Thoughts? (4 Viewers)

[fishbulb]

I think the titre volume (23.45mL) was for the Na2CO3 ---- while the 25mL was for the HCl samples.

Yeh it was and i had the sodium carbonate concentration at 0.05M.

 

[friction]

I got 0.1057 i think.

Also what did u guys get for that tablet question it was fucked.

 

[uaiof100.00]

I GOT 100% easily im so smart!

 

[fishbulb]

I got 0.1057 i think.

Also what did u guys get for that tablet question it was fucked.

I put 1.7mg and then divided it by 4 because i thought they used 4 tablets in the sample. oh well thats 1 mark gone

 

[fishbulb]

I got 0.1057 i think.

Also what did u guys get for that tablet question it was fucked.

Did u keep all your numbers in your calculator, that could be the cause of a slightly off answer.

 

[friction]

Danz you are right just your are 10 x what you need. @ Fish bulb you did not factor in the 2:1 ratio therefor you got that not 0.1.

It is looking up now.

 

[fishbulb]

Danz you are right just your are 10 x what you need. @ Fish bulb you did not factor in the 2:1 ratio therefor you got that not 0.1.

It is looking up now.

Yeh i did bud.

Moles of Sodium Carbonate:

0.05 = Moles/0.02345

Therefore 2.345 x 10^-3 moles of HCL (Multiplying above answer by 2)

Molarity of HCL = (2.345 x 10^-3)/0.025

= 0.0938Mol/L

 

[Zippora]

i do calculation questions in past papers, and then i somehow always manage to not be able to do them in the exam itself.



i found calculations in today's paper so confusing.

plus zinc tablets - wtf?
plus what were the 2 redox reactions in the shipwrecks option. one was ag+ + 2e- --> ag(s)
what was the other?
i made up alot.

the other was the oxidation of water

 

[friction]

Yeh i did bud.

Moles of Sodium Carbonate:

0.05 = Moles/0.02345

Therefore 2.345 x 10^-3 moles of HCL (Multiplying above answer by 2)

Molarity of HCL = (2.345 x 10^-3)/0.025

= 0.0938Mol/L

I think i did it very much differently to you. I cant remember the question however.

 

[Zephyrio]

Fishbulb is correct for the titration question.

 

[Daniel-08]

Yeh i did bud.

Moles of Sodium Carbonate:

0.05 = Moles/0.02345

Therefore 2.345 x 10^-3 moles of HCL (Multiplying above answer by 2)

Molarity of HCL = (2.345 x 10^-3)/0.025

= 0.0938Mol/L

100% correct i remember i got this aswell
hey i think with the zinc question were meant to divide by 4 because there were 4 tablets dissolved?

 

[fishbulb]

100% correct i remember i got this aswell
hey i think with the zinc question were meant to divide by 4 because there were 4 tablets dissolved?

sweet, i thought we might of, i didnt read the question well so i dunno.

 

[Loganrah]

The answer for the zinc question was 0.17 they clearly said AVERAGE absorbency, meaning that they had dissolved each of the tablets separately, measured their absorbency and averaged it.

 

[advanced sam]

Jesus... I thought the extended response bits were pretty good but as for the calculations and the options? let's just not mention them

yeah i agree with that

 

[Kearnzo]

Ahhhhhh I forgot to times by 2!! How could I forget that. I haven't got a moles questions wrong since like day 1. Pretty funny actually.

Anyone know what the question was out of? 2? Meh, they better carry errors in chem lol.

 

[P.T.F.E]

I think the titre volume (23.45mL) was for the Na2CO3 ---- while the 25mL was for the HCl samples.

the titre was HCl the acid therefore it was 23.45 the Na2CO3 was 25ml so the answer was the .1057 or whatever

 

[danz90]

Confuciousnesss over that stupid titration.

If we used the wrong titre volume, then its 1 mark lost - big deal. lol

 

[timmiitippii]

I'm not gonna make some people feel bad by commenting on the difficulty of the paper. But i have to say that was the best paper i could of hoped for. No long winded essay questions that worth 6, 7 or 8 marks. Was so surprised when had to go through the paper to see so many calculations since i love calculations and only one single 7 marker in the industrial chem section. No 6 marker whatsoever.
I'm really glad we were given this paper though, after the shitness that was maths 3u adn 4u hsc exams and to an extent physics.

 

[mick135]

yeah it seemed um... too good to be true!

hopefully that doesn't mean it"ll scale badly...

yeah it was very good i thought.
considering i was unprepared and not really ready for it.
they couldn't of made it easier on me

for industrial - do you think they'll notice i made up a cationic detergent "Mr Sheens carwash"?
haha i had no idea what to write for that 7 mark question.
i somehow related it back to solvay process and blabbed about that
anyways - good luck to you's all!

 

[timmiitippii]

i noticed most of u got 1.6V for the galvanic cell question...
But i'm certain u meant to use the 1/2 Cl2 (aq) + e- ---> Cl- 1.4V
rather than the 1/2 Cl2 (g) + e- ---> Cl- 1.36V

Because the Cl2 gas is bubbled into the water, forming aqueous solution. The gas one is used if it a sealed chamber filled with the gas in gaseous states im pretty sure.
So the voltage would be .24 + 1.40 = 1.64V