some chem questions.. need help

[Notoriousjae]

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These questions are from the Barrons book.
1.The diffusion rate of methane gas compared to that of helium is
A.twice as fast
B.4 times as slow
C.4 times as fast
D.half as fast
E.12 times as fast
the answer to this in the book is D. although i think B is correct.

2. Catalysts have which of the following effects on chemical reaction?
A.lower activation energy
B.increase free energy
C.cause the reaction to proceed spontaneously
D.increase the rate at which the product is formed
E.shift the equilibrium so the reverse reaction proceeds more easily.

The answer to this in the book is B, although i picked A and think A is right.


3. how many atoms of carbon are in 4.00*10^-8 grams of C3H8?
A. 1.64*10^15
B. 2.34*10^11
C. 1.89*10^14
D. 4.66*10^2
E. 4.00*10^4

Answer= A dunno

4. If all of the chloride in a 5 gram sample of unknown metal chloride is precipitated as AgCl with 70.9mL of 0.201M AgNO3 what is the percentage of chloride in the sample?
A. 50.55%
B. 20.22%
C. 1.43%
D. 10.10%
E. none of the above

Answer= B dunno

 

[who knows...]

1.The diffusion rate of methane gas compared to that of helium is
A.twice as fast
B.4 times as slow
C.4 times as fast
D.half as fast
E.12 times as fast

is it B?

2. Catalysts have which of the following effects on chemical reaction?
A.lower activation energy
B.increase free energy
C.cause the reaction to proceed spontaneously
D.increase the rate at which the product is formed
E.shift the equilibrium so the reverse reaction proceeds more easily.

A?

3. how many atoms of carbon are in 4.00*10^-8 grams of C3H8?
A. 1.64*10^15
B. 2.34*10^11
C. 1.89*10^14
D. 4.66*10^2
E. 4.00*10^4

something having to do with avogadro's number maybe?
maybe if you fine the moles first...so 4.00*10^-8 grams / 44g/mol
then multiply by avogadro's number...so = 5.47*10^14 molecules...
thats the number of molecules of the whole thing...so now multiply by three since there are three carbons per one whole thing...so = 1.64*10^15
so A?

4. If all of the chloride in a 5 gram sample of unknown metal chloride is precipitated as AgCl with 70.9mL of 0.201M AgNO3 what is the percentage of chloride in the sample?
A. 50.55%
B. 20.22%
C. 1.43%
D. 10.10%
E. none of the above

thats one of thos solubility questions eh? grrrrr!

 

[Djapprentice]

1.The diffusion rate of methane gas compared to that of helium is
A.twice as fast
B.4 times as slow
C.4 times as fast
D.half as fast
E.12 times as fast


2. Catalysts have which of the following effects on chemical reaction?
A.lower activation energy
B.increase free energy
C.cause the reaction to proceed spontaneously
D.increase the rate at which the product is formed
E.shift the equilibrium so the reverse reaction proceeds more easily.

3. how many atoms of carbon are in 4.00*10^-8 grams of C3H8?
A. 1.64*10^15
B. 2.34*10^11
C. 1.89*10^14
D. 4.66*10^2
E. 4.00*10^4

4. If all of the chloride in a 5 gram sample of unknown metal chloride is precipitated as AgCl with 70.9mL of 0.201M AgNO3 what is the percentage of chloride in the sample?
A. 50.55%
B. 20.22%
C. 1.43%
D. 10.10%
E. none of the above

I'm not quite sure about the percentage of Cl- problem, you might want to double check this one:

AgCl ---> Ag+ + Cl-

you're kind of working backwards. First we'll figure out the moles of Cl- in the sample, using the given info, then mult it by the mass of the sample 5 grams, to get the original grams of the Cl-, then take Mass of Cl- / Mass of sample * 100%

Vol = 70.9 = 0.0709

Molarity = moles / volume, therefore, we get n (moles) = MV (0.201)(0.0709) = 0.01 / 1/100

Now to find the mass of the Cl- in the sample:

1/100 mol Cl- X 5 grams of Cl- sample / 1 mol Cl- = 5/100 g, which can be reduced to 1/20 g

Now you use the percent equation:

Mass of Cl- / Mass of Sample * 100%

= 1/20 * 1/5 = 1/100 * 100 = 1%, so I get E, but again, the gang might wanna double check this!

 

[egots]

first one

(4/16)^(1/2)= 1/2 ....half as fast

not sure..hey where are these questions from?

 

[Notoriousjae]

these questions are from the barrons test A although i think i shud stop going through it because it seems to have a descent amount of errors in the answer key.

 

[kayo]

I got 1/2 on the first question, too.
rate of methane/ rate of he=(molecular weight of He/MW of methane)^1/2
=1/2

the second question is B?

correct me if im wrong..

kayo

 

[egots]

yeaa i have the barrons book i never even used it...there are to many sources...i figure kaplan 5 subject tests...midterm final, exam on cd, two natural science sections, 3 topscore exams, and dat achiever are more then enough...

 

[who knows...]

first one

(4/16)^(1/2)= 1/2 ....half as fast

not sure..hey where are these questions from?

wow...I never even knew there was a formula for this kinda stuff...i was just trying to think about it...
thanx! 😳 now I know 😀

 

[who knows...]

I got 1/2 on the first question, too.
rate of methane/ rate of he=(molecular weight of He/MW of methane)^1/2
=1/2

the second question is B?

correct me if im wrong..

kayo

"The activation energy can be reduced by the addition of a catalyst. The catalyst works by increasing the frequency of collisions in the forward reaction and similarly in the reverse reaction"
Barron's page 140
I dont think catalysts can actually increase the amount of free enrgy...how would they do it...they can only orient the molecules reacting in a favorable way or allow for another reaction path to take place...

 

[Notoriousjae]

first one

(4/16)^(1/2)= 1/2 ....half as fast

not sure..hey where are these questions from?

yeah you might be right on the first one. dang i completely blanked out on the effusion/diffusion formula. this means time for more review.
the answer to the second question is A which i am positive about. who knows came up with the right answer for the third one. Can anyone take a stab at the last question? 😕 😕