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Section I: Multiple Choice (1 Viewer)

shakky15

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tau281290 said:
Question 14 is the only one im unsure of...

I put pH 7.0 but that's pretty wrong.
yeah i put 4.0 but that was an educated guess - i was at pains on that fukn qn spent ages on it..

otherwise 14/15 so im pretty happy.
 

tommykins

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berna007 said:
GUYS 7 IS NOT D!!!!!
if u add reactant, it will shift to the product side..and therefore shift left..however, if u increase temp, it will go to the reactant side and shift RIGHT
THEREFORE ANSWER TO 7 IS C NOT D!!!!!
lmaoo
Exothermic.

Increase temp -> want to decrease -> left reaction as that is endothermic.
Product side = right side.
 

berna007

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tommykins said:
Exothermic.

Increase temp -> want to decrease -> left reaction as that is endothermic.
Product side = right side.
product side = left side actually..
just look at the question okay!!!!
 
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berna007 said:
product side = left side actually..
just look at the question okay!!!!
got the question in fron of me.

It is an exothermic reaction ( enthalpy is negative ).
increasing the temperature causes it to shift left towards the reactant. Therefore c) is not correct, d) is the answer.............

it is asking what will make it shift to the right.......
 

shakky15

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tommykins said:
Exothermic.

Increase temp -> want to decrease -> left reaction as that is endothermic.
Product side = right side.
thank god..

that kid above almost gave me a heart attack.
 

adosh

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berna007 said:
product side = left side actually..
just look at the question okay!!!!
i think you have misunderstood a part of chemistry. that is, the LHS HI chemical is a reactant andt he H2 and I2 chemicals are products, it does not have to be ther other way around,,,,that is its not as if reactants can only be two compunds and when they reactot hey form one compund which does not have to be a product.....so the lhs is the ractant adn the rhs is the product....
 

danz90

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tommykins said:
lol the 4 d's put me off so bad.
lol dittooo hardddddddd

I just kept looking over the 4 D's questions.. and i'm like, i've obviously screwed up a question I don't realise.
 

P.T.F.E

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I really fuked this part up. i got 12 a. which one was the water one?
 

ewasemma

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hoochiscrazy said:
whoaaaaa 15/15 (according to the answers posted on here) pretty happy with that since yesterday did a whole bunch on board of studies website and got 82%!! lol

1-A
2-B
3-A
4-C
5-C
6-D
7-D
8-D
9-D
10-B
11-A
12-A
13-C
14-B
15-B
hey guys im not sure if this is right but i think that 7 is c because adding more of the reactant would shift to the left in this case. well thats what i thought... probs wrong...
 

Kearnzo

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ewasemma said:
hey guys im not sure if this is right but i think that 7 is c because adding more of the reactant would shift to the left in this case. well thats what i thought... probs wrong...
Left <> right
Reactants <> products

adding more reactants, means that more products must be formed to counteract the change. Ie as the question asks for, a shift to the right.

15/15 for me. Somehow!
 

Propadanda

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I thought that in an exothermic reaction, the "heat" was written on the LHS, on the reactant side, so by adding heat to the reaction, that will shift away from the heat to the products side, the RHS? That's probably wrong though.

And can someone please explain to me how the fuck you can add a strong acid to a strong base, in relatively similar proportions (not identical, but similar), and end up with a very acidic final pH? I thought Q14 had to be at least C or D. I thought because the weren't in proportion it would be C but then I thought that was still too acidic to have a pH of 4...so I put D, even though it obviously wasn't a perfect neutralisation but ehhh.
 

chewnugm

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shakky15 said:
yeah i put 4.0 but that was an educated guess - i was at pains on that fukn qn spent ages on it..

otherwise 14/15 so im pretty happy.
To do q 14 :
0.02*0.08-0.03*0.05=1*10^-4
therefore 1*10^-4 H+ ions are excess. pH=-log[H+]
pH=-log[1*10^-4]
pH=4
 

Kearnzo

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chewnugm said:
To do q 14 :
0.02*0.08-0.03*0.05=1*10^-4
therefore 1*10^-4 H+ ions are excess. pH=-log[H+]
pH=-log[1*10^-4]
pH=4
Thats moles.

Ph = -log10[H+]

c = n/v

volume is 20mls +30 mls = 0.05 L

therefore [H+] is 0.0001/0.05 = 0.002

Ph = -log10[0.002] is equal to 2.69... = 2.7
 

Propadanda

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Kearnzo said:
Thats moles.

Ph = -log10[H+]

c = n/v

volume is 20mls +30 mls = 0.05 L

therefore [H+] is 0.0001/0.05 = 0.002

Ph = -log10[0.002] is equal to 2.69... = 2.7
Maybe this sounds stupid, but that just doesn't seem logical that a strong base and a strong acid could mix to make a solution with a pH that low.

I'd say C lol, even though I put D.
 

Aaron.Judd

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I'm not sure of the exxact working, but the answer is defiantly C accorging to my Chem teacher at James Ruse and shes extrememly smart.
 

independantz

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Aaron.Judd said:
I'm not sure of the exxact working, but the answer is defiantly C accorging to my Chem teacher at James Ruse and shes extrememly smart.
who cares if shes a teacher from JRAHS? teachers make mistakes and the answer is B, i put C aswell however you're loging the amount of moles not the concentration of H30+
 

Kearnzo

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Propadanda said:
Maybe this sounds stupid, but that just doesn't seem logical that a strong base and a strong acid could mix to make a solution with a pH that low.

I'd say C lol, even though I put D.
>.< how can you argue when I just did all the working?

Its a very small amount, so even if there arn't many moles in it. The amount of H+ per volume is high.

The pH of the HCL alone is 1.1

1 ph higher corresponds to a 10 times lower concentration of H+.

so 1.1 to 2.7 is nearly 100 times lower. Thats a big difference
 

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