Post by: lacoste on June 05, 2009, 07:59:56 pm
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
Post by: lacoste on June 05, 2009, 08:03:48 pm
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!!!!!!!!!!!
Post by: TrueTears on June 05, 2009, 08:06:32 pm
Add those 2
Answer is B
Post by: TrueTears on June 05, 2009, 08:09:47 pm
Question 5Answer is C
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!!!!!!!!!!!
Molar mass of
% of Zn is
Post by: lacoste on June 05, 2009, 08:10:15 pm
Post by: TrueTears on June 05, 2009, 08:12:34 pm
Why is it 1:0.67?We require n(Cu) to be 1.
So we know
divide both by 3 yields:
Post by: lacoste on June 05, 2009, 08:15:31 pm
Post by: lacoste on June 05, 2009, 08:23:57 pm
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
Post by: TrueTears on June 05, 2009, 08:27:04 pm
Question 12What is the weak acid? :P
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
Post by: lacoste on June 05, 2009, 08:40:17 pm
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?
Post by: TrueTears on June 05, 2009, 08:44:21 pm
Questions 11 and 12 refer to the following information.Oxidation number of
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?
In
Now acting as an oxidant means
In
Hence it decreased
As you can see in
Post by: lacoste on June 05, 2009, 08:45:21 pm
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?
Post by: TrueTears on June 05, 2009, 08:46:38 pm
Post by: lacoste on June 05, 2009, 08:49:28 pm
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!!! :)
Post by: lacoste on June 05, 2009, 08:52:15 pm
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
Post by: arthurk on June 05, 2009, 08:57:00 pm
lol truetears your knowledge extends beyond the required knowledge of the course
Post by: lacoste on June 05, 2009, 09:02:37 pm
their solution:
C10 H22 (g) ----> C5H10 (g) + C5H10 (g)
Is this right? I thought it would be C10 H22 (g) ----> C5H10 (g) + C5H12 (g)
also with the states, are they gases, i thought C1 to C4 are gases then liquids to about C14?
Post by: TrueTears on June 05, 2009, 09:02:45 pm
Thanks Truetears again, wish i could + karma you again.Spot on :)
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!!! :)
Post by: TrueTears on June 05, 2009, 09:03:23 pm
cracking? wont be on exam i thoughtI think thermal and catalyst cracking are on the course (they help your understanding of fractional distillation, such as the fractional distillation of crude oil), although it's not a very big part of the course.
lol truetears your knowledge extends beyond the required knowledge of the course
Post by: TrueTears on June 05, 2009, 09:05:38 pm
Write a balanced chemical equation for the cracking of decane to produce pentene.As a normal guideline:
their solution:
C10 H22 (g) ----> C5H10 (g) + C5H10 (g)
Is this right? I thought it would be C10 H22 (g) ----> C5H10 (g) + C5H12 (g)
also with the states, are they gases, i thought C1 to C4 are gases then liquids to about C14?
1-4 are gases
5-20 are liquids
21+ are solids
Post by: lacoste on June 05, 2009, 10:00:23 pm
A student attempted to determine the amount of sodium chloride in a brand of chicken stock powder. The label on the bottle states the sodium content at 17050 mg per 100 g.
a. Calculate the amount of sodium, in mol, in 3.0 g of the chicken stock, assuming the label is accurate.
b. The student dissolves 3.0 g of stock in distilled water and makes up the solution to 250 mL in a volumetric flask. A 20.00 mL sample is taken and excess silver nitrate solution is added to precipitated all the chloride ions as silver chloride.
Using the information provided, calculate the mass of silver chloride that the students would expect to precipitate.
Not too sure with part B.
thanks!!!!!
Post by: Samira on June 05, 2009, 10:52:08 pm
a)m(Na)= 17050 mg in 100g
xmg in 3g
x= (17050*3)/100 = 5115.5 mg
so m(Na)=5115.5 mg in 3g
n(Na) .51155/23= 0.02224 mole
b) Nacl+AgNO3---> NaNO3 + AgCl
n(Na)=(Nacl)= 0.0224 mole
However thats in 250 ml...we need the mole in 20 ml as we took 20 ml sample
so... n(NaCl)= 0.0224*(20/250) =0.001779 mole
n(nacl)=n(AgCl)= 0.001779 mole
m(Agcl) =0.001779* (107.9+35.45) =.255g or 255mg
hope that helps
Post by: lacoste on June 06, 2009, 12:56:25 am
a)m(Na)= 17050 mg in 100g
xmg in 3g
x= (17050*3)/100 = 5115.5 mg
i got 5115mg not 5115.5 mg, how did you get the extra decimal?
Post by: arthurk on June 06, 2009, 01:02:06 am
if u think about it 4 x 5115 = 20460 mg in 12 g (4 times ur 3 gram calculation) thats already over the mass you had originally.
i think it should be 511.5mg
Post by: Samira on June 06, 2009, 08:04:52 am
Post by: lacoste on June 06, 2009, 11:26:11 am
another q:
IR spectroscopy is an analytical tool often used in organic chemistry. Which of the following
could be distinguished from the other molecules due to the absence of a particularly distinct
and characteristic peak?
(a) CH3CH2CH2CH2OH
(b) CH3CH2CH2COOH
(c) CH3CH2CH2CH2NH2
(d) CH3CH2CH2COOCH3
Post by: d0minicz on June 06, 2009, 11:31:44 am
the others have OH's and an NH2 soo thats around 3000cm^-1 right
Post by: lacoste on June 06, 2009, 11:57:41 am
cheers!! got it now!!
Is it possible for a different molecule apart from glucose to undergo fermentation?
Post by: TrueTears on June 06, 2009, 12:56:10 pm
Yes, its D, thanks dominicz.Fermentation is the process of deriving energy from the oxidation of organic compounds. For VCE all you have to know is glucose undergoes fermentation.
cheers!! got it now!!
Is it possible for a different molecule apart from glucose to undergo fermentation?
Post by: lacoste on June 06, 2009, 02:52:33 pm
another one :
QUESTION 3
300 ml of a 0.35 M Ca(OH)2 solution was reacted with 300 ml of 0.40 M HCl .
The pH of the resultant solution is closest to:
A 0.82
B 1.12
C 12.88
D 13.18
Please show working too. thanks!
Post by: NE2000 on June 06, 2009, 03:09:17 pm
n(OH-) = 2 x 0.105 = 0.21 mol
n(H+) = 0.12 mol
OH- in excess = 0.09 mol
Total volume = 0.6L
c(OH-) = 0.15 M
pOH = 0.82
Therefore pH=14-0.82=13.18
Therefore answer is D
Alternatively you can avoid the pOH thing and use [H+][OH-] = 10^-14.
Post by: TrueTears on June 06, 2009, 03:10:27 pm
Ratio
We have
so we only need
therefore
Hence D
Post by: lacoste on June 06, 2009, 04:36:36 pm
Which of the following reactions is NOT a redox process?
A 2K(s) + Br2(l) →2KBr(s)
B Ca(OH)2(s) + CO2(g) →CaCO3(s) + H2O(l)
C NH4NO3(s) → N2O(g) + 2H2O(g)
D CH4(g) + Cl2(g) →CH3Cl(g) + HCl(g)
Post by: chem-nerd on June 06, 2009, 04:41:57 pm
Post by: lacoste on June 06, 2009, 04:45:58 pm
also, how do you know which reductant goes to which oxidant etc?
thanks .
Post by: Gloamglozer on June 06, 2009, 04:46:21 pm
It cannot be A because it has an element (K). All equations containing an element is redox.
I'd say B because the oxidation numbers haven't changed.
EDIT: Damn, beaten to it. lol
Post by: lacoste on June 06, 2009, 11:24:50 pm
??
Post by: TrueTears on June 06, 2009, 11:27:09 pm
Can monosaccharides be hyrdolysed to form smaller units?no.
??
Post by: Mao on June 07, 2009, 02:55:23 am
Can monosaccharides be hyrdolysed to form smaller units?
??
not into smaller units, but it can be hydrolysed into a straight chain, which is slightly heavier that the ringed structure (extra H2O).
Post by: lacoste on June 07, 2009, 10:29:09 am
Fe2O3 (s) + 3CO (g) ---------> 2Fe (l) + 3 CO2 (g)
Post by: hyperblade01 on June 07, 2009, 10:49:39 am
If 3 mol of Fe2O3 and 2 mole of CO are mixe together, 1 mole of Fe is produced. Calculate the amount of substance remaining at the end of the reaction.
Fe2O3 (s) + 3CO (g) ---------> 2Fe (l) + 3 CO2 (g)
Looks like a limiting/excess reactant question but actually this one is double excess :p
Using the equation it reads:
1 Mole Fe2O3 and 3 Mole CO gives 2 Mole Fe and 3 Mole CO2
Therefore if 1 mole of Fe is produced, you divide everything by 2:
0.5 Mole Fe2O3 and 1.5 Mole CO gives 1 Mole Fe and 1.5 Mole CO2
3 - 0.5 = 2.5 Mole Fe2O3 remaining
2 - 1.5 = 0.5 Mole Co remaining
Post by: lacoste on June 07, 2009, 11:48:27 am
Thats what I did to work it out but the solution is 5.5 mol, is there a different way to work it out?
Also, another q, the pH of a 10^-9M solution of HCl is closest to :
A 4
B 6
C 7
D 9
thanks.
Post by: Mao on June 07, 2009, 12:04:10 pm
Post by: lacoste on June 07, 2009, 12:08:30 pm
a 10^-9 M HCl would completely ionise and form 10^-9 M of [H+]. However, water at neutral point has [H+] of 10^-7. Hence addition of 10^-9 M HCl makes neutral water 1% more acidic, pH is still 7.
man, thats confusing. is this likely to be on the exam?
If 3 mol of Fe2O3 and 2 mole of CO are mixe together, 1 mole of Fe is produced. Calculate the amount of substance remaining at the end of the reaction.
Fe2O3 (s) + 3CO (g) ---------> 2Fe (l) + 3 CO2 (g)
Why is the answer 5.5 mol? Is this wrong?
Thanks.
Post by: chem-nerd on June 07, 2009, 12:08:51 pm
If 3 mol of Fe2O3 and 2 mole of CO are mixe together, 1 mole of Fe is produced. Calculate the amount of substance remaining at the end of the reaction.
Fe2O3 (s) + 3CO (g) ---------> 2Fe (l) + 3 CO2 (g)
Looks like a limiting/excess reactant question but actually this one is double excess :p
Using the equation it reads:
1 Mole Fe2O3 and 3 Mole CO gives 2 Mole Fe and 3 Mole CO2
Therefore if 1 mole of Fe is produced, you divide everything by 2:
0.5 Mole Fe2O3 and 1.5 Mole CO gives 1 Mole Fe and 1.5 Mole CO2
3 - 0.5 = 2.5 Mole Fe2O3 remaining
2 - 1.5 = 0.5 Mole Co remaining
Thats what I did to work it out but the solution is 5.5 mol, is there a different way to work it out?
amount of substance at the end of the reaction would be from the reactants AND products. so you have to add the 1 mol of Fe and 1.5 mol of CO2 produced
Post by: lacoste on June 08, 2009, 01:56:50 pm
SA: Q 1B, regarding triprotic
I got a different answer to the solutions, can someone please explain, how to work this one out.
Thanks!!
Post by: monokekie on June 08, 2009, 02:06:40 pm
Just assume this triprotic acid is H3PO4, then you will get a mole ration of 2 h3po4 : 3 Na2CO3
then use stochiometry to solve this question :)
Post by: lacoste on June 08, 2009, 02:12:41 pm
=]