General Chemistry Question Thread

[QofQuimica]

All users may post questions about MCAT, DAT, OAT, or PCAT general chemistry here. We will answer the questions as soon as we reasonably can. If you would like to know what general chemistry topics appear on the MCAT, you should check the MCAT Student Manual (http://www.aamc.org/students/mcat/studentmanual/start.htm)

Acceptable topics:
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Unacceptable topics:
-actual MCAT questions or passages, or close paraphrasings thereof
-anything you know to be beyond the scope of the MCAT

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If you really know your gen chem, I can use your help. If you are willing to help answer questions on this thread, please let me know. Here are the current members of the General Chemistry Team:

-QofQuimica (thread moderator): I have my M.S. in organic chemistry and I'm currently finishing my Ph.D., also in organic chemistry. I have several years of university general chemistry TA teaching experience. In addition, I teach general chemistry classes through Kaplan for their MCAT, DAT, OAT, and PCAT courses. On the MCAT, I scored 14 on PS, 43 overall.

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[hmcalley]

question deleted

 

[hmcalley]

This is a partial repost of post#765 on the previous page. (The original post hasn't been answered yet.) I also do not understand how to arrive at the answer. Can someone help? Thanks.

Partial repost:
[Passage ]

Reaction 1
15.0 mL of 0.300 M Pb(NO3)2(aq) was mixed with 15.0 mL of 0.300 M Na2SO4(aq). All the Pb(NO3)2 reacted to form Compound A, a white precipitate. Compound A was removed by filtration.

Reaction 2
15.0 mL of 0.300 M KI(aq) was added to Compound A. The mixture was agitated and some of Compound A dissolved. In addition, a yellow precipitate of PbI2(s) was formed.

Reaction 3
The PbI2(s) was separated and mixed with 15.0 mL of 0.300 M Na2CO3(aq). A white precipitate of PbCO3(s) formed. All of the PbI2(s) was converted into PbCO3(s).

Reaction 4
The PbCO3(s) was removed by filtration and a small sample gave off a gas when treated with dilute HCl.


Question:
A soluble form of Pb2+ can be carefully added to a solution to sequentially precipitate and separate anions present in the solution. When Pb2+ is added, in what order will the following anions be precipitated?

A) SO42- then I-

**B) CO32- then I-
The reactions described in the passage show that lead(II) is successively precipitated as PbSO4, PbI2, and PbCO3. This sequence shows (assuming equal anion concentrations, as must be done here) that PbCO3 is less soluble than PbI2, and PbI2 is less soluble than PbSO4. The order in which the anions precipitate Pb2+ is: CO32- then I- then SO42-. When this sequence is applied to the question, answer choice B is in the correct order, and answers A, C, and D are all in the opposite order. Thus, answer choice B is the best answer.

C) SO42- then CO32-

D) I- then CO32-

 

[dochoov]

Hello! I have a question on V=IR…

The galvanic cell is made up of “a copper strip… partially immersed in a 1M solution of Cu(NO3)2 and a silver strip into a 1M solution of AgNO3… The two solutions are connected by a salt bridge… The two metal strips are connected by wires to a voltmeter which measures the potential difference between the metal strips. The voltmeter has infinite resistance, and it allows negligible amount of current to pass through…” The overall equation is 2Ag(s) + Cu2+(aq) 2Ag+ (aq) + Cu(s). The scientist measures 0.46 V cell voltage.

Question: The voltmeter shorts; its resistance becomes finite and some current passes through the wires. How will the reading of the voltmeter change?

A. The reading will increase because the current will increase the voltage.
B. The reading will increase because the current will increase the resistance of the circuit.
C. The reading will decrease because the current will decrease the resistance of the circuit.
D. The reading will decrease because the voltage decreases as the current increases.

The correct answer is D.

Can someone please explain why and what is going on? Why does the voltmeter have infinite resistance?

Thanks in advance

Voltmeters are connected in parallel, and in order for them to not draw current from the circuit, they have very high resistances. If a voltmeter has a high resistance, it's not drawing current out of the circuit. Why is this beneficial? Because your measurement should not interfere with what you're actually measuring!

D is correct. The measured voltage will decrease as the current through the voltmeter increases.

 

[hmcalley]

Voltmeters are connected in parallel, and in order for them to not draw current from the circuit, they have very high resistances. If a voltmeter has a high resistance, it's not drawing current out of the circuit. Why is this beneficial? Because your measurement should not interfere with what you're actually measuring!

D is correct. The measured voltage will decrease as the current through the voltmeter increases.

Thanks Dochoov!

 

[BrokenGlass]

This question comes from a passage. The following reaction has reached equilibrium at 1200K:

CO2 (g) + C (s) <--> 2CO (g)

The question:

When the system stabilized at 1200K, a sample of helium was injected into the furnace. What should happen to the amount of carbon dioxide in the system?

A. It should increase.
B. It should decrease.
C. It should be completely converted to carbon monoxide
D. It will remain the same.

Kaplan answer: D. Explanation: You should know that helium, a noble gas, is very unreactive and would almost certainly not react with any of the species in the furnace. Because the helium does not react with any of the species that participate in the equilibrium, the equilibrium is unaffected by the addition of helium. Even though it increases the total pressure inside the system, the partial pressures of the reacting gases are unchanged (Dalton's law) and therefore they keep on behaving as if the helium weren't present. The correct selection is therefore choice D.

The bold is what is really confusing me. I thought the answer should be A because an increase in pressure would cause the reaction to shift to the left (Le Chat.'s principle). And this whole thing about the partial pressures being the same... well, I'm confused.

La Chatelier's principle applies only when equilibrium is disturbed. Since helium is neither a reactant, nor a product, the equilibrium is not disturbed when we add helium. So the reaction has no reason to shift in order to preserve K (in this case K would be given in terms of partial pressures because reactans and products are gases and we don't include pure solids and pure liquids in the equilibrium expression) because adding helium would not affect K (since helium doesn't appear in the equilibrium expression for this reaction).

Note: even though it doesn't directly apply to this problem, you can think of heat as a reactant in an endothermic reaction and as a product in an exotherimic reaction, so you can see how adding heat or removing heat would disturb K.

 

[dochoov]

La Chatelier's principle applies only when equilibrium is disturbed. Since helium is neither a reactant, nor a product, the equilibrium is not disturbed when we add helium. So the reaction has no reason to shift in order to preserve K (in this case K would be given in terms of partial pressures because reactans and products are gases and we don't include pure solids and pure liquids in the equilibrium expression) because adding helium would not affect K (since helium doesn't appear in the equilibrium expression for this reaction).

Note: even though it doesn't directly apply to this problem, you can think of heat as a reactant in an endothermic reaction and as a product in an exotherimic reaction, so you can see how adding heat or removing heat would disturb K.

Ok. I understand that helium is inert, but wouldn't its addition increase the total pressure?

P=nRT/V more moles of gas leads to an increase in pressure?

 

[minsoox1983]

Q. Which of the following expressions gives an approximate value of the equilibrium constant (Ka) of the acid in Titration of HOAc with NaOH?

A) log([H3O+]2 x [HOAc])

B) [H3O+]2 x [HOAc]

C) log([H3O+]^2 /[HOAc])

**D) [H3O+]^2 / [HOAc]

Ka for the dissociation of HOAc is given by Ka = [H3O+][AcO&#8211;]/[HOAc]. Because [H3O+] = [OAc&#8211;], Ka = [H3O+]2/[HOAc]. Thus, D is the best answer.

------------------------------------
Even though I got this right by eliminating other choices,, I would like to get this straight..

Can anyone explain me why [H+] = [OAc-] ?
All i know for the titration is that..

At the half-equivalent pt (which helps us to determine Ka), [acid] = [conjugate base] therefore.. if Ka = [H+][OAc-] / [HOAc] and [HOAc] = [OAc-] , you get Ka = [H+].

At the equivalent pt, [acid] = [base], therefore [HOAc] = [NaOH]

Where do we get the reasoning that [H+] = [OAc-] ?

Thanks in advance, =)

 

[BrokenGlass]

Ok. I understand that helium is inert, but wouldn't its addition increase the total pressure?

P=nRT/V more moles of gas leads to an increase in pressure?

Yes, the total pressure would go up. Lets make up some numbers to see that it's possible for total pressure to go up and for partial
pressures to stay the same.

partial pressure = mole fraction * total pressure.

Suppose we have 3 moles of gas 1 and 2 moles of gas 2 (5 moles total). Lets assume total pressure is 10 before helium is added.

3/5 * 10 = 6 (partial pressure of gas 1 before helium is added)
2/5 * 10 = 4 (partial pressure of gas 2 before helium is added)

Now suppose we add 5 moles of helium and the total pressure becomes 20 after helium is added.

3/10 * 20 = 6 (partial pressure of gas 1 after helium is added)
2/10 * 20 = 4 (partial pressure of gas 2 after helium is added)

 

[BrokenGlass]

Q. Which of the following expressions gives an approximate value of the equilibrium constant (Ka) of the acid in Titration of HOAc with NaOH?

A) log([H3O+]2 x [HOAc])

B) [H3O+]2 x [HOAc]

C) log([H3O+]^2 /[HOAc])

**D) [H3O+]^2 / [HOAc]

Ka for the dissociation of HOAc is given by Ka = [H3O+][AcO&#8211;]/[HOAc]. Because [H3O+] = [OAc&#8211;], Ka = [H3O+]2/[HOAc]. Thus, D is the best answer.

------------------------------------
Even though I got this right by eliminating other choices,, I would like to get this straight..

Can anyone explain me why [H+] = [OAc-] ?
All i know for the titration is that..

At the half-equivalent pt (which helps us to determine Ka), [acid] = [conjugate base] therefore.. if Ka = [H+][OAc-] / [HOAc] and [HOAc] = [OAc-] , you get Ka = [H+].

At the equivalent pt, [acid] = [base], therefore [HOAc] = [NaOH]

Where do we get the reasoning that [H+] = [OAc-] ?

Thanks in advance, =)

The equation for the reaction of acetic acid solution with sodium hydroxide solution is:

HOAc + NaOH <--> NaOAc + H2O

The assumption here seems to be that at the end of titration, there is some acetic acid left over. In other words,
moles of NaOH added < moles HOAc initially present. This means that acetic acid that remains after the titration is
over will dissociate in water.

Even if moles of NaOH added = moles HOAc initially present (meaning that the equivalence point is reached,
HOAc + NaOH <--> NaOAc + H2O is reversible, so there still will be a VERY SMALL amount of acetic acid.

Either way, we need to use the reaction below:

HOAc + H2O = AcO&#8211; + H3O+ (acid dissociation)

Ka is acid dissociation equilibrium constant, Ka = [H3O+][AcO&#8211;]/[HOAc]

Because AcO&#8211; and H3O+ are produced together in a 1:1 molar ratio and both species are in the same solution (meaning that their volumes
are the same), [H3O+] = [OAc&#8211;]. So Ka = [H3O+]^2/[HOAc].

 

[dochoov]

Yes, the total pressure would go up. Lets make up some numbers to see that it's possible for total pressure to go up and for partial
pressures to stay the same.

partial pressure = mole fraction * total pressure.

Suppose we have 3 moles of gas 1 and 2 moles of gas 2 (5 moles total). Lets assume total pressure is 10 before helium is added.

3/5 * 10 = 6 (partial pressure of gas 1 before helium is added)
2/5 * 10 = 4 (partial pressure of gas 2 before helium is added)

Now suppose we add 5 moles of helium and the total pressure becomes 20 after helium is added.

3/10 * 20 = 6 (partial pressure of gas 1 after helium is added)
2/10 * 20 = 4 (partial pressure of gas 2 after helium is added)

Ok! It was helpful to actually see that with some numbers. And now I understand their rational much more.

I'm still wondering about the effects of pressure on equilibrium. If the total pressure was increased in that system while it was at equilibrium without adding another gas, the partial pressure of each gas would increase, right? And that would force the equilibrium towards the left?

 

[BrokenGlass]

Ok! It was helpful to actually see that with some numbers. And now I understand their rational much more.

I'm still wondering about the effects of pressure on equilibrium. If the total pressure was increased in that system while it was at equilibrium without adding another gas, the partial pressure of each gas would increase, right? And that would force the equilibrium towards the left?

Yep. That could happen if volume decreases while n and T stay constant.

 

[BrokenGlass]

Voltmeters are connected in parallel, and in order for them to not draw current from the circuit, they have very high resistances. If a voltmeter has a high resistance, it's not drawing current out of the circuit. Why is this beneficial? Because your measurement should not interfere with what you're actually measuring!

D is correct. The measured voltage will decrease as the current through the voltmeter increases.

We use a voltmeter to measure voltage (potential difference) between 2 points in a circuit. The voltmeter is connected in parallel with the
component across which potential difference is to be measured so that the current divides and passes through both the voltmeter and the
component at the same time.

A voltmeter has a high internal resistance so that very little current passes through it so that the voltmeter does not appreciably change
the circuit it is measuring. So we want the voltmeter to disturb the circuit as little as possible.

From Kirchoff's loop rule we know that the potential difference between any 2 points in the circuit is the same, regardless of path taken by
the current. By connecting the voltmeter and the circuit element whose voltage drop we are trying to measure IN PARALLEL, we are
providing an alternative path for current to follow. Even though to voltmeter shows voltage drop across itself, this voltage drop (approximately) equals the voltage drop across the circuit element whose voltage drop we are trying to measure.

If the current is drawn away from the circuit element whose voltage drop we are trying to measure (i.e. if I goes down), it follow from V=I*R
that V would go down, i.e. the voltage reported by the voltmeter would be too low. The solution is to have very high resistance across voltmeter so that we can meet both goals simultaneously: (1) provide an alternative path for the current and (2) have very little current diverted from the circuit element whose voltage we need to meausure.

If resistance across voltmeter decreases, more current will be diverted AWAY the circuit element whose voltage drop we are trying to measure and TOWARDS the voltmeter. Since current through the circuit element whose V we are trying to measure goes down, it follows from V=I*R that V across this circuit element goes down (R across this circuit element stays the same). So the voltage reported by the voltmeter would decrease after the voltmeter shorts. Again, the voltage reported by the voltmeter is the same as voltage across any circuit element connected in parallel with this voltmeter.

 

[unsung]

Okay, I have a question regarding the effect of the size of an anion on acidity based on 1 & 2 below:

1) Acidity increases going down a group, as the size of elements increases, such that:

HF < HCl < HBR < HI

This is becaus a larger anion can spread out its negative charge more, resulting in greater stability (as a conjugate base), and thus a stronger acid.

2) Acidity increases going across a period, as the electronegativity of elements increases, such that:

HCH3 < HNH2 < HOH < HF

The rationale being that a more electronegative element is able to "bear" the negative charge more easily, so again, the anion is more stable as a conjugate base, resulting in a stronger acid.

BUT, the contradiction here is that size actually decreases (as electronegativity increases) going across a period!

So, since the atoms are getting smaller, the negative charge of the anion should be LESS stable going across a period.

The question is: why does acidity increase going down a column due to increasing size, but increase going across a period despite decreasing size?

And a related question is: what does it mean that a more electronegative element is able to "bear a negative charge more easiy" ? That is a very vague statement.

Thanks in advance!

 

[BrokenGlass]

Okay, I have a question regarding the effect of the size of an anion on acidity based on 1 & 2 below:

1) Acidity increases going down a group, as the size of elements increases, such that:

HF < HCl < HBR < HI

This is becaus a larger anion can spread out its negative charge more, resulting in greater stability (as a conjugate base), and thus a stronger acid.

2) Acidity increases going across a period, as the electronegativity of elements increases, such that:

HCH3 < HNH2 < HOH < HF

The rationale being that a more electronegative element is able to "bear" the negative charge more easily, so again, the anion is more stable as a conjugate base, resulting in a stronger acid.

BUT, the contradiction here is that size actually decreases (as electronegativity increases) going across a period!

So, since the atoms are getting smaller, the negative charge of the anion should be LESS stable going across a period.

The question is: why does acidity increase going down a column due to increasing size, but increase going across a column despite decreasing size?

And a related question is: what does it mean that a more electronegative element is able to "bear a negative charge more easiy" ? That is a very vague statement.

Thanks in advance!

The atomic radius doesn't change all that much across a row (period) in the periodic table. So we usually assume that atomic radius is the same within a period. But if you still want to take atomic radius into account, the answer to your question is that electronegativity is the dominanant consideration. Of the two competing effects (electronegativity vs. atomic radius) electronegativity wins out.

 

[pezzang]

deleted

 

[pezzang]

post deleted due to repetition

 

[midn]

I had an electrochem question based on a problem off AAMC 7.

Say I have two half reactions given:

2e + X(2+) -> X(0) E1 = +1
4Y(0) -> 4Y(1+) + 4e- E2 = -3

Then, it asks for the E of the reaction:

2X(2+) + 4Y(0) -> 4Y(1+) + 2X(0)

Wouldn't E= 2*E1 + E2 since we took two equivalents of X?

If so, I think there maybe an error in the AAMC test. If not, I'm missing some electrochem concept here.

Sorry for the convoluted question with the made up elements. I don't want to put the exact problem down since that's not allowed.

 

[humerus]

Therefore, you do not multiply the potentials or the overall potential by coefficients. However, to get a balanced redox reaction, you do multiply by coefficients for the half-reactions.

 

[poppytart]

I was just wondering...do values of Ka, Keq, and Ksp change only when the temperature is changed?

Are there any other factors (such as pH) that can alter the K?

Thanks for any help!

 

[pezzang]

I was just wondering...do values of Ka, Keq, and Ksp change only when the temperature is changed?

Are there any other factors (such as pH) that can alter the K?

Thanks for any help!

Here is what i know:
K is equivalent to k(f)/k(r). Since k(f or r) is dependent upon catalyst, pressure and temperature, any change in P, T or catalyst will affect its rate constant. As for catalyst, it lowers both k(f) and k(r), so K stays constant. The effect of pressure on K is minimal; its change is irrelevant to the change in K. Hence, the only other factor that affects K is temperature. Increase in T increses the rate constant and the rate of reaction. Increasing/decreasing effect of K upon the change in T depends on the value of Keq. Hope it helped.

 

[novawildcat]

I was just wondering...do values of Ka, Keq, and Ksp change only when the temperature is changed?

Are there any other factors (such as pH) that can alter the K?

Thanks for any help!

I think you know the answer already.


lets say you have a reaction between two gases to give your product.



Take a look at the ideal gas law.

PV=nRT. Rearrange to get n/V=P/RT. What is n/V???? ITS CONCENTRATION. thus you can express Keq in terms of pressure and temperature rather than using concentration for the Keq expression. Changes in pressure and temperature affect Keq.

 

[BrokenGlass]

message deleted

 

[BrokenGlass]

message deleted

 

[BrokenGlass]

I think you know the answer already.


lets say you have a reaction between two gases to give your product.



Take a look at the ideal gas law.

PV=nRT. Rearrange to get n/V=P/RT. What is n/V???? ITS CONCENTRATION. thus you can express Keq in terms of pressure and temperature rather than using concentration for the Keq expression. Changes in pressure and temperature affect Keq.

Changes in pressure do NOT affect Keq. Keq depends on temperature, but not on pressure. Changes in pressure and concentration
do NOT affect Keq.

For every reaction which can go forward and backward, one direction is endothermic and the other is exothermic. A reaction is endothermic if it takes heat from its surroundings (i.e. heat is a reactant). A reaction is exothermic if it gives heat to the surroundings (i.e. heat is a product).

The effect of temperature on K

Exothermic reactions: if temperature rises then position of equilibrium moves to the left (this follows from Le Chatelier's principle; raising temperature amounts to adding heat; heat is a product in exothermic reactions; so reaction shifts towards reactants) and K becomes smaller. If temperature falls then position of equilibrium moves to the right (this follows from Le Chatelier's principle )and K becomes bigger.

Endothermic reactions:if temperature rises then position of equilibrium moves to the right (this follows from Le Chatelier's principle) and K becomes bigger.
If temperature falls then position of equilibrium moves to the left (this follows from Le Chatelier's principle) and K become smaller.

Thus, raising the temperature will cause an endothermic reaction to shift in the direction of more products, while it will cause an exothermic reaction to shift toward reactants. So If you increase the temperature, then the endothermic reaction will be favored because that will take in some of the excess heat. If you decrease the temperature, the exothermic reaction will be favored because it will produce
the heat that was lost.


Lets consider the equation &#8710;G° = –RTlnK.

We can solve this equation for K; lnK = - &#916;G°/RT => K = e-&#916;G°/RT

So clearly K depends on T.

We also know that &#916;G° = &#916;H° - T &#916;S°

After a little algebra we get: K = e-&#916;H°/RT *e&#916;S°/R


If the reaction is endothermic, &#916;H° >0, and so K will be small compared to 1, as expected for an endothermic reaction (In an endothermic reaction reactants are more stable than products, and since equilibrium favors the more stable side, reactants are favored over products in an endothermic reaction).

If T is increased, then &#916;H°/RT decreases, and the exponential involving &#916;H° increases, so that K increases. The effect is that the reaction shifts towards products. If T is decreased, K decreases, and the reaction shifts toward reactants. If, on the other hand, the reaction is exothermic, &#916;H° < 0 and K is a much larger than 1. But if T is increased, &#916;H°/RT decreases, and the exponential decreases, so that K decreases. This has the effect of shifting the reaction towards reactants. Similarly, when T is decreased, K increases for an exothermic reaction, and the reaction shifts toward products.

 

[cloosh]

i'm confused about vapor pressure, boiling point, and freezing point. if i add more solute to the solution, wouldn't this increase the boiling point, thus increase the vapor pressure? also, wouldn't this increase of solute decrease the freezing point? i'm confused because earlier someone told me that vapor pressure decreases with solute addition.

also, could someone clear up what exactly vapor pressure means? thanks.

 

[BrokenGlass]

i'm confused about vapor pressure, boiling point, and freezing point. if i add more solute to the solution, wouldn't this increase the boiling point, thus increase the vapor pressure? also, wouldn't this increase of solute decrease the freezing point? i'm confused because earlier someone told me that vapor pressure decreases with solute addition.

also, could someone clear up what exactly vapor pressure means? thanks.

See #460 and #461 in the Organic Chemistry Questions Thread. That should be enough to get you started.

 

[The Force]

i'm confused about vapor pressure, boiling point, and freezing point. if i add more solute to the solution, wouldn't this increase the boiling point, thus increase the vapor pressure? also, wouldn't this increase of solute decrease the freezing point? i'm confused because earlier someone told me that vapor pressure decreases with solute addition.

also, could someone clear up what exactly vapor pressure means? thanks.

A portion of particles on the surface of a liquid have enough energy to break free. The pressure created by these particles is the vapor pressure. Vapor pressure increases with temperature because a larger number of particles have the energer to break free from the surface. A liquid boils whens its vapor pressure is equal to atmospheric pressure. When a nonvolatile solute is introduced, vapor pressure decreases. Since the vapor pressure decreases, a higher temperature must be reached before the solution boils, thus, boiling point is elevated.
The increase in solute also depresses the freezing point because it interrupts the formation of the creation of the crystal structure.

Hope that helps

 

[5moreminutes]

This is a question from EK 1001, I couldnt figure out...
It is question # 189 in the 1001 book of General Chemistry..
In an 11.2 liter container the partial pressure of nitrogen gas is 0.5 atm at 25 degree celcius. What is tth emass of nitrogen in the container?

The answer is 7 grams..


Thanks!

 

[cloosh]

A portion of particles on the surface of a liquid have enough energy to break free. The pressure created by these particles is the vapor pressure. Vapor pressure increases with temperature because a larger number of particles have the energer to break free from the surface. A liquid boils whens its vapor pressure is equal to atmospheric pressure. When a nonvolatile solute is introduced, vapor pressure decreases. Since the vapor pressure decreases, a higher temperature must be reached before the solution boils, thus, boiling point is elevated.
The increase in solute also depresses the freezing point because it interrupts the formation of the creation of the crystal structure.

Hope that helps

hey yeah i read through that and it helped a bunch.

 

[BrokenGlass]

This is a question from EK 1001, I couldnt figure out...
It is question # 189 in the 1001 book of General Chemistry..
In an 11.2 liter container the partial pressure of nitrogen gas is 0.5 atm at 25 degree celcius. What is tth emass of nitrogen in the container?

The answer is 7 grams..


Thanks!

PV = nRT

V = 11.2L
P = 0.5 atm
T = 25 degrees Celcius = (25 + 273) Kelvins = 298 Kelvins
R = 0.082 L &#183; atm &#183; K-1 &#183; mol-1 (note: there are many different values for R. Use units of other variables in
the problem to determine which value of R to use).

http://en.wikipedia.org/wiki/Gas_constant

n = PV/RT = (0.5 * 11.2)/(0.082 * 298) = 5.6 / 24.436 = 0.23 moles

Molecular Weight (or more precisely molecular mass) of N2 is 28 g/mol (remember that nitrogen is a diatomic molecule)

0.23 moles * 28 g/mol ~ 7 grams

 

[pezzang]

Hi,

I have a question about partial pressure:
4NH3(g) + 7O2(g) -> 4NO2 (g) + 6H2O(l)
It's EK G Chem Ch2 (pg28) #27. We start with 8 mol NH3 and 14 O2 in 11 atm (fixed box and const T) and have to find the pp of NO2. I realize why it's 4 atm but I am wondering what happened to the rest 8 atm once the reaction runs to completion? I mean we only produce 4 atm pp from NO2 and since water is not a gas, it doesn't have pp. So what happens to 7 atm in the box?

 

[BrokenGlass]

Hi,

I have a question about partial pressure:
4NH3(g) + 7O2(g) -> 4NO2 (g) + 6H2O(l)
It's EK G Chem Ch2 (pg28) #27. We start with 8 mol NH3 and 14 O2 in 11 atm (fixed box and const T) and have to find the pp of NO2. I realize why it's 4 atm but I am wondering what happened to the rest 8 atm once the reaction runs to completion? I mean we only produce 4 atm pp from NO2 and since water is not a gas, it doesn't have pp. So what happens to 7 atm in the box?

PV = nRT => P/n = RT/V = constant (because R, T, and V are constant)

P initial = 11 atm
n initial = 8 + 14 = 22 moles

P final = ?
n final = 8 moles (this follows from stoichiometry)

P initial / n initial = P final / n final = constant

11 / 22 = P final / 8

P final = 4 atm

I am not sure I undestand your question correctly, but in this problem
P and n are directly proportianal to each other, so if n goes down (which
it does according to reaction stoichiometry), P goes down.

 

[PalmettoGuy]

In TPR Cracking the MCAT book, there is the following question:

Would methane or ethane be expected to have the higher ionization energy?

The answer:
Ethane, because a primary cation is more stable than a methyl cation.

I understand that a primary cation is more stable than a methyl cation, however, would this stability not make it easier to remove an electron from ethane? And if it is easier to remove an electron from ethane, wouldn't that make the ionization energy of ethane lower?

Thanks

 

[Green Pirate]

In TPR Cracking the MCAT book, there is the following question:

Would methane or ethane be expected to have the higher ionization energy?

The answer:
Ethane, because a primary cation is more stable than a methyl cation.

I understand that a primary cation is more stable than a methyl cation, however, would this stability not make it easier to remove an electron from ethane? And if it is easier to remove an electron from ethane, wouldn't that make the ionization energy of ethane lower?

Thanks

this confuses me too. I thought that, generally speaking, molecules with more atoms, and thus more electron sheilding, had lower ionization energies.

 

[5moreminutes]

I know that alkaline substances are basic. So CH3OH and KOH are both basic, so then, how it is possible that only KOH is alkaline and CH3OH is not? WHat makes CH3OH neutral?

thanks

 

[HippocratesX]

I know that alkaline substances are basic. So CH3OH and KOH are both basic, so then, how it is possible that only KOH is alkaline and CH3OH is not? WHat makes CH3OH neutral?

thanks

Alkalinity is the measure of a substance to "neurtralize" acids. KOH is not an acid. It's a base. CH3OH is not a base...its an acid. So it cannot be alkaline since an acid cannot neutralize an acid (i don't think?). This link should explain it better than I did and in more detail

http://en.wikipedia.org/wiki/Alkalinity

 

[shevie]

can someone help me with this problem?
zinc hydroxide is insoluble in water but dissolves when nitric acid soln is added, write balanced total ionic and net ionic equations.

 

[BrokenGlass]

can someone help me with this problem?
zinc hydroxide is insoluble in water but dissolves when nitric acid soln is added, write balanced total ionic and net ionic equations.

Some acid-base reactions occur between insoluble hydroxides and acids. These reactions cause the hydroxide to dissolve.

The balanced equation is:
Zn(OH)2 (s) + 2HNO3 (aq) -> Zn(NO3)2 (aq) + 2H2O (l)


Balanced total ionic equation is:
Zn(OH)2 (s) + 2H+(aq) + 2NO3-(aq) -> Zn2+(aq) + 2NO3-(aq) + 2H2O(l)


Net ionic equation is (After canceling the spectator ions to leave the net ionic equation):
Zn(OH)2 (s) + 2H+(aq) -> Zn2+(aq) + 2H2O(l)


Note: For net ionic equations substances that dissociate extensively should be written as ions (strong acids and bases, soluble salts)
Also, leave out any molecules or ions that are not changed in the reaction (spectator ions), which in this case are 2NO3-(aq) because
they appear on both sides of the equation.

This doesn't seem to be an MCAT type of question though...

 

[shevie]

This doesn't seem to be an MCAT type of question though...

Thats REALLY GOOD TO KNOW. Because thats more in depth than Im studying, and if each question took me that long to complete, id never finish the section. My friend is taking Gen chem and asked me and I was right, I told her just what you wrote. But I havent taken Gen chem in 3 years, and I wanted to make sure ...

thanks for your help

 

[nata723]

Hi,

I was wondering if, when a reaction in solution is not in equilibrium, adding products will necessarily decrease the rate of the reaction. This was sort of a question in AAMC Test 8 #14 of Physical Sciences.

I'd like to have someone's explanation for it.

Thanks!

 

[Playmakur42]

I was wondering if, when a reaction in solution is not in equilibrium, adding products will necessarily decrease the rate of the reaction. This was sort of a question in AAMC Test 8 #14 of Physical Sciences.

The question you're asking about actually has a simple answer that is easy to miss when rushing through the timed test.

Starch is only an indicator of I2 in the reactions and not an actual reactant, so that's the best answer.

 

[Sapneil Parikh]

I am having a difficult time with Newton's Law of Universal Gravitation. Is there any one that can help me figure out this problem: I understand the problem but need to understand how to use the ratios that are involved. Can someone show me as it applies to these two problems.

1. Planet B has twice the mass of Planet A. Planet A has a radius half as large as Planet B. A 5kg mass is dropped 10 m above the surface of Planet B and at the same time a 10 kg mass is dropped 10m above the surface of Planet A. If the mass on Planet B strikes the ground in 10 sec,the mass on Planet A strikes the ground in approx:
a.7, b. 10, c. 14, d.20 ans is a.
2. Planet A and B have the same mass . Planet A has a radius half as large as Planet B. A 5kg mass is dropped 10m above the surface of Planet A and at the same a 5kg mass is dropped 10m above the surface of Planet B. If the mass on Planet B strikes the ground in 10 sec, the mass of Planet A strikes the ground in :
a. 2.5s b.5s c.10s d. 20s ans is b.

thanks

 

[MileyCyrus]

Well, as you know it ths day is fially gonna come up anwaz.... soooooo like me I am gonna talk my MCAT's this MAy on the 11. I know a superstar doing MCAT's!!! BUt let me give you advice to how I am leraning alot is that i go on www.kaptests.com. Get ur own account and then do all you can for the practice quizzes, and then you can do better, take an online quiz and get graded check it and each time u do u will learn more! see SO GOOD LUCK ON THE MCAT'S!

 

[mm6284]

Couple of questions:
1) Are catalysts included in the rate law?
2) Is the pH of the half equivalance pt half of the difference between the pH of the weak acid or base and the pH at the equivalence pt?
3) Can we use any units in the dilution equation as long as the units are the same on both sides?

 

[BrokenGlass]

Couple of questions:
1) Are catalysts included in the rate law?
2) Is the pH of the half equivalance pt half of the difference between the pH of the weak acid or base and the pH at the equivalence pt?
3) Can we use any units in the dilution equation as long as the units are the same on both sides?

1) Arrhenius equation says k=A*exp(-Ea/R*T). So rate constant and therefore the rate itself vary with activation energy. Catalysts lower activation energy, so their effects are factored into rate law via the rate constant k.

2) Henderson-Hasselbalch equation says pH = pKa + log([A-]/[HA]).

HA + OH- -> A- + H2O

pH at half equivalence point = pKa (because at half equivalence point [A-] = [HA]).

pH at equivalence point = (pKa + pH titrant)/2. Basically, you are taking the average of 2 values. This is a graphical approximation.

3) I don't understand your question.

 

[BeatrixKiddo]

how do you know which subjects are more important than others?

 

[HippocratesX]

how do you know which subjects are more important than others?

Ini Minni Myni Moe...

 

[MSTPbound]

Ini Minni Myni Moe...

DITTO.

Go to the AAMC web site, find the list of topics tested. Know everything. EVERYTHING.

 

[BrokenGlass]

I am having a difficult time with Newton's Law of Universal Gravitation. Is there any one that can help me figure out this problem: I understand the problem but need to understand how to use the ratios that are involved. Can someone show me as it applies to these two problems.

1. Planet B has twice the mass of Planet A. Planet A has a radius half as large as Planet B. A 5kg mass is dropped 10 m above the surface of Planet B and at the same time a 10 kg mass is dropped 10m above the surface of Planet A. If the mass on Planet B strikes the ground in 10 sec,the mass on Planet A strikes the ground in approx:
a.7, b. 10, c. 14, d.20 ans is a.
2. Planet A and B have the same mass . Planet A has a radius half as large as Planet B. A 5kg mass is dropped 10m above the surface of Planet A and at the same a 5kg mass is dropped 10m above the surface of Planet B. If the mass on Planet B strikes the ground in 10 sec, the mass of Planet A strikes the ground in :
a. 2.5s b.5s c.10s d. 20s ans is b.

thanks

Ma = mass of planet a
Mb = mass of planet b
Ra = radius of planet a
Rb = radius of planet b



F = (G * M * m) / R^2
F = m*a
a = F/m
a = (G * M) / R^2

1)

Mb = 2*Ma
Rb = 2*Ra

In the formula a = (G * M) / R^2 if we double the mass and double the radius, we decrease the acceleration by a factor of 2.

=> acceleratino on planet b = 1/2 acceleration on planet a
=> acceleration on planet a = 2 * acceleartion on planet b

So time on planet a < time on planet b, so 7 sec is the right answer (no need to do any more work, we can stop right here).

2)

Mb = Ma
Rb = 2*Ra

=> acceleratino on planet b = 1/4 acceleration on planet a
=> acceleartion on planet a = 4 * acceleration on planet b

x = x0 + v0*t + 1/2 * a * t^2

x = 1/2 * a * t^2 (because initial displacement and initial velocity are both zero)

t^2 = (2*x)/a

t = sqrt((2*x)/a), where sqrt stands for square root

t ~ sqrt(1/a)

In the formula t ~ sqrt(1/a) if acceleration goes up by a factor of 4, t goes down by a factor of sqrt(4) = 2

10sec /2 = 5 sec


Is this an AAMC problem?

 

[wongb18c]

I don't know if someone can help me with an actual MCAT practice question on AAMC #3 exam?

I having difficulty understanding question #35 on the exam. It's a question relavent to molality. Although I've read the answer, I'm still having trouble comprehending how they calculated that oxalic acid solution provided the greatest number of moles of solute as compared to the other solutions.

I was wondering if anyone can give a step by step thought process explanation...