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CSSA chem trial (1 Viewer)

CM_Tutor

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Li0n said:
i put c for that one
fucken had no idea...

also that combustion of carbon with sulfur
i just realised i stuffed it!

i did
CS + O2 --> C + SO2

and just now i realise its CS2 + 2O2 --> C + 2SO2

my first answer for L of Sulfur was like 6 million or 6 thousand L i think, some shit like that :/, CM_Tutor help us out yo :)
I haven't seen the paper, so I can't really comment definitively. Everything I post on this thread should thus be taken with that caveat in mind, and I reserve the right to change my opinion as further information becomes available.

That having been said, I have serious doubts that CS<sub>2</sub> would be relevant. It seems (from other posts) that the question mentions coal. The compound CS<sub>2</sub> is an extremely dangerous liquid, which readily burns and is a component in napalm. I doubt there is much of it in coal. If the question is discussing sulfur content of coal, then I would be inclined to consider its combustion as the combustion of elemental S. I have never heard of CS, so I don't even know if such a compound actually exists.
 

CM_Tutor

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sig said:
pH of H2SO4 is C - 1.6

Diprotic therefore -log[2 x 0.0115] = 1.64
Heinz said:
hmm.... I still chose 1.9 knowing full well that it was diprotic. The reason being that the second ionisation stage is in equilibrium, so we cant assume complete ionisation like the first stage. Guess we have to wait. :confused:
I am inferring that the question was pH of 0.0115 M sulfuric acid.

Assuming complete ionisation of the both steps, you get a pH of 1.638.

Considering only the first step (ie treating it is a strong monoprotic acid), you get a pH of 1.939

Heinz - you are correct that the second ionisation is incomplete, so let's see what happens if we take that fact into account. The second ionisation has a pK<sub>a</sub> of 1.99. This means that if we take the actual concentration of H<sub>3</sub>O<sup>+</sup> due to this ionisation to be x, and ignore the fact that there is H<sub>3</sub>O<sup>+</sup> present from the first ionisation, then x is given by the equation:

x<sup>2</sup> / (0.0115 - x) = 10<sup>-1.99</sup>

x > 0, and the actual [H<sub>3</sub>O<sup>+</sup>] = 0.0115 + x

This approach gives pH = 1.736 and suggests that the hydrogensulfate ion is 59.8 % ionised. This suggests that the pH is closer to 1.6 than it is to 1.9.

(Note - the usual approximation, that [H<sub>3</sub>O<sup>+</sup>] = sqrt(K<sub>a</sub> * orig [HA]), is not valid in this case, as the assumption it requires is not satisfied. However, calculating pH with this method would give pH = 1.651 and indicate 94.3 % ionisation)

Now, what if we take into account the hydronium present from the initial ionisation. In that case, [H<sub>3</sub>O<sup>+</sup>] = 0.0115 + x where x is the positive solution of the equation:

x(0.0115 + x) / (0.0115 - x) = 10<sup>-1.99</sup>

in which case I get pH = 1.796 and the hydrogensulfate ion is 39.0 % ionised.

So...

I think you have a case, but I don't like your chances of winning. The problem that you have is that the answer 1.9 is the answer you'd get if you thought that it was monoprotic, and you haven't covered these calculations that I have used. So, how is a teacher supposed to know which approach you took in the exam? (You should also check I've done the maths correctly, as it is late. :))

PS: I do agree that this is not a good question, but the intended approach is definitely to say that it is diprotic and to take both ionisations as going to completion.
 
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za

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we got our marks today.

i got 13/15
57/60
24/25
total: 94/100

came first... cheering!
 

Enigmatic Eve

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za said:
we got our marks today.

i got 13/15
57/60
24/25
total: 94/100

came first... cheering!
Hey!!
Congratulations on a great mark (and a mad ranking!! Lol!!)
Wish I can score that!!
Anyways, take care dude!!
 

snoopwogg

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Wo0ah Za!! well done on such a great mark.. u must b such and admirable character.. maybe i should sit next to you more in class... i only got 87.. pretty goood considering i didnt study! :D anyway, ill congratulate you in person when i see you next babe ;) Have Fun Dude!!
 

za

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How about the heat of combustion (6 mark) of alkanols question?
u just talk about how u used cupper tin instead of a beaker, as its a better conductor. also how u used reflecting tiles to minimise heat loss etc...
and then u go on about repetition, collaboration. u just talk about how it was effective in carryin out its purpose.
 

snoopwogg

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SO WHAT IF I AM GAY? DO U HAVE A PROBLEM WITH THAT? Gay people are still people, they shouldnt be discriminated against. We must not lament that they don't fit into the boxes of "male" and "female," and therefore "homosexual" and "heterosexual," too. Those boxes weren't made for them. They don't constitute an adequate theory of human sexuality. The theory is what's insufficient and malformed -- not them. Although the dominant paradigm may not have a vocabulary that recognizes them, they do exist. I shall send an email to the moderator of this forum to get you reprimanded. Have a good day.
 

withoutaface

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I got 80/100. I coulda done better but I was pretty happy.
 

jam3s

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snoopwogg said:
SO WHAT IF I AM GAY? DO U HAVE A PROBLEM WITH THAT? Gay people are still people, they shouldnt be discriminated against. We must not lament that they don't fit into the boxes of "male" and "female," and therefore "homosexual" and "heterosexual," too. Those boxes weren't made for them. They don't constitute an adequate theory of human sexuality. The theory is what's insufficient and malformed -- not them. Although the dominant paradigm may not have a vocabulary that recognizes them, they do exist. I shall send an email to the moderator of this forum to get you reprimanded. Have a good day.
i said u must be Gay i didn;t say i hate gays did i?
 

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