HSC Chemistry Question Thread

[RuiAce]

I say that sucrose from sugar cane is hydrolysed to form glucose, then glucose ia fermented to get ethanol. Where does the cellulose come in?

Now that's going beyond the syllabus.

To briefly summarise it though (for sake of completeness), sucrose is the DIMER formed in condensation polymerisation. Two beta-glucose monomers form this sucrose molecules, and it's them which 'polymerise' into cellulose. So what actually happens is that you have your cellulose, which you break down into sucrose first, which then gets further broken down into glucose.

[wesadora]

So I was looking through my trial paper:
"A solution was made by mixing sodium acetate solid and a solution of acetic acid. Use an equation to predict and explain the impact on the pH of the solution of adding a very small volume of hydrochloric acid to this mixture".

I said: The impact of the mixture is negligible due to the buffer solution of CH3COOH and CH3COONa, as the CH3COO- ion is a weak conjugate base of the weak acid CH3COONa (as per required for a buffer solution). When HCl is added: HCl(aq) + CH3COONa(aq) --> NaCl (aq) + CH3COOH(aq), it is neutralised. pH change will thus be resisted, creating a weaker acid and neutral salt.

I got 1/2...is this wrong/how could I have gotten full marks? There was a cross on top of my equation - was this where I went wrong?

Thanks! 😊

[RuiAce]

So I was looking through my trial paper:
"A solution was made by mixing sodium acetate solid and a solution of acetic acid. Use an equation to predict and explain the impact on the pH of the solution of adding a very small volume of hydrochloric acid to this mixture".

I said: The impact of the mixture is negligible due to the buffer solution of CH3COOH and CH3COONa, as the CH3COO- ion is a weak conjugate base of the weak acid CH3COONa (as per required for a buffer solution). When HCl is added: HCl(aq) + CH3COONa(aq) --> NaCl (aq) + CH3COOH(aq), it is neutralised. pH change will thus be resisted, creating a weaker acid and neutral salt.

I got 1/2...is this wrong/how could I have gotten full marks? There was a cross on top of my equation - was this where I went wrong?

Thanks! 😊

1. It is not negligible, because something small happens and it gets recovered but something still happens that's not minuscule.
2. You did not properly relate to Le Chatelier's principle
3. You did not write the equation of the buffer system (identification that there was a buffer was probably the mark)

When the buffer system was formed, the following equilibrium was established (noting that the Na⁺ is a neutral, spectator cation.
CH₃COO^- + H₃O⁺ <-> CH₃COOH_(aq) + H₂O_(l)
Because the solution was made by combining the substances, both of which are a weak base/acid respectively, the above buffer system is formed. The buffer will resist small changes in pH, which you mentioned but not explained in the correct way.

When a small quantity of HCl is added, we introduce some H₃O⁺ into the system. So by LCP, the equilibrium will shift to the right, producing more acetic acid and water.

Now the purpose of a buffer to resist pH changes is like so. There are two things to keep in mind.
1. By shifting the equilibrium to the right, we have eliminated some of the strong HCl and introduced more weak CH₃COOH.
2. The fact that there is both CH₃COO^- AND CH₃COOH imply that the pH of the mixture is quite stable. There's both acidic and basic substances in there, and hence the concentrations of acidic and basic substances are pretty stable, implying the pH doesn't change so easily.

In adding the strong acid, note that by LCP it just gets converted to a weak acid, which is a part of the buffer. Hence the pH will increase, however by a smaller amount than if the buffer was not there.

You can determine where you lost the marks and what you would've needed to get them. Obviously since it was 2 marks not AS much depth is required here, but you can determine where the mark lost was.

[WLalex]

Im not getting any of these answers...

[wesadora]

I see....I think it was because I didn't have the equation for buffer tbh. Also, I'll be more careful with my wording next time. Maybe "significant pH change is resisted" would do better. Thanks!

[RuiAce]

Im not getting any of these answers...

If citric acid is triprotic then 3 moles of NaOH will react with 1 mol of citric acid.

So n_{C3H4OH(COOH)3} = 1/3 * n_NaOH = 1/3 *CV = 3 * 0.550 * 0.02950 = 5.408...*10^-3 mol

Converting the moles to mass using m=nMM we have m = 5.408...*10^-3 *192.12 = 1.039049 g

Then just divide by 0.025

Edit: Cheers for picking the mistake up below me

[wesadora]

Im not getting any of these answers...

make sense?

[wesadora]

If citric acid is triprotic then 3 moles of NaOH will react with 1 mol of citric acid.

So n_{C3H4OH(COOH)3} = 3 * n_NaOH = 3 *CV = 3 * 0.550 * 0.02950 = 0.048675 mol

Converting the moles to mass using m=nMM we have m = 0.048675*192.12 = 9.351441 g

Then just divide by 0.025

sorry Rui...aren't you meant to divide by 3, not times by 3? 3 moles of NaOH will react with Citric. Therefore number of moles of citric will be 1/3 of NaOH.
edit: lol that's ok

[massive]

You lost me on this one. Example please

Like you know how one of the condition for an acidic oxide is that it neutralises bases to form salt and water. How can you show that using an equation i.e. what's the general equation?

[RuiAce]

Like you know how one of the condition for an acidic oxide is that it neutralises bases to form salt and water. How can you show that using an equation i.e. what's the general equation?

There isn't a "general equation" per se. Because take aluminium oxide, which is amphoteric. If you look at how it reacts with an acid that's pretty straight forward but it reacts in the most peculiar ways possible with an acid. (Or other way around, I don't remember. This is mostly independent research)

Pretty sure Jake would have some. He literally made a whole list of chemical equations for you to use

Otherwise I'm certain that the Jacaranda textbook has a few. (But yeah, they're all examples. There is no "general" rule of thumb here)

[RuiAce]

hey, would anyone happen to know when PLA was first introduced as a plastic and statistics of how commonly it is used?

Not required.as per the syllabus.

use available evidence to gather and
present data from secondary sources
and analyse progress in the recent
development and use of a named
biopolymer. This analysis should
name the specific enzyme(s) used
or organism used to synthesise the
material and an evaluation of the
use or potential use of the polymer
produced related to its properties

[Furrrball]

I don't understand why the answer is d..

[RuiAce]

I don't understand why the answer is d..(Image removed from quote.)

A is just wrong because those aren't even -OH groups, they're just hydrogens. That's not going to be a site for polymerisation

B is wrong because that double bond shouldn't be there. Looking at the polymer clearly we only have single bonds there and two hydrogens leaning off each carbon.

C has the same reason as A

So in a way D is correct by process of elimination. You can also check D is correct by trying to join  the two monomers to form the dimer. The fact that -OH groups are hanging off each side guarantee that we have a site for condensation polymerisation (recall: a small molecule such as water falls off during the polymerisation process). The monomers are right - the carbons off the alkane have no double bond and the carbons hanging off the benzene ring are indeed bonded with the oxygens

[sherlockian]

Does anyone know the vanadium redox cell cathode reaction? I just realised that I had 2e- on the LHS (which is what it says in my textbook) but all other sources are stating there is only one electron on the LHS. Thanks!

[RuiAce]

Does anyone know the vanadium redox cell cathode reaction? I just realised that I had 2e- on the LHS (which is what it says in my textbook) but all other sources are stating there is only one electron on the LHS. Thanks!

Everything says 1.

Check your charges. Do they balance out? If not, then you know where the mistake is.