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Section I: Multiple Choice (1 Viewer)

independantz

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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
This is correct.
 

Darrow

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For 12 I got B
Explain why it is A
 

Kearnzo

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Darrow said:
For 12 I got B
Explain why it is A
The sample at Site 1 is just after the sewage plant discharge, this would increase the BOD and E. Coli, for obvious reasons.

This would disperse by the Estuary sample.

EDIT:

The TDS at site 3 is very high, this would be because of the salty water of the sea, compared to the fresh water of the river.

Site 4 (mine) increases TDS by 205, and the sewage by a bit aswell.

And thats really all you need to look at.
 
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Propadanda

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I'ts definitely 2.7, I just did the calc.

EDIT: Just realised I fucked the calc haha.
 
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Darrow

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Doesnt that make it B?
BOD would be high in an estuary
So would TDS
Gah I need the question

All water flows to the estuary remember..
 

Propadanda

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n[H+] = .02 x .08
= 1.6 x 10^-3
n[OH-] = .05 x .03
= 1.5 x 10^-3

Therefore, H+ is in excess, and 1 x 10^-4 is left over after mix.

New volume = .05 L

Therefore [H+] = 1 x 10^-4 / .05

= 2.69

Well I'm an idiot.

I skipped a step there somewhere but that's the gist of it.
 

Propadanda

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Kearnzo said:
The sample at Site 1 is just after the sewage plant discharge, this would increase the BOD and E. Coli, for obvious reasons.

This would disperse by the Estuary sample.

The TDS at site 3 is very high, this would occurs because of copper mine discharge. TDS would be greatly increased due to the ions in the water from mining.

And thats really all you need to look at.
But there isn't even the option of Site 3 being the mine?

I put A..
 

friction

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independantz said:
1-A
2-B
3-A
4-C
5-C
6-D
7-D
8-D
9-D
10-B
11-A
12-D- but ppl are saying the answer is A not entirely sure
13-C
14-C- answer is B forgot to divide by the volume to find concentration then log it >.>
15-B

That's what i got, so far 2 wrong ...>.>
I got exactly the same except either A or B for question 12. I think A.
 

shakky15

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Propadanda said:
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.
that was my reasoning. spent ages with plenty of calculations to no avail. had to move on so picked C. however, if u remember the titration topic, the whole equivalence point thing, u notice how once equivalence is reached, it very quickly reaches the extreme, hence 2.7 is believable.
 

shakky15

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Propadanda said:
n[H+] = .02 x .08
= 1.6 x 10^-3
n[OH-] = .05 x .03
= 1.5 x 10^-3

Therefore, H+ is in excess, and 1 x 10^-4 is left over after mix.

New volume = .05 L

Therefore [H+] = 1 x 10^-4 / .05

= 2.69

Well I'm an idiot.

I skipped a step there somewhere but that's the gist of it.
my god i was SO CLOSE! i got the 0.1 x 10^-3 difference, but stalled there!! ahh dammit!! coulda got 15/15 :(
 

tommykins

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berna007 said:
product side = left side actually..
just look at the question okay!!!!
lol r u fo' srs?

a chemical reaction is when new chemicals are made, it doesn't have to be the bonding of two reactants to produce a product.
 

hayyleyy

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did anyone else find multiple choice REALLY hard?!?!?
it was a bad start to the paper except it got better after that i think :)
 

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