Collins Chemistry Questions Help

[xNeenax]

1) How do you solve this:

Chem Test #3 Ques. 4

2LiClO3 ----> 3O2 + 2LiCl

What is the mole ratio of LiClO3 to O2 to form 30g of O2?
Answer: 2:3


2) Chem Test #3 Ques. 13
I don't get that. I thought it was [A]^2 because the rate doubles when B is held constant, but then [A] went from .2 to .3 so I don't know what to think of that


3) Chem Test #4 Ques. 19
30ml of H2SO4 neutralized by 60mL of 1M NaOH what is molarity of H2SO4
(1M)(.06)= (.03)x
x= 2M
the answer is 1 M, where did I go wrong?


4) Chem Test #5 Ques. 9
The valance e- are held tightly so we can expect:
Answer: high ionization energy and low e- affinity.

Why low e- affinity? Isn't high e- affinity the ability to gain electrons and low affinity the ability to lose electron, if the elections are held tightly then it wouldn't lose electrons. I don't understand why the answer is low e- affinity.

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[Kansas Pharmer]

Lets see if I can help.

1. The 30g is extra info and not needed. The equation shows a balanced 2:3 ratio so that is the correct answer.

2. I am not sure. Rates was my only weakness in Chem II.

3. NaOH makes 1 ion (OH-) but the acid H2SO4 makes 2 ions (H+) so you only need half the molarity.

4. Ionization energy and electron affinity are 2 entirely different things. Ionization has to do with how hard it is to lose an electron and affinity only deals with how easy it is to gain an electron. So, tightly bound valence electrons have high ionization and if they are tightly bound it is probably a full shell making it hard to add one therefore low affinity.

 

[DH1987]

1) How do you solve this:

Chem Test #3 Ques. 4

2LiClO3 ----> 3O2 + 2LiCl

What is the mole ratio of LiClO3 to O2 to form 30g of O2?
Answer: 2:3


2) Chem Test #3 Ques. 13
I don't get that. I thought it was [A]^2 because the rate doubles when B is held constant, but then [A] went from .2 to .3 so I don't know what to think of that


3) Chem Test #4 Ques. 19
30ml of H2SO4 neutralized by 60mL of 1M NaOH what is molarity of H2SO4
(1M)(.06)= (.03)x
x= 2M
the answer is 1 M, where did I go wrong?


4) Chem Test #5 Ques. 9
The valance e- are held tightly so we can expect:
Answer: high ionization energy and low e- affinity.

Why low e- affinity? Isn't high e- affinity the ability to gain electrons and low affinity the ability to lose electron, if the elections are held tightly then it wouldn't lose electrons. I don't understand why the answer is low e- affinity.

Yeah, Question 2 was weird. I got it correct though, not because I was able to compute the answer, but because I realized that there was only one possible answer. The rate can't be just k[A]^2 or k^2, because both concentrations have an effect on it. It also can't be [A]^2, because there's no constant. Therefore, the only remaining choice is k[A].

But I suppose we could go through the math of it:

.060/.040 = 1.5
.3/.2 = 1.5
1.5^x = 1.5, x =1

So [A] operates on first-order kinetics. At this point, you can already deduce the answer, but you can also do the other one:

.04/.01 = 4
.02/.01 = 2
Since both concentrations were doubled:
(2^x)(2^y) = 4

We already determined x to be 1.

(2)(2^y) = 4
4^y = 4
y = 1

So also operates on first order kinetics. Rate = k[A]

Question 3 is not just a dilution problem. It wants to know what you need to neutralize the solution. If you look in the appropriate section of the Chemistry packet, you'll see that it mentions a "shortcut" to those kind of problems.

nA * mA * vA = nB * mB * vB

Where n is the "strength" of the acid (A) or base (B), m is the molarity of the acid or base, and v is the volume of the acid or base.

How you determine the strength of the acid or base is determine how many protons it can donate or accept. H2SO4 can donate 2 protons (2 H's), so it has a n value of 2. NaOH can accept one proton (one OH), so it has a value of 1.

Therefore:

(2)(30ml)(x) = (1)(60ml)(1M)
60x=60
x = 1M

The above poster answered the last question more accurately than I could.

 

[xNeenax]

Thank you my smart, beautiful people

 

[Kansas Pharmer]

Thank you my smart, beautiful people

How dare you call me that! I am neither of those; not that smart and especially not beautiful! If you have any other questions I would be happy to help if I can. It is nice to talk to people dealing with the same thing (PCAT)... It seems to lessen the pain a little...

 

[xNeenax]

I have a question on the Collins QA section
On Test 3 number 15:
f(x)=x^3+ 2x, g is inverse of f. what is g'(3)

how do you find out that f(1)=3?
why do you use 1 for x when you differentiate?

 

[DH1987]

I have a question on the Collins QA section
On Test 3 number 15:
f(x)=x^3+ 2x, g is inverse of f. what is g'(3)

how do you find out that f(1)=3?
why do you use 1 for x when you differentiate?

That is a very strange problem. Here is how I interpret it...

Theorem: If g is the inverse of f, and f(a) = b or g(b) = a, then g'(b) = 1/f'(a)

g'(3) is given to you for the question. To satisfy the theorem above, you have to find the value of f(a) which will become b, which will then be applied into the final part of the theorem.

b = 3
a = ?

f(a) = b
f(a) = 3
3 = x^3 + 2x
x = 1 = a

Now plug it into the final part of the theorem:
g'(b) = 1/f'(a)
g'(3) = 1/f'(1)

You then just differentiate the given function: 3x^2 + 2, which is f'(a).
Then solve for f'(1), which is 5, then you get the answer:

g'(3) = 1/5

 

[xNeenax]

Cellular Respiration in Eukaryotes:

Glycolysis: 2 ATP (net production)
Decarb.: 6 ATP
Krebs: 24 ATP
election transport Chain: 4 ATP

Total 36 ATP production, right?

 

[chemguy79]

I have a question on the Collins QA section
On Test 3 number 15:
f(x)=x^3+ 2x, g is inverse of f. what is g'(3)

how do you find out that f(1)=3?
why do you use 1 for x when you differentiate?

There is already another thread on the board for this exact question.

 

[xNeenax]

There is already another thread on the board for this exact question.

meh

 

[DH1987]

Cellular Respiration in Eukaryotes:

Glycolysis: 2 ATP (net production)
Decarb.: 6 ATP
Krebs: 24 ATP
election transport Chain: 4 ATP

Total 36 ATP production, right?

Between 32-36, if my memory serves correctly. However, I am pretty sure the majority of the ATP comes from the ETC via chemiosmosis, not from the Kreb's Cycle.

But I am not completely sure about that. Guess that's something for me to review...

 

[kys2000]

Cellular Respiration in Eukaryotes:

Glycolysis: 2 ATP (net production)
Decarb.: 6 ATP
Krebs: 24 ATP
election transport Chain: 4 ATP

Total 36 ATP production, right?

Glycolysis:
+ 4 ATP
- 2 ATP
= 2 ATP Net + 2 Pyruvate + 2 NADH

Decarb:
+1 NADH
+ 1 CO2
= +2 NADH for two pyruvate per glucose molecule

Krebs:
+2 ATP
+ 6 NADH
+ 2 FADH2

ETC: Produce 3 ATP for each NADH and 2 ATP for each FADH2 (FADH is doesn't produce as much as it enters ETC further down the chain (I think at a cytochrome) hence less protons are pumped out and produces less ATP than NADP from the ATP synthase.

In total:
Glycolysis: + 2 ATP
Krebs: +2 ATP
ETC: converts a total of 10 NADH and 2 FADH for a total of 34 ATP
Final total ATP: 38

Eukaryotic cells consume two 2 ATP to move the pyruvate to in mitochondria for a total of 36 ATP
Bacterial cells produces 38 ATP as ETC is accomplished on cell membrane

 

[luckyrx]

On chem test 7,

Na and Cl readily form NaCl...

Why is the answer: Na has low IE, and Cl has low EA?

 

[kys2000]

On chem test 7,

Na and Cl readily form NaCl...

Why is the answer: Na has low IE, and Cl has low EA?

Na being a Group IA element (along with Li and K) haa one valence electron and will easily give it up forming a +1 charge and a full octet. Thus it is has low (first) ionization energy (the amount of energy needed to remove an electron). Cl being a group 7A element (along Br, F), really needs another electron for a full octet, in recieving another electron, the energy release is high in adding another electron (lower negative value) thus Cl has a low electron affinity.


Simply put:
Na has low ionization energy but high electron affinity
Cl has high ionization energy but low electron affinity

 

[frankdahtank]

Na being a Group IA element (along with Li and K) haa one valence electron and will easily give it up forming a +1 charge and a full octet. Thus it is has low (first) ionization energy (the amount of energy needed to remove an electron). Cl being a group 7A element (along Br, F), really needs another electron for a full octet, in recieving another electron, the energy change is low in adding another electron thus Cl has a low electron affinity.


Simply put:
Na has low ionization energy but high electron affinity
Cl has high ionization energy but low electron affinity

That answer doesn't make sense at all. As you said, Cl really needs another electron meaning it would have a high electron affinity. Electron affinity and ionization energy are directly related when looking at the periodic table. The answer doesn't make sense and I attributed it to a typo.

 

[chemguy79]

On chem test 7,

Na and Cl readily form NaCl...

Why is the answer: Na has low IE, and Cl has low EA?

This question has been answered and discussed on the board a few times. Search for Collins Chemistry and you will see a similar discussion to what is being performed below.

FYI, this question is one that I would know, from personal experience.

 

[kys2000]

That answer doesn't make sense at all. As you said, Cl really needs another electron meaning it would have a high electron affinity. Electron affinity and ionization energy are directly related when looking at the periodic table. The answer doesn't make sense and I attributed it to a typo.

Sorry I did have a typo, meant that it has a high energy release but there is two ways to look at electron affinity, as a positive value or negative value for the energy release. For the PCAT, I took electron affinity as a negative value thus the more negative, the lower.

 

[frankdahtank]

Sorry I did have a typo, meant that it has a high energy release but there is two ways to look at electron affinity, as a positive value or negative value for the energy release. For the PCAT, I took electron affinity as a negative value thus the more negative, the lower.

Hmm, that is an interesting way to look at it. I missed a few of those questions on the Collin's practice tests and maybe looking at it like this will solve the problems. I KNOW electron affinity as I've been taught it and the closer to F, the more electronegative. Also, when looking at it from a valence electron prospective the closer you are to an octet the more electronegative (valence of 6 has more EA than valence of 2). I don't know, can anyone else shed some light on this.

 

[krishna1990]

is everyone memorizing the solubility rules for the exam?
sorry i knw im lil off the topic here but didnt want to start another new thread lol

 

[DH1987]

is everyone memorizing the solubility rules for the exam?
sorry i knw im lil off the topic here but didnt want to start another new thread lol

Strangely enough, I haven't seen a lot of questions regarding solubility rules on the Pearson Practice exams. Still probably wouldn't hurt to know it, though.

 

[newmember123]

I think this is a typo in Dr. Collin chemistry packets but I just want to double check:

According to the kinetic-molecular theory, if the volume of a gas doubles the temperature must have:
A) doubled B)been reduced by half C) remained unchanged D) decreased exponentially

the answer key says its B but I think its A, right? Since volume and temp are directly proportional to each other.

 

[hushpuppies]

^ I believe you are correct. doubling V is compensated with doubling T.