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Author Topic: sos, may day! help  (Read 10661 times)  Share 

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lacoste

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sos, may day! help
« on: June 05, 2009, 07:59:56 pm »
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Some questions that I don't quite get... more will be coming  :)

Q3).
Copper metal can be oxidised by nitric acid according to the following half-equations
Cu (s) ---> Cu2+ (aq) + 2 e–
NO3– (aq) + 4 H+ (aq) + 3 e– ---> NO (g) + 2 H2O (l)
The number of moles of nitrate ions reduced by one mole of copper is closest to
A.   0.33
B.   0.67
C.   1.5
D.   2.0

lacoste

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Re: sos, may day! help
« Reply #1 on: June 05, 2009, 08:03:48 pm »
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Question 5
A chemist working for a mining company found that a mineral sample contained 67% zinc.
A possible chemical formula for the mineral sample is
A.   ZnO
B.   Zn2S
C.   ZnS
D.   FeZnS

What is your working out?

THANKS!!!!!!!!!!!

TrueTears

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Re: sos, may day! help
« Reply #2 on: June 05, 2009, 08:06:32 pm »
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(Did not put in states because I cbf lol)





Add those 2





Answer is B
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TrueTears

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Re: sos, may day! help
« Reply #3 on: June 05, 2009, 08:09:47 pm »
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Question 5
A chemist working for a mining company found that a mineral sample contained 67% zinc.
A possible chemical formula for the mineral sample is
A.   ZnO
B.   Zn2S
C.   ZnS
D.   FeZnS

What is your working out?

THANKS!!!!!!!!!!!

Answer is C

Molar mass of

% of Zn is

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Interested in asset pricing, econometrics, and social choice theory.

lacoste

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Re: sos, may day! help
« Reply #4 on: June 05, 2009, 08:10:15 pm »
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Why is it 1:0.67?

TrueTears

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Re: sos, may day! help
« Reply #5 on: June 05, 2009, 08:12:34 pm »
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Why is it 1:0.67?
We require n(Cu) to be 1.

So we know

divide both by 3 yields:







« Last Edit: June 05, 2009, 08:14:09 pm by TrueTears »
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lacoste

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Re: sos, may day! help
« Reply #6 on: June 05, 2009, 08:15:31 pm »
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Thanks Truetears! you are great!!!!!!!!!!!!!!!!!!!

lacoste

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Re: sos, may day! help
« Reply #7 on: June 05, 2009, 08:23:57 pm »
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Question 12
The volume of 0.20 mol L-1 KOH needed to neutralise the weak acid is
A.   100 mL
B.   50 mL
C.   10 mL
D.   5.0 mL


TrueTears

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Re: sos, may day! help
« Reply #8 on: June 05, 2009, 08:27:04 pm »
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Question 12
The volume of 0.20 mol L-1 KOH needed to neutralise the weak acid is
A.   100 mL
B.   50 mL
C.   10 mL
D.   5.0 mL


What is the weak acid? :P
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lacoste

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Re: sos, may day! help
« Reply #9 on: June 05, 2009, 08:40:17 pm »
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Questions 11 and 12 refer to the following information.
A 100 mL sample of an aqueous solution of a weak monoprotic acid has a concentration of 0.10 mol L-1 and a pH of 2.0.


I got the answer, thanks, true tears again. I was looking at the MC question by itself without referring to the above given info.
LOL. No wonder I had no clue at first. I thought i was going nuts




next q.
Question 19
In which of the following reactions is sulfuric acid, H2SO4, acting as an oxidant?
A.   H2SO4 (aq) + CuO (s) ---> CuSO4 (aq) + H2O (l)
B.   2 H2SO4 (aq) + Cu (s) ---> CuSO4 (aq) + SO2 (g) + H2O (l)
C.   H2SO4 (aq) + CaCO3 (s) ---> CaSO4 (aq) + CO2 (g) + H2O (l)
D.   H2SO4 (aq) + 2 NaCl (s) ----> 2 HCl (g) + Na2SO4 (aq)

How do you know that in B, H2SO4 goes to CuSO4 or SO2?

TrueTears

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Re: sos, may day! help
« Reply #10 on: June 05, 2009, 08:44:21 pm »
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Questions 11 and 12 refer to the following information.
A 100 mL sample of an aqueous solution of a weak monoprotic acid has a concentration of 0.10 mol L-1 and a pH of 2.0.


I got the answer, thanks, true tears again. I was looking at the MC question by itself without referring to the above given info.
LOL. No wonder I had no clue at first.




next q.
Question 19
In which of the following reactions is sulfuric acid, H2SO4, acting as an oxidant?
A.   H2SO4 (aq) + CuO (s) ---> CuSO4 (aq) + H2O (l)
B.   2 H2SO4 (aq) + Cu (s) ---> CuSO4 (aq) + SO2 (g) + H2O (l)
C.   H2SO4 (aq) + CaCO3 (s) ---> CaSO4 (aq) + CO2 (g) + H2O (l)
D.   H2SO4 (aq) + 2 NaCl (s) ----> 2 HCl (g) + Na2SO4 (aq)

How do you know that in B, H2SO4 goes to CuSO4 or SO2?
Oxidation number of from left to right is : +1 , +6 and -2

In Oxidation numbers from left to right is : +4 , -2

Now acting as an oxidant means must be reduced. This means it's oxidation number must decrease.

In was +6 but in it changed to +4

Hence it decreased it acts as an oxidant

As you can see in , the oxidation of S does not change, hence it must go to
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lacoste

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Re: sos, may day! help
« Reply #11 on: June 05, 2009, 08:45:21 pm »
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Question 2
Natural gas consisting of the alkanes containing one to four carbon atoms can undergo thermal cracking in a cracker.
 
Why is steam added into the cracker?

TrueTears

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Re: sos, may day! help
« Reply #12 on: June 05, 2009, 08:46:38 pm »
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Hydrocarbons ranging from 1 - 4 carbon atoms are gases. I think using steam helps carry the gases up the cracker. It kind of acts like a "carrier".
« Last Edit: June 05, 2009, 09:10:40 pm by TrueTears »
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lacoste

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Re: sos, may day! help
« Reply #13 on: June 05, 2009, 08:49:28 pm »
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Thanks Truetears again, wish i could + karma you again.

Would it be wise to go through every MC option A to D and check for the oxidation numbers, or go through until you get one ie like B the one above and then stop and say its B?    

So if it doesnt change we just move onto the other "SO2 option" and find a decrease?

Cheers!!! :)

lacoste

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Re: sos, may day! help
« Reply #14 on: June 05, 2009, 08:52:15 pm »
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Hydrocarbons ranging from 1 - 4 carbon atoms are gases. Using steam heaps carry the gases up the cracker. It kind of acts like a "carrier".

I like your answer as it makes more sense than the solution.

the solution is:(a)   Steam is added to the thermal cracker to reduce carbon build-up during thermal cracking.   
[The chemical equation for the process is: H2O (g) + C (s) ---> CO (g) + H2 (g)]

What does it mean by reduce carbon build up?


actually your answer answers this i think, helps carry the gases c1 to c4 hydrocarbons up