Author Topic: HSC Chemistry Question Thread (Read 1295066 times) Tweet Share

Mate2425 · « **Reply #3390 on:** April 02, 2018, 01:04:44 am »

Hello,
Does anyone know where i can find a link to the reasoning behind multiple choice questions in the HSC Chemistry paper 2015.
Thanks

RuiAce · « **Reply #3391 on:** April 02, 2018, 07:06:29 am »

Unfortunately NESA has only ever provided the answers to the multiple choice and no solutions behind them. If you need some help with specific questions please link to the paper and say which questions you're having trouble with (or attach a screenshot)

Mate2425 · « **Reply #3392 on:** April 02, 2018, 01:07:03 pm »

Hey could i please have some help with the calculations for this question. Thanks

RuiAce · « **Reply #3393 on:** April 02, 2018, 01:29:39 pm »

Quote from: Mate2425 on April 02, 2018, 01:07:03 pm

Hey could i please have some help with the calculations for this question. Thanks

Where's the question?

Mate2425 · « **Reply #3394 on:** April 02, 2018, 02:02:07 pm »

Whoops.. HSC Question 2017 MC Q20 thanks

https://educationstandards.nsw.edu.au/wps/wcm/connect/cb65ca25-fc28-4e21-8f2d-4886bf176ddd/2017-hsc-chemistry.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-cb65ca25-fc28-4e21-8f2d-4886bf176ddd-lZ.o-ZW

RuiAce · « **Reply #3395 on:** April 02, 2018, 02:10:41 pm »

$\text{Using }\boxed{n=CV}\\ n_{Ba(OH)_2} = 0.020 \times 0.020 = 0.0004\text{ mol}.\\ n_{HCl} = 0.040 \times 0.050 = 0.002\text{ mol}.$
$\text{The relevant chemical equation is}\\ Ba(OH)_{2(aq)} + 2\, HCl_{(aq) }\to BaCl_{2(aq)} + 2\, H_2O_{(l)},\\ \text{so we infer that 1 mol of barium hydroxide}\\ \text{reacts with 2 mol of hydrochloric acid.}$
$\text{So in our case, barium hydroxide is the limiting reagent}\\ \text{as 0.0004 mol of barium hydroxide will react with}\\ \text{0.0008 mol of hydrochloric acid only.}$
$\text{Hence, we have }n_{HCl} = 0.002 - 0.0008 = 0.0012 \text{ mol of excess }HCl.$
___________________________________________________
$\text{The combined volume is now 70 mL, so using }\boxed{C = \frac{n}{V}}\\ [HCl] = \frac{0.0012}{0.070} = \frac{3}{175}\text{ mol L}^{-1}.$
$\text{Of course, since HCl is a strong acid and fully ionises, this implies that}\\ [H_3O^+] = \frac{3}{175}\text{ mol L}^{-1}\\ \text{and therefore }pH = -\log_{10} \frac{3}{175} = 1.7659\dots$
Hence C.

Mate2425 · « **Reply #3396 on:** April 02, 2018, 02:19:04 pm »

Quote from: RuiAce on April 02, 2018, 02:10:41 pm

$\text{Using }\boxed{n=CV}\\ n_{Ba(OH)_2} = 0.020 \times 0.020 = 0.0004\text{ mol}.\\ n_{HCl} = 0.040 \times 0.050 = 0.002\text{ mol}.$
$\text{The relevant chemical equation is}\\ Ba(OH)_{2(aq)} + 2\, HCl_{(aq) }\to BaCl_{2(aq)} + 2\, H_2O_{(l)},\\ \text{so we infer that 1 mol of barium hydroxide}\\ \text{reacts with 2 mol of hydrochloric acid.}$
$\text{So in our case, barium hydroxide is the limiting reagent}\\ \text{as 0.0004 mol of barium hydroxide will react with}\\ \text{0.0008 mol of hydrochloric acid only.}$
$\text{Hence, we have }n_{HCl} = 0.002 - 0.0008 = 0.0012 \text{ mol of excess }HCl.$
___________________________________________________
$\text{The combined volume is now 70 mL, so using }\boxed{C = \frac{n}{V}}\\ [HCl] = \frac{0.0012}{0.070} = \frac{3}{175}\text{ mol L}^{-1}.$
$\text{Of course, since HCl is a strong acid and fully ionises, this implies that}\\ [H_3O^+] = \frac{3}{175}\text{ mol L}^{-1}\\ \text{and therefore }pH = -\log_{10} \frac{3}{175} = 1.7659\dots$
Hence C.

Thank You very much

3.14159265359 · « **Reply #3397 on:** April 02, 2018, 07:09:38 pm »

hey, what would the answer to this be?

kiwiberry · « **Reply #3398 on:** April 02, 2018, 07:29:11 pm »

Quote from: 3.14159265359 on April 02, 2018, 07:09:38 pm

hey, what would the answer to this be?

Hey!!
When bromine water is added to hexane, no reaction will occur, and the bromine water will float above it in another layer. Hexene will be more reactive than hexane due to its double bond, and when bromine water is added to it, a reaction occurs, decolourising the bromine water. Hence the answer is C

3.14159265359 · « **Reply #3399 on:** April 02, 2018, 07:36:08 pm »

Quote from: kiwiberry on April 02, 2018, 07:29:11 pm

Hey!!
When bromine water is added to hexane, no reaction will occur, and the bromine water will float above it in another layer. Hexene will be more reactive than hexane due to its double bond, and when bromine water is added to it, a reaction occurs, decolourising the bromine water. Hence the answer is C

that makes so much sense! thank you

3.14159265359 · « **Reply #3400 on:** April 02, 2018, 09:28:29 pm »

Hey, you see when we calcualte E° we add the E(oxidation) And E(reduction) and for the oxidation if it’s -ve we flip the sign to be positive because we flip the eqn. How about if E(oxidation) was already positive, do we still change like above (so from positive to negative)

RuiAce · « **Reply #3401 on:** April 02, 2018, 09:37:56 pm »

Quote from: 3.14159265359 on April 02, 2018, 09:28:29 pm

Hey, you see when we calcualte E° we add the E(oxidation) And E(reduction) and for the oxidation if it’s -ve we flip the sign to be positive because we flip the eqn. How about if E(oxidation) was already positive, do we still change like above (so from positive to negative)

Yeah. The sign change always occurs regardless.

If you don't flip the sign then you'll get an E° value that's too high (when they're both +'ve). The cell actually produces relatively small voltage when they're both positive.

3.14159265359 · « **Reply #3402 on:** April 02, 2018, 09:47:13 pm »

Quote from: RuiAce on April 02, 2018, 09:37:56 pm

Yeah. The sign change always occurs regardless.

If you don't flip the sign then you'll get an E° value that's too high (when they're both +'ve). The cell actually produces relatively small voltage when they're both positive.

Hmmm makes sense. Thanks!!

Mate2425 · « **Reply #3403 on:** April 02, 2018, 10:17:40 pm »

Hi could someone please help me with wondering why the answer to Q11 2016 MC is D and not A.
http://boardofstudies.nsw.edu.au/hsc_exams/2016/exams/2016-hsc-chemistry.pdf
Thank you.

3.14159265359 · « **Reply #3404 on:** April 03, 2018, 07:10:08 am »

Quote from: Mate2425 on April 02, 2018, 10:17:40 pm

Hi could someone please help me with wondering why the answer to Q11 2016 MC is D and not A.
http://boardofstudies.nsw.edu.au/hsc_exams/2016/exams/2016-hsc-chemistry.pdf
Thank you.

Hey, I think it’s because you have to add all the numbers and you pick the one with the lowest numbers.
Like for a:
There is a 1+1+2+2+2= 8

And for b:
There is a 2+2+1+1+1= 7

So that’s why it’s d because 7<8.

I’m not too sure tho, so just wait for one of the moderators to confirm

Search the forums now!

We have moved!

Author Topic: HSC Chemistry Question Thread (Read 1295066 times) Tweet Share

Mate2425

Re: Chemistry Question Thread

RuiAce

Re: Chemistry Question Thread

Mate2425

Re: Chemistry Question Thread

RuiAce

Re: Chemistry Question Thread

Mate2425

Re: Chemistry Question Thread

RuiAce

Re: Chemistry Question Thread

Mate2425

Re: Chemistry Question Thread

3.14159265359

Re: Chemistry Question Thread

kiwiberry

Re: Chemistry Question Thread

3.14159265359

Re: Chemistry Question Thread

3.14159265359

Re: Chemistry Question Thread

RuiAce

Re: Chemistry Question Thread

3.14159265359

Re: Chemistry Question Thread

Mate2425

Re: Chemistry Question Thread

3.14159265359

Re: Chemistry Question Thread

Recent Posts

User:
Password: