which is the strongest acid?
A. Hydrofluoric acid
B. Sulfurous acid
C. Nitric acid
D. Hydrocyanic acid
E. Phosphoric acid

What is the pH of 1.0 M NH3 (Kb = 2 x 10 –5)?
A. 9.6
B. 10.8
C. 11.7
D. 12.4
E. 13.1

An indicator has Ka equal to 8 x 10 –7
. What is the ratio of the concentration of the acid form of the indicator
to the concentration of the basic form, [HIn] / [In–], in a buffer solution in which [H+] = 10–5 M?
A. 1/0.08
B. 8/1
C. 1/0.8
D. 1/8
E. 0.08/1

The primary reason for the nonexistence of NaCl 2 is that:
A. Cl has a low electron affinity.
B. the Cl2–ion has been observed only as a ligand in coordination complexes of transition metal ions.
C. NaCl2 would have very high lattice energy.
D. the expected repulsion between the Cl atoms is strong.
E. Na has a high second ionization energy.

If hypochlorous acid is titrated with sodium hydroxide, the pH at the equivalence point is:
A. equal to one-half the pH at the half-equivalence point.
B. less than 7.
C. equal to 7.
D. greater than 7.

Sodium chloride readily dissolves in water even though the reaction is endothermic. Which of the following
statements best explains this phenomenon?
A. Heat is released when the sodium and chloride ions are hydrated.
B. Heat is required for the sodium and chloride ions to be hydrated.
C. The negative entropy change of dissolution results in a net decrease in Gibbs free energy.
D. The positive entropy change of dissolution results in a net decrease in Gibbs free energy.

Of the following pairs of atoms, which have the greatest difference in size?
A. Ni and Cu
B. F and I
C. K and Br
D. Cs and F
E. Li and Br

NO2 is soluble in carbon tetrachloride. As it dissolves, it dimerizes into N2O4. If the equilibrium concentration
of NO2 is 0.5 M at 20 C, what is the equilibrium concentration of N 2O4 given that Kc = 28?
A. (0.05)(28) M
B. (0.05)2(28) M
C. (0.05)2/28 M
D. 28/(0.05) M
E. 28/(0.05)2 M


Given the following half-cell potentials,

Zn2+ + 2 e– &#8594; Zn (E ° = –0.76 V)
Cr3+ + 3e– &#8594; Cr (E ° = –0.74 V)

Determine the cell potential whose overall redox reaction under standard conditions is
3 Zn(s) + 2 Cr3+(aq) &#8594; 3 Zn2+(aq) + 2 Cr(s)
A. –0.80 V
B. –0.02 V
C. +0.02 V
D. +0.80 V

Where did you find these questions?! I recently took a test that had all these questions on it.

 

[Czarcasm]

Where did you find these questions?! I recently took a test that had all these questions on it.

What test was it that you took?