HSC Chemistry Question Thread

[jakesilove]

Could i have help with q11 a?
How can you tell if its endo or exo without the change of enthalpy being indicated?

If the answer is a), I really don't know how you were supposed to get that without a delta H value!

I was also wondering how would you do e for this question?

Again, I really don't know. Look for whichever bond is 'strongest'; which that is, I'm not sure! Falls outside the HSC curriculum. However, I reckon it's the second one: Oxygen and Hydrogen bonds are very strong, as one is very electro negative and one is very electro positive

[Fahim486]

Hi everyone,

I was just wondering if anyone had notes on interpreting equilibrium graphs.

Thanks!

[f_tan]

Hi! I need help with these past HSC questions, I'm not sure why/how to get the answer:

1. A household cleaning agent contains a weak base of general formula NaX. 1.00 g of this compound was dissolved in 100.0 mL of water. A 20.0 mL sample of the solution was titrated with 0.1000 mol L–1 hydrochloric acid and required 24.4 mL of the acid for neutralisation. What is the molar mass of this base?

2. The following equation represents a chemical system in equilibrium:
OCl−(aq) + H2O(l) <--> HOCl(aq) + OH−(aq)
Which of the following is an acid/base conjugate pair?
(A) H2O / HOCl
(B) HOCl / OH−
(C) HOCl / OCl−
(D) OCl− / H2O

Thank you!

[kiwiberry]

Hi! I need help with these past HSC questions, I'm not sure why/how to get the answer:

1. A household cleaning agent contains a weak base of general formula NaX. 1.00 g of this compound was dissolved in 100.0 mL of water. A 20.0 mL sample of the solution was titrated with 0.1000 mol L–1 hydrochloric acid and required 24.4 mL of the acid for neutralisation. What is the molar mass of this base?

2. The following equation represents a chemical system in equilibrium:
OCl−(aq) + H2O(l) <--> HOCl(aq) + OH−(aq)
Which of the following is an acid/base conjugate pair?
(A) H2O / HOCl
(B) HOCl / OH−
(C) HOCl / OCl−
(D) OCl− / H2O

Thank you!

1) Finding the moles of HCl used first:
n(HCl) = 0.1*0.0244 = 0.00244 mol

Because the reaction is NaX +HCl --> NaCl + HX
When neutralisation occurs, number of moles of NaX = number of moles of HCl
Therefore n(NaX) = 0.00244 mol

Because only 20mL out of the 100mL solution of NaX was titrated, m(NaX) = 1*0.2 = 0.2g
So there are 0.00244 moles in 0.2g of NaX!
0.00244 mol = 0.2 g
1 mol = 0.2/0.00244 = 81.97 g

So the molar mass is 81.97 g/mol

2) When an acid donates a proton, it forms its conjugate base, and when a base accepts a proton, it forms its conjugate acid
Looking at OCl^-, it accepts a proton from water to form HOCl. Thus, the conjugate acid-base pair is OCl^-/HOCl
Looking at H₂O, it donates a proton to OCl^- to form OH^-. So the pair is H₂O/OH^-
Therefore the answer is C!

[RuiAce]

Hi everyone,

I was just wondering if anyone had notes on interpreting equilibrium graphs.

Thanks!

As rules of thumb:

Sudden spike in ALL substances - change in pressure
Sudden spike in ONE substance - change in concentration of just one thing
No spike but the graph suddenly changes - change in temperature

Reasoning:
The graph is concentration v.s. time and we know that C=n/V. So a spike, i.e. a sudden change in concentration, only occurs as a result of a change in the moles or the volume.

If the volume gets changed, then the concentration of EVERYTHING is changed. Hence this relates to a change in pressure.
If the moles gets changed, well then the moles of only one thing changes.

If no spike happens, but the concentrations start gradually changing afterwards, well the last thing left on the list is temperature. Which makes sense; temperature still causes the equilibrium to shift, but we haven't abruptly affected the concentration.


You must then use LCP as appropriate to determine which side the equilibrium shifted to and etc.

[Fahim486]

As rules of thumb:

Sudden spike in ALL substances - change in pressure
Sudden spike in ONE substance - change in concentration of just one thing
No spike but the graph suddenly changes - change in temperature

Reasoning:
The graph is concentration v.s. time and we know that C=n/V. So a spike, i.e. a sudden change in concentration, only occurs as a result of a change in the moles or the volume.

If the volume gets changed, then the concentration of EVERYTHING is changed. Hence this relates to a change in pressure.
If the moles gets changed, well then the moles of only one thing changes.

If no spike happens, but the concentrations start gradually changing afterwards, well the last thing left on the list is temperature. Which makes sense; temperature still causes the equilibrium to shift, but we haven't abruptly affected the concentration.


You must then use LCP as appropriate to determine which side the equilibrium shifted to and etc.

Thank you so much for your help RuiAce 

[kylesara]

Hi, i was wondering if you have to number your steps in the method of a report or if dot points are alright. Thanks.

[jakesilove]

Hi, i was wondering if you have to number your steps in the method of a report or if dot points are alright. Thanks.

Hey! Either works fine; I much prefer to number though.

[J.B]

Hi I was just doing Q22 b) from the 2004 chemistry paper. And the question is:

Write Two chemical equations to show that the dihydrogen phosphate ion is amphiprotic. I can't seem to find any worked solutions for this so I was wondering if my answer was correct:

As an acid: H2PO4- (aq)    +     OH-  (aq)   >     HPO4 2- (aq)   + H20  (l)

As a base:   H2PO4-  (aq)   +    H3O+  (aq)   >   H3PO4  (aq)    +  H2O  (l)

Thank you.

[kiwiberry]

Hi I was just doing Q22 b) from the 2004 chemistry paper. And the question is:

Write Two chemical equations to show that the dihydrogen phosphate ion is amphiprotic. I can't seem to find any worked solutions for this so I was wondering if my answer was correct:

As an acid: H2PO4- (aq)    +     OH-  (aq)   >     HPO4 2- (aq)   + H20  (l)

As a base:   H2PO4-  (aq)   +    H3O+  (aq)   >   H3PO4  (aq)    +  H2O  (l)

Thank you.

Looks good!!

[RuiAce]

Can confirm. You basically demonstrated with suitable equations that it is both a proton donor and a proton acceptor, so the equations are valid and correct.

[bananna]

Hi
I have a q from the 2012 chem paper

All the lead ions present in a 50.0 mL solution were precipitated by reaction with excess chloride ions. The mass of the dried precipitate was 0.595 g.
What was the concentration of lead in the original solution?

thanks!

[bananna]

Hi again (sorry)

I also have this question

The heat of combustion of propan-1-ol is 2021kJmol–1. Combustion takes place according to the equation:
2C3H7OH(l) + 9O2(g) → 6CO2(g) + 8H2O(l)
What mass of water is formed when 1530 kJ of energy is released?

thanks again

[bananna]

Again, I am sooooo sorry

can you please check my response to this question its out of 4 marks

An indicator is placed in water. The resulting solution contains the green ion, Ind −, and 4
the red molecule, HInd.
Explain why this solution can be used as an indicator. In your response, include a
suitable chemical equation that uses Ind − and Hind.
HSC 2013

Hind<--> H^+  +  Ind^-
red                        green

This solution is an indicator because it has a differentiating power (colour) between pHs. It will recognise the lower pH substance as red, and will change to green when a substance of higher pH is added. Thus, it is a valid indicator.

I wasn't really sure how to answer this q
thank you

[kiwiberry]

Hi
I have a q from the 2012 chem paper

All the lead ions present in a 50.0 mL solution were precipitated by reaction with excess chloride ions. The mass of the dried precipitate was 0.595 g.
What was the concentration of lead in the original solution?

thanks!

Pb²⁺+ 2Cl^- --> PbCl₂

We know that v(Pb²⁺) = 0.05 L and m(PbCl₂) = 0.595 g. Our goal is to find c(Pb²⁺), so somehow we need to find n(Pb²⁺)!

We can find n(PbCl₂) since we are given its mass
n(PbCl₂) = 0.595/(207.2+2*35.45) = 0.0021395… mol
Since n(PbCl₂) = n(Pb²⁺),
n(Pb²⁺) = 0.0021395 mol
m(Pb²⁺) = 0.0021395*207.2 = 0.44308… g

Now we can find concentration!
c(Pb²⁺) = 0.44308.../0.05 = 8.866.. g/L = 8.87 g/L