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July 02, 2020, 03:45:29 pm

Author Topic: HSC Chemistry Question Thread  (Read 595649 times)  Share 

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classof2019

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Re: Chemistry Question Thread
« Reply #3780 on: October 30, 2019, 08:48:47 am »
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Hi there!

I did the question (working below) and got the answer of pH = 11.1

I'm not 100% sure if this is correct so if someone could check it and let me (and classof2019) know, that would be great!

Thanks!
 :)
(ps. Sorry if the image quality is bad!)

Hey,

I really appreciate your help, however, I've just realised that the correct answer is D. 11.6. I'm not sure how they would have gotten this.

I've attached my working (which still doesn't align with their answer) in the spoiler - let me know what you think.

Spoiler

25.0mL acetic acid has pH - 1.30, therefore [H+]=10-1.30

So n(H+) = 0.025 x 10-1.30

20mL Ca(OH)2 --> n(Ca(OH)2) = 0.130 x 0.020 = 0.0026

So n(OH-) = 0.0026 X 2 = 0.0052 moles

H+(aq) + OH-(aq)--> H2O(l)

So n(OH-) excess = 0.0052 - (0.0025 x 10-1.30) = 0.00394703191 ...

So [OH-] = 0.00394703191 ... / 0.045 = 0.08771182035

Therefore, pOH = -log (0.08771182035...) = 1.056941876 ...

So pH = 12.94

I think this may be problematic in the sense that it doesn't account for any further dissociation of acetic acid (which decreases pH) - but I don't really know any alternative.


Thanks.


classof2019

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Re: Chemistry Question Thread
« Reply #3781 on: October 30, 2019, 08:53:11 am »
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Hi, I might be wrong. There are some point in your solution that I don't understand.
Since the ratio between calcium hydroxide and acetate acid is 1:2, 1.25*10-3 mols of CH3COOH should react with 6.25*10-4 mols of Ca(OH)2. So,Ca(OH)2 left would be 1.975*10-3 mols.
And I think we should do 1.975*10-3 / volume which is 4.5*10-2 to get concentration of OH- first and substitute it into -log formula.
But I still did not get the answer though...

Hm, I think if you multiply the [Ca(OH)2] you found by 2 to find [OH-], you eventually end up with a result of 12.94.

Perhaps the question itself has an error.

worldno1

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Re: Chemistry Question Thread
« Reply #3782 on: October 30, 2019, 02:32:43 pm »
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Hm, I think if you multiply the [Ca(OH)2] you found by 2 to find [OH-], you eventually end up with a result of 12.94.

Perhaps the question itself has an error.
hi, i just did the question independently (without looking at your solution) and i landed on the same answer of 12.94. i'm pretty sure that is the correct answer! :) not sure how 11.6 came about.

classof2019

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Re: Chemistry Question Thread
« Reply #3783 on: October 30, 2019, 04:41:59 pm »
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hi, i just did the question independently (without looking at your solution) and i landed on the same answer of 12.94. i'm pretty sure that is the correct answer! :) not sure how 11.6 came about.

Awesome, thanks for that!

Just another q (for anyone to answer) - how many carbon environments does hexane have? Because I always thought it was 2 but a spectrum I saw had three peaks.

studyingg

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Re: Chemistry Question Thread
« Reply #3784 on: October 30, 2019, 04:58:40 pm »
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Awesome, thanks for that!

Just another q (for anyone to answer) - how many carbon environments does hexane have? Because I always thought it was 2 but a spectrum I saw had three peaks.

I think It would be 3, if you draw a hexane structure and rule a line of symmetry there are 3 carbons on each side of the line, hence, 3 C environments I believe. 

InnererSchweinehund

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Re: Chemistry Question Thread
« Reply #3785 on: October 30, 2019, 08:40:09 pm »
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Hi, I might be wrong. There are some point in your solution that I don't understand.
Since the ratio between calcium hydroxide and acetate acid is 1:2, 1.25*10-3 mols of CH3COOH should react with 6.25*10-4 mols of Ca(OH)2. So,Ca(OH)2 left would be 1.975*10-3 mols.
And I think we should do 1.975*10-3 / volume which is 4.5*10-2 to get concentration of OH- first and substitute it into -log formula.
But I still did not get the answer though...

Hi!

I checked with a chemistry teacher and she seemed to think the way I did it would be correct, if the answer I got was correct.
She also did the question and couldn't get pH 11.6

Not really sure what's going on...
 :o
« Last Edit: October 30, 2019, 08:43:11 pm by InnererSchweinehund »

worldno1

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Re: Chemistry Question Thread
« Reply #3786 on: October 30, 2019, 11:07:44 pm »
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Hi!

I checked with a chemistry teacher and she seemed to think the way I did it would be correct, if the answer I got was correct.
She also did the question and couldn't get pH 11.6

Not really sure what's going on...
 :o

hey there! when you were calculating pOH, you didn't convert the number of moles of Ca2+ to concentration (by dividing by 0.045 L), since pOH = log [OH-]. additionally, the number of moles of OH- is twice the number of moles of Ca(OH)2, so you need to multiply by 2. :)

Hawraa

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Re: Chemistry Question Thread
« Reply #3787 on: November 02, 2019, 12:08:56 pm »
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Hi everyone,
With the haber process equilibrium graph:
N2+3H2<>2NH3
If the reaction vessel is doubled          (pressure decreased), isn't like we show that as a sharp decrease in the concentration of all species (reactants and products) and then there will be a decrease in the product curve and increase in the reactants curve?
Also would this be different if we graph it as (moles vs. time) or (concentration vs. time) graphs? Thanks.

fun_jirachi

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Re: Chemistry Question Thread
« Reply #3788 on: November 02, 2019, 01:29:53 pm »
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Hi there!

You're 100% correct! If you're given a graph of concentration/time with an actual scale though, don't forget the sharp decrease ends when each substance has its concentration halved ie. if there were 6 mol/L of nitrogen, after the halved pressure, the sharp drop would drop the concentration to 3 mol/L. Afterwards, there would be a shift towards the left side which has more moles of gas, by LCP as you say :) Make sure the curve shows proportional increases/decreases as well ie. in a 1:3:2 ratio.

The above is for concentration versus time. When we change the pressure or volume, the number of moles of gas shouldn't change; consider the equation PV=nRT - by increasing/decreasing pressure, we increase/decrease volume by the inverse of that factor, and vice versa ie. nRT remains constant (all this means is that provided everything is at a constant temperature, changes in volume or pressure have no effect on the no. of moles of gas!). However, after equilibrium shifts, you should note a change in the number of moles, as the system shifts one way or another to adjust to the change in volume or pressure :)

Hope this helps :)
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worldno1

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Re: Chemistry Question Thread
« Reply #3789 on: November 02, 2019, 06:15:56 pm »
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Hi there!

You're 100% correct! If you're given a graph of concentration/time with an actual scale though, don't forget the sharp decrease ends when each substance has its concentration halved ie. if there were 6 mol/L of nitrogen, after the halved pressure, the sharp drop would drop the concentration to 3 mol/L. Afterwards, there would be a shift towards the left side which has more moles of gas, by LCP as you say :) Make sure the curve shows proportional increases/decreases as well ie. in a 1:3:2 ratio.

The above is for concentration versus time. When we change the pressure or volume, the number of moles of gas shouldn't change; consider the equation PV=nRT - by increasing/decreasing pressure, we increase/decrease volume by the inverse of that factor, and vice versa ie. nRT remains constant (all this means is that provided everything is at a constant temperature, changes in volume or pressure have no effect on the no. of moles of gas!). However, after equilibrium shifts, you should note a change in the number of moles, as the system shifts one way or another to adjust to the change in volume or pressure :)

Hope this helps :)

so in other words, when you're graphing moles vs time, there'd be no immediate drop (or sharp decrease as you said) but afterwards, there'd be a gradual increase/decrease of moles in whichever direction equilibrium moves?

RACHEL1111

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Re: Chemistry Question Thread
« Reply #3790 on: November 03, 2019, 11:12:14 am »
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Hi, all!
I have seen this definition of oxidation reaction: ' Oxidation of alcohols will create a double bond from the carbon that the alcohol was originally on to another carbon molecule by taking away two hydrogen atoms- 1 per carbon'. However, it contradicts to my understanding of oxidation reaction. In my drawing, when the alcohol is oxidised, two circled hydrogen from the same carbon, rather than one per carbon,  are removed to form hydrogen gas.
So, which one is correct? Any help is appreciated!

Hawraa

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Re: Chemistry Question Thread
« Reply #3791 on: November 03, 2019, 03:52:30 pm »
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Hi everyone,
Is ammonium a strong or weak acid?  So when we write its ionisation equation to produce hydrogen ion and ammonia (conjugate base) do we use the normal arrow or the equilibrium arrow?
The same thing with the floride ion and the hydrofluoric acid, is it strong or weak and which arrow do we use?
Thanks.

r1ckworthy

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Re: Chemistry Question Thread
« Reply #3792 on: November 03, 2019, 04:01:16 pm »
+2
Hi everyone,
Is ammonium a strong or weak acid?  So when we write its ionisation equation to produce hydrogen ion and ammonia (conjugate base) do we use the normal arrow or the equilibrium arrow?
The same thing with the floride ion and the hydrofluoric acid, is it strong or weak and which arrow do we use?
Thanks.

From researching, it seems that while ammonium is a stronger acid than ammonia (which is a weak base), it is not totally strong (does not fully dissociate). Check out this link for a more in-depth explanation.

So I would say, still use the equilibrium arrow when writing these reactions.

Hydrofluoric acid is a weak acid. While it dissociates in water, the hydroxide ions form a strong bond to the dissociated fluoride ion, which limits it's strength. Check out this link for a better explanation.

Hope that helps!
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Hawraa

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Re: Chemistry Question Thread
« Reply #3793 on: November 04, 2019, 01:24:45 pm »
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Hi everyone,
Can someone please explain this question. I thought it would be A but the answer is C.

louisaaa01

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Re: Chemistry Question Thread
« Reply #3794 on: November 04, 2019, 01:48:03 pm »
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Hi everyone,
Can someone please explain this question. I thought it would be A but the answer is C.

Hi Hawraa,

We can rule out option A because nitrous acid will partially dissociate to produce H+ ions and NO2- ions. The presence of H+ ions will reduce pH (noting that pH = -log10[H+], so greater [H+] means lower pH). Option B is ruled out for a similar reason.

Now, to discern between C and D we look at pKa. Since nitrous acid has a higher pKa than chlorous acid, this means that nitrous acid is a weaker acid - so its conjugate base (NO2-) is a stronger base. This means a solution containing NO2- ions has a greater pH than a solution containing the same concentration of ClO2 ions, so C must be the answer. Note also that Na+ ions don't really react with water - it's the anion in this instance that governs pH.

So to sum up, C has a higher pH than A simply because there are no H+ ions in its structure. C has the highest pH since its anion is the strongest base present.
« Last Edit: November 04, 2019, 01:50:22 pm by louisaaa01 »
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