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CSSA chemistry (4 Viewers)

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Riproot

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Dylanamali said:
umz, how did you go? sounds like you went well...

I found it quite challenging..

What was your answer for the HCl being added to BaOH and the resulting pH of the solution.

Also what was the answer of what is the use of the Geiger counter.

And just a checking thing: in MCs there was a question comparing sulfuric acid and HCl - my answer was they have the same H+ and the same pH. Is that what you got?
Click to expand...
sulfuric acid had more h+ and lower pH because it can partially dissociate another h+ ion.
 
Hermes1

Hermes1

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Re: CSSA Chemistry 2011

Dylanamali said:
umz, how did you go? sounds like you went well...

I found it quite challenging..

What was your answer for the HCl being added to BaOH and the resulting pH of the solution.

Also what was the answer of what is the use of the Geiger counter.

And just a checking thing: in MCs there was a question comparing sulfuric acid and HCl - my answer was they have the same H+ and the same pH. Is that what you got?
Click to expand...
no wouldnt sulfuric acid have a lower pH bcuz it is diprotic
 
D

Dylanamali

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Re: CSSA Chemistry 2011

Riproot said:
sulfuric acid had more h+ and lower pH because it can partially dissociate another h+ ion.
Click to expand...
BUT. the concentration of the sulfuric was 0.25 and the conc of the HCl was 0.5

so if you do the calculation pH=-log(0.25x2), they have the same pH and also the same H+conc
 
Hermes1

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Re: CSSA Chemistry 2011

Riproot said:
sulfuric acid had more h+ and lower pH because it can partially dissociate another h+ ion.
Click to expand...
lol riproot u were all depressed b4 this exam and now u look like u did quite well judging by ur answers.
 
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umz93

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Re: CSSA Chemistry 2011

Riproot said:
sulfuric acid had more h+ and lower pH because it can partially dissociate another h+ ion.
Click to expand...
Dude r u srs? thats what i got
But people in my school said it was the same conc...
 
Hermes1

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Re: CSSA Chemistry 2011

Dylanamali said:
BUT. the concentration of the sulfuric was 0.25 and the conc of the HCl was 0.5

so if you do the calculation pH=-log(0.25x2), they have the same pH and also the same H+conc
Click to expand...
oh ok i was going by the fact they had equal concentrations. but now i can see wif new info that u were right.
 
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umz93

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Re: CSSA Chemistry 2011

Dylanamali said:
umz, how did you go? sounds like you went well...

I found it quite challenging..

What was your answer for the HCl being added to BaOH and the resulting pH of the solution.

Also what was the answer of what is the use of the Geiger counter.

And just a checking thing: in MCs there was a question comparing sulfuric acid and HCl - my answer was they have the same H+ and the same pH. Is that what you got?
Click to expand...
Yeah i went alrite, hopefully didnt make too many stupid mistakes :) Wbu?
 
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umz93

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Re: CSSA Chemistry 2011

!@#$% well whats done is done.
 
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Dylanamali

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Re: CSSA Chemistry 2011

umz93 said:
Yeah i went alrite, hopefully didnt make too many stupid mistakes :) Wbu?
Click to expand...
Very challenging I must say.. like the calculations were relatively difficult, some MC's were tricky, other than that e.g. the worded questions were quite simple and basic. But still dont think I did that great, minimum of 80% I'm thinking atm.
 
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umz93

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Re: CSSA Chemistry 2011

Yeah im about the same. Only question i couldnt do was the eqbm experiment cos i didnt study it :/
 
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Dylanamali

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Re: CSSA Chemistry 2011

How did people do the octane question???

This is how I went about doing it..
They provided us with the octane density e.g. 780g/L and they also provided us with the molar mass e.g. 144g/moL
So I said, in 1L of octane there is 780g of octane. No. of mols of octane = m/M = 780/144 = what ever the answer is.

And then the next question was like... how much energy is released from 1L of octane.
I then used my answer from part i) the no of mols of octane in 1L
In the equation it provided us a energy value e.g. 5000kJ/mol
so I just did 5000/no. of mols of octane in 1L

Next part I did a similar thing to ethanol e.g. got the no. of mols in 1L by doing the same method.
Then I did ratios ... e.g. ___no. of mols of octane = ____kJ = 10km
therefore this ____no. of mols of ethanol = _____kJ = how ever long.
 
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