Login

Welcome, Guest. Please login or register.

March 29, 2024, 05:22:21 am

Author Topic: HSC Chemistry Question Thread  (Read 1040612 times)  Share 

0 Members and 1 Guest are viewing this topic.

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: Chemistry Question Thread
« Reply #495 on: July 25, 2016, 11:40:03 pm »
0

Lol i made a dumb. Of course the final products must stay there.

In (1) and (2) there is exactly 1 mol of Cl(g) there. Thus, we can immediately perform (1) + (2)

CFCl3(g) + Cl(g) + O3(g) -> CFCl2(g) + Cl(g) + ClO(g) + O2(g)

Cancel out the Cl on both sides:
CFCl3(g) + O3(g) -> CFCl2(g) + ClO(g) + O2(g)

This new equation, and equation 3, both have exactly 1 mol of ClO present. Thus we can perform (3) + the above

CFCl3(g) + O3(g) + ClO(g) + O(g) -> CFCl2(g) + ClO(g) + O2(g) + Cl(g) + O2(g)

Cancel out the ClO:

CFCl3(g) + O3(g) + O(g) -> CFCl2(g) + 2 O2(g) + Cl(g)

(Image removed from quote.)

Hey guys im not sure to do q 15 hehe thanks
This is just your ordinary rigorous calculation.

Write out relevant equations:
A) K2CO3 + 2 HCl -> CO2 + H2O + 2 KCl
B) KHCO3 + HCl -> CO2 + H2O + KCl
C) Na2CO3 + 2 HCl -> CO2 + H2O + 2 KCl
D) NaHCO3 + HCl -> CO2 + H2O + KCl

Clearly, 1 mol of reactant _ yields 1 mol of CO2 in all four equations.

Now, determine the moles of each substance present using n=m/MM:
A) nCO2 = nK2CO3 = 7.23 * 10-3 mol
B) nCO2 = nKHCO3 = 1.05 * 10-2 mol
C) nCO2 = nNa2CO3 = 9.43 * 10-3 mol
D) nCO2 = nNaHCO3 = 1.19 * 10-2 mol

So since n = V/VM, clearly the answer must be D.

And now for the smart way

However obvious it was that the moles of the reactant and the moles of CO2 were the same varies for each person. But without even writing out the full equation you should've might've been able to see it. This is because CO32- is being decomposed down to CO2 by HCl, which has no carbons in it!

Then, because m=1, we have that the moles is INVERSELY proportional to the molar mass. Thus the one with the smallest molar mass (NaHCO3) will yield the largest moles.
« Last Edit: July 26, 2016, 12:26:07 am by RuiAce »

Aliceyyy98

  • Trailblazer
  • *
  • Posts: 43
  • Respect: 0
  • School: Willoughby Girls High School
  • School Grad Year: 2016
Re: Chemistry Question Thread
« Reply #496 on: July 26, 2016, 12:57:50 am »
0
Hey human
Hopefully, I can help a bit...
Buffers are solutions with the ability to resist pH change when small quantities of acid/base are added to them. Buffers are generally a solution which contains a weak acid-base conjugate pair (a weak acid and its conjugate base or vice-versa). These acids and bases and co-exist without neutralising each other, as neutralisation generally occurs in an acid-base reaction, yet they can still react to neutralise any strong acid or strong base added to the buffer.
An example of a buffer solution could include a 2mol ethanoic acid solution, and 2 mol of sodium ethanoate; then making the solution u to 1L. The buffer contains a high concentration of both weak acid CH3COOH and its conjugate weak base, CH3COO-.
The equilibrium reaction is shown below:
CH3COOH (aq) + H20 (l)   <---->   CH3COO- +H3O+(aq)
Buffer behaviour can be predicted via Le Chateliers Principle (LCP). Eg, if you add a base to the buffer will neutralise some of the H3O+ present, causing its concentration to fall. This falling concentration causes the equilibrium reaction to shift right, as explained via LCP, (forward reaction favoured) o replace some of the lost hydronium concentration, and preventing too much of a significant fall. This is the property of a buffer being able to resist some changes in pH. If you flood the buffer with acid/base this will not work, and a pH change will result (each buffer has a buffer capacity)

Thanx a bunch!

Alexander23

  • Fresh Poster
  • *
  • Posts: 4
  • Respect: 0
Re: Chemistry Question Thread
« Reply #497 on: July 26, 2016, 08:56:38 am »
0
I've been told that the addition of an inert gas has no effect on a reaction at equilibrium. I was wondering why it has no effect on equilibrium direction, even if you are adding in this non reaction gas in a closed system which seems to decrease volume and hence increase pressure.

Also, I was hoping for a clearer explanation as to why temperature is the only factor which effects the K expression of an equilibrium reaction.

Thanks.   :)

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: Chemistry Question Thread
« Reply #498 on: July 26, 2016, 09:14:02 am »
0
I've been told that the addition of an inert gas has no effect on a reaction at equilibrium. I was wondering why it has no effect on equilibrium direction, even if you are adding in this non reaction gas in a closed system which seems to decrease volume and hence increase pressure.

Also, I was hoping for a clearer explanation as to why temperature is the only factor which effects the K expression of an equilibrium reaction.

Thanks.   :)
Adding an inert gas is essentially doing the opposite of the effect of a catalyst - it slows down the rate equilibrium is attained however does nothing to the equilibrium itself.

This is because by adding a catalyst, you are not hindering just ONE of the forward/reverse reactions. You are actually hindering BOTH of the forward/reverse reactions.

If you change the pressure by pumping in more equilibrium mixture you are causing an imbalance in the concentrations of products and reactants. This is the same as varying the volume of the vessel, because the amount of space you have will promote either the forward or reverse reaction to allow the moles of gas to be reasonable. (Recall: Avogadro's law states that at equal temperatures and pressures, systems will rather having an equal number of molecules of gases.)

Now, temperature is an extrinsic factor to the position of the equilibrium. All other factors such as pressure and varying concentrations are intrinsic. When you alter the pressure by pumping equilibrium mixture, change the volume of the vessel or increase/decrease concentrations of just one substance, as a consequence of Le Chatelier's Principle the system will try to bring back the equilibrium it started with.

Note that K is just a rough indicator of where the equilibrium actually is to begin with! The concentrations of your reactants/products readjust so that the reaction quotient Q returns to K.

When you alter the temperature, however, you didn't actually alter the concentrations of the reactants and products. Instead, you manipulated the fact that the reaction is exothermic or endothermic. From LCP we do know that the equilibrium does indeed adjust if you supply/take out heat, however in doing so you're forcing the ratio of products and reactants to change themselves WITHOUT actually having added/taken away any in advance! Hence, K will actually change for different temperatures
« Last Edit: July 26, 2016, 09:28:26 am by RuiAce »

liiz

  • Trailblazer
  • *
  • Posts: 25
  • Respect: 0
  • School: Our Lady of the Sacred Heart
  • School Grad Year: 2016
Re: Chemistry Question Thread
« Reply #499 on: July 26, 2016, 02:55:45 pm »
0
Hey there, just wondering if someone would be please able to help me with this question. I find calculation questions super hard so any explanation would be awesome, thankyou!!  ;D ;D"Phosphorus pentoxide reacts with water to form phosphoric acid according to the following equation P2O5(s)  + 3H2O(l) → 2H3PO4 (aq) Phosphoric acid reacts with sodium hydroxide according to the following equation. H3PO4 (aq) + 3NaOH(aq) → Na3PO4 (aq) + 3H2O(l). A student reacted 1.42 g of phosphorus pentoxide with excess water. What volume of 0.30 mol L-1 sodium hydroxide would be required to neutralise all the phosphoric acid produced?"

sweetcheeks

  • Forum Obsessive
  • ***
  • Posts: 496
  • Respect: +83
  • School: ---
  • School Grad Year: 2016
Re: Chemistry Question Thread
« Reply #500 on: July 26, 2016, 03:52:49 pm »
0
Hey there, just wondering if someone would be please able to help me with this question. I find calculation questions super hard so any explanation would be awesome, thankyou!!  ;D ;D"Phosphorus pentoxide reacts with water to form phosphoric acid according to the following equation P2O5(s)  + 3H2O(l) → 2H3PO4 (aq) Phosphoric acid reacts with sodium hydroxide according to the following equation. H3PO4 (aq) + 3NaOH(aq) → Na3PO4 (aq) + 3H2O(l). A student reacted 1.42 g of phosphorus pentoxide with excess water. What volume of 0.30 mol L-1 sodium hydroxide would be required to neutralise all the phosphoric acid produced?"

Initially we start with 1.42g of phosphorus pentoxide. First we must find the amount of moles of phosphorus pentoxide. Each molecule of phosphorus pentoxide produces two mole of phosphoric acid. The phosphoric acid produced will then react with Sodium hydroxide in a 1:3 ratio, that is you need 3 molecules of NaOH per molecule of phosphoric acid.

Here are the steps
1. Find moles of phosphorus pentoxide, using the formula n=m/M where n=number moles m=grams M=molecular weight
2. Ratio this to moles of Phosphoric acid produced (2:1)
3. Substitute the mole of phosphoric acid into the second equation and ratio to NaOH to find the amount (in mole) of NaOH required. Using the formula n/C=v (rearranged C=n/v), where n=moles of NaOH, C= concentration of the NaOH solution and V is the volume in litres.

It helps to draw a diagram when doing questions like this. Also, write out both equations and work in columns under the substances of interest (i.e. when finding moles of phosphorus pentoxide, work underneath it on the formula.)


Is there anything in particular that you struggle with when it comes to these type of questions?

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: Chemistry Question Thread
« Reply #501 on: July 26, 2016, 05:01:10 pm »
0
Hey there, just wondering if someone would be please able to help me with this question. I find calculation questions super hard so any explanation would be awesome, thankyou!!  ;D ;D"Phosphorus pentoxide reacts with water to form phosphoric acid according to the following equation P2O5(s)  + 3H2O(l) → 2H3PO4 (aq) Phosphoric acid reacts with sodium hydroxide according to the following equation. H3PO4 (aq) + 3NaOH(aq) → Na3PO4 (aq) + 3H2O(l). A student reacted 1.42 g of phosphorus pentoxide with excess water. What volume of 0.30 mol L-1 sodium hydroxide would be required to neutralise all the phosphoric acid produced?"
Because the equations are given to you, there's one less thing to do. Because normally when you work between reactants/products and other reactants/products you always need a chemical equation.

You also always work in moles. According to n=m/MM

nP2O5 = 1.42/(2*30.97 + 5*16.00) = 0.010004227... mol

Look at the equation and compare the mole ratio. 1 mol of diphosphorus pentoxide produces 2 moles of phosphoric acid

nH3PO4 = 2 nP2O5 = 0.02000845427... mol

So we now know the moles of phosphoric acid we must neutralise.

Look at the second equation for the neutralisation. Clearly three moles of NaOH is required to neutralise one mole of H3PO4

nNaOH = 3 nH3PO4 = 0.06002536282... mol

We are given the concentration of NaOH. Because we know that C=n/V

V = n/C

VNaOH =  0.06002536282... / 0.30
=  0.06002536282... L

The lowest amount of significant figures used in this question is three (from the mass of diphosphorus pentoxide. So rounding to 3 s.f.

V ≈ 6.0 * 10-2 L

Key strategy: Break the question apart and do not try to do multiple things at once. Chemistry calculations are always one-way.

It helps to draw a diagram when doing questions like this. Also, write out both equations and work in columns under the substances of interest (i.e. when finding moles of phosphorus pentoxide, work underneath it on the formula.)
May I ask what diagram?

sweetcheeks

  • Forum Obsessive
  • ***
  • Posts: 496
  • Respect: +83
  • School: ---
  • School Grad Year: 2016
Re: Chemistry Question Thread
« Reply #502 on: July 26, 2016, 05:22:31 pm »
0
@RuiAce a diagram that shows the steps of the process. It helps visualize what is going on. for example, you draw the phosphorus pentoxide being weighed out, being added to water, the solution being reacted with NaOH. Add the information underneath to help keep track of what is going on and what where calculations need to be made. It works especially great for things like titrations involving samples and dilutions. My teacher swears by this and always did it when he worked in the field.

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: Chemistry Question Thread
« Reply #503 on: July 26, 2016, 05:30:43 pm »
0
@RuiAce a diagram that shows the steps of the process. It helps visualize what is going on. for example, you draw the phosphorus pentoxide being weighed out, being added to water, the solution being reacted with NaOH. Add the information underneath to help keep track of what is going on and what where calculations need to be made. It works especially great for things like titrations involving samples and dilutions. My teacher swears by this and always did it when he worked in the field.
A flow chart?

MysteryMarker

  • Forum Regular
  • **
  • Posts: 82
  • Respect: 0
Re: Chemistry Question Thread
« Reply #504 on: July 26, 2016, 05:35:04 pm »
0
Industrial Chemistry Question:

During the production of soap in a school laboratory, why is saturated salt solution added to aid in the precipitation of soap from the mixture?

Thanks guys.

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: Chemistry Question Thread
« Reply #505 on: July 26, 2016, 05:46:48 pm »
0
Industrial Chemistry Question:

During the production of soap in a school laboratory, why is saturated salt solution added to aid in the precipitation of soap from the mixture?

Thanks guys.
Haven't done this in ages so I don't remember much at all but I'm pretty sure you do this both in the laboratory and in industrial preparation.

The idea is to manipulate a precipitation procedure known as 'salting out'. The solubility of the soap is forced down because the other ions (Na+ and Cl-), which are more soluble, dissolve in their place. It makes it easier for us to collect the soap by using this method.

leila_ameli

  • Adventurer
  • *
  • Posts: 5
  • Respect: 0
Re: Chemistry Question Thread
« Reply #506 on: July 26, 2016, 08:20:30 pm »
0
Hey!

Essentially, all I would be talking about to answer this question is the reactivity of each alloptrope. As Ozone has a lower bond energy, it will be easier to break/react. Oxygen will be more difficult to break/react. Clearly, Ozone is more reactive than Oxygen. We can explain this by saying that coordinate covalent bonds are easier to break than normal covalent bonds. This explains why Ozone only really forms in higher levels of the atmosphere, where even more reactive Oxygen free radicals are present to react. I think this is as much as an answer would require in the HSC!

Jake


thankyouu!!

liiz

  • Trailblazer
  • *
  • Posts: 25
  • Respect: 0
  • School: Our Lady of the Sacred Heart
  • School Grad Year: 2016
Re: Chemistry Question Thread
« Reply #507 on: July 26, 2016, 08:47:03 pm »
0
Initially we start with 1.42g of phosphorus pentoxide. First we must find the amount of moles of phosphorus pentoxide. Each molecule of phosphorus pentoxide produces two mole of phosphoric acid. The phosphoric acid produced will then react with Sodium hydroxide in a 1:3 ratio, that is you need 3 molecules of NaOH per molecule of phosphoric acid.

Here are the steps
1. Find moles of phosphorus pentoxide, using the formula n=m/M where n=number moles m=grams M=molecular weight
2. Ratio this to moles of Phosphoric acid produced (2:1)
3. Substitute the mole of phosphoric acid into the second equation and ratio to NaOH to find the amount (in mole) of NaOH required. Using the formula n/C=v (rearranged C=n/v), where n=moles of NaOH, C= concentration of the NaOH solution and V is the volume in litres.

It helps to draw a diagram when doing questions like this. Also, write out both equations and work in columns under the substances of interest (i.e. when finding moles of phosphorus pentoxide, work underneath it on the formula.)


Is there anything in particular that you struggle with when it comes to these type of questions?
Ah, awesome I get you. Those steps really helped to guide me through working it out :) And the diagram suggestion is really awesome too!! I've never thought to do diagrams like that before, and as a more visual learner it's something I can definitely incorporate now!! Thanks so much for your help :)) I think I just struggle to simply comprehend what the question is actually asking me and relate it to the formulas I know and what info they've given. I guess just doing heaps of practice, I'll (hopefully) get there haha

liiz

  • Trailblazer
  • *
  • Posts: 25
  • Respect: 0
  • School: Our Lady of the Sacred Heart
  • School Grad Year: 2016
Re: Chemistry Question Thread
« Reply #508 on: July 26, 2016, 08:49:14 pm »
0
Because the equations are given to you, there's one less thing to do. Because normally when you work between reactants/products and other reactants/products you always need a chemical equation.

You also always work in moles. According to n=m/MM

nP2O5 = 1.42/(2*30.97 + 5*16.00) = 0.010004227... mol

Look at the equation and compare the mole ratio. 1 mol of diphosphorus pentoxide produces 2 moles of phosphoric acid

nH3PO4 = 2 nP2O5 = 0.02000845427... mol

So we now know the moles of phosphoric acid we must neutralise.

Look at the second equation for the neutralisation. Clearly three moles of NaOH is required to neutralise one mole of H3PO4

nNaOH = 3 nH3PO4 = 0.06002536282... mol

We are given the concentration of NaOH. Because we know that C=n/V

V = n/C

VNaOH =  0.06002536282... / 0.30
=  0.06002536282... L

The lowest amount of significant figures used in this question is three (from the mass of diphosphorus pentoxide. So rounding to 3 s.f.

V ≈ 6.0 * 10-2 L

Key strategy: Break the question apart and do not try to do multiple things at once. Chemistry calculations are always one-way.
May I ask what diagram?
Thanks Ruiace for your input too!! I was able to do it from Sweetcheeks and then check my working with yours which was really helpful!! Awesome tip as well to not do multiple things at once - will stick by it a bit more now!! Thanks again :))

MysteryMarker

  • Forum Regular
  • **
  • Posts: 82
  • Respect: 0
Re: Chemistry Question Thread
« Reply #509 on: July 27, 2016, 04:22:56 pm »
0
Assess the significance of the industrial development of the Haber process to society at the beginning on the 1900s. Include a relevant chemical equation in your answer. 6 marks.

Hey guys, just wondering what would the recommended structure be for this/ some main points i would need to discuss to get the 6 marks.

Thanks