HSC 2012-2015 Chemistry Marathon (archive) (5 Viewers)

[abdog]

re: HSC Chemistry Marathon Archive

Find the H+ conc. Minus it from the molarity of the solution. then divide it by 0.1 because that's the original number of moles to find the ratio.

Can you show me your working? Cause I still dont really understand.....

 

[bangladesh]

re: HSC Chemistry Marathon Archive

Can you show me your working? Cause I still dont really understand.....

I'll show you after MX1 tomorrow

 

[Frie]

re: HSC Chemistry Marathon Archive

The pH of a 0.1 mol L-1 solution of a monoprotic acid was measured by a student and found to be close to 3. What proportion of the acid molecules remains unconverted to ions?

Degree of Ionisation (DOI) = [H+ ions in solution] / [Original Solution] x 100
[H+] = 10^-pH
[H+ ions in solution] = 10^-3 = 0.001 M
DOI = 0.001 / 0.1 x 100 = 1%

Since Degree of Ionisation is 1%, that means that 1% of acid molecules are converted to ions, whilst 99% remains unconverted.

(Sorry, don't know how to use the really good formatting for working out, hope its not hard to follow.)

 

[abdog]

re: HSC Chemistry Marathon Archive

Thanks! That will do!

 

[babberz]

re: HSC Chemistry Marathon Archive

n(Cu) = 0.325/63.55
= 5.114x10^-3 moles
Mol ratio of solid copper : CuSO4 = 1:1
therefore n(CuSO4 'lost') = 5.114x10^-3 mol

initial amount (using n=cV) was 0.250 x 0.05 = 0.0125 mol
final amount = 0.0125 - 5.114x10^-3
= 6.886x10^-3 mol

therefore final concentration = 6.886x10^-3/0.250
=0.0275 mol/L

i can't see where you got this from, i tested exact values and sig fig and still couldnt get this

 

[abdog]

re: HSC Chemistry Marathon Archive

A student wanted to determine the phosphate content in a certain brand of washing detergent. The phosphorus was precipitated as Ca2P2)7 and then filtered. A 4.42g sample of washing detergent resulted in a precipitate of mass 0.232g. Calculate the percentage of phosphorus, by mass, in the sample.

So this is what I did:
Molar mass of phosphorus/Molar mass of Ca2P2O7(as percentage) = (30.97/254.1) x 100, which gives 12.2%.

So 12.2% of 0.232(precipitate) is phosphorus, which is 0.028g.

Mass of phosphorus/Mass of sample(as percentage) = (0.028/4.42) x 100 = 0.633%, which is not even close to the answer, which is 1.28%.

Where am I doing wrong?

 

[Frie]

re: HSC Chemistry Marathon Archive

There are Ca2P2O7

So Percentage of Phosphorus in Ca2P2O7 = Molar mass of phosphorus/Molar mass of Ca2P2O7 = (30.97x2/254.1) x 100, which gives 24.38%.

So 24.38% of 0.232(precipitate) is phosphorus, which is 0.0566g.

Mass of phosphorus/Mass of sample(as percentage) = (0.0566/4.42) x 100 = 1.279%

I'm guessing that's how you do it, since I've only come across working out sulfate% in fertiliser.

 

[bangladesh]

re: HSC Chemistry Marathon Archive

alright, lets do hardcore chemistry from today guys. Keep the questions coming.

 

[Frie]

re: HSC Chemistry Marathon Archive

i can't see where you got this from, i tested exact values and sig fig and still couldnt get this

His answer is wrong.

He made a mistake on that step that you bolded.

Shoulda been:

initial amount (using n=cV) was 0.250 x 0.05 = 0.0125 mol
final amount = 0.0125 - 5.114x10^-3
= 7.386x10^-3 mol

therefore final concentration = 7.386x10^-3/0.250
=0.029544 mol/L
= 0.03 mol/L (2 sig figs)

 

[abdog]

re: HSC Chemistry Marathon Archive

Which of the following will cause the greatest change in pH to a 100mL sample of 0.1 molL-1 HCL?
A. adding 1g of CaCO3 powder
B. adding 10mL of 0.1 molL-1 NaOH
C. adding 100mL of 0.1 molL-1 HCL
D. diluting the sample to 1000mL

Please show your working if possible, otherwise there's no point

 

[bangladesh]

re: HSC Chemistry Marathon Archive

Which of the following will cause the greatest change in pH to a 100mL sample of 0.1 molL-1 HCL?
A. adding 1g of CaCO3 powder
B. adding 10mL of 0.1 molL-1 NaOH
C. adding 100mL of 0.1 molL-1 HCL
D. diluting the sample to 1000mL

Please show your working if possible, otherwise there's no point

Is the answer A? I have solutions and i'll upload it but i just want to make sure my answer is right.

 

[abdog]

re: HSC Chemistry Marathon Archive

Is the answer A? I have solutions and i'll upload it but i just want to make sure my answer is right.

Yes it is A

 

[bangladesh]

re: HSC Chemistry Marathon Archive

Yes it is A

https://dl-web.dropbox.com/get/Came...AA9mAidIb6FyGFYn6poHRQUIWzDWzl6Z_vUw2e0TKUjhw

This is the worked solution, Sorry it's a bit messy. However, I don't know where you've gotten this question from but this will not show up in the HSC as you do not need to know the degree of ionisation of Carbonic acid. You could still guess A without knowing the degree of ionisation of carbonic acid as the other choices have very limited pH change, if any. Hope it helps. Keep posting questions.

 

[abdog]

re: HSC Chemistry Marathon Archive

https://dl-web.dropbox.com/get/Came...AA9mAidIb6FyGFYn6poHRQUIWzDWzl6Z_vUw2e0TKUjhw

This is the worked solution, Sorry it's a bit messy. However, I don't know where you've gotten this question from but this will not show up in the HSC as you do not need to know the degree of ionisation of Carbonic acid. You could still guess A without knowing the degree of ionisation of carbonic acid as the other choices have very limited pH change, if any. Hope it helps. Keep posting questions.

Oh that's great news! Anyway, I got it off a CSSA paper, so that explains it

 

[abdog]

re: HSC Chemistry Marathon Archive

What mass of sulfur is formed when 2L of sulfur dioxide is mixed with 2L of hydrogen sulfide, under standard conditions(273K and 100kPa)?

SO2 + 2H2S --> 3S + 2H20

What I got so far:
Let X be moles of SO2 or H2S, so X=2/24.79=0.08moles.
My question is, do we use S02 or H2S? So I did both, got 8g for SO2 and 4g for H2S. Which one's right?

 

[babberz]

re: HSC Chemistry Marathon Archive

His answer is wrong.

He made a mistake on that step that you bolded.

Shoulda been:

initial amount (using n=cV) was 0.250 x 0.05 = 0.0125 mol
final amount = 0.0125 - 5.114x10^-3
= 7.386x10^-3 mol

therefore final concentration = 7.386x10^-3/0.250
=0.029544 mol/L
= 0.03 mol/L (2 sig figs)

ah thx for clearing that up

 

[babberz]

re: HSC Chemistry Marathon Archive

What mass of sulfur is formed when 2L of sulfur dioxide is mixed with 2L of hydrogen sulfide, under standard conditions(273K and 100kPa)?

SO2 + 2H2S --> 3S + 2H20

What I got so far:
Let X be moles of SO2 or H2S, so X=2/24.79=0.08moles.
My question is, do we use S02 or H2S? So I did both, got 8g for SO2 and 4g for H2S. Which one's right?

this, since H2S is the limiting reagent

 

[abdog]

re: HSC Chemistry Marathon Archive

this, since H2S is the limiting reagent

S02 and H2S both have the same amount of moles. How do you work out which one is the limiting reagent?

 

[Frie]

re: HSC Chemistry Marathon Archive

What mass of sulfur is formed when 2L of sulfur dioxide is mixed with 2L of hydrogen sulfide, under standard conditions(273K and 100kPa)?

SO2 + 2H2S --> 3S + 2H20

What I got so far:
Let X be moles of SO2 or H2S, so X=2/24.79=0.08moles.
My question is, do we use S02 or H2S? So I did both, got 8g for SO2 and 4g for H2S. Which one's right?

I'm pretty sure that since there is 2L of both SO2 and H2S, and they have different molar ratios, H2S becomes the limiting reagent. There would be 0.044 moles of SO2 leftover because 0.088 moles of H2S and 0.044 moles of SO2 is used up in the reaction.
Therefore, 3 x 0.044 = 0.132 moles of Sulfur is produced.
Mass of sulfur = Molar mass x moles of sulfur = 32.07 x 0.132 = 4.23324g

 

[babberz]

re: HSC Chemistry Marathon Archive

S02 and H2S both have the same amount of moles. How do you work out which one is the limiting reagent?

look at the equation for every 1 mole of SO2 used, 2 moles of H2S is consumed, so in the end 4 moles of SO2 will be in excess