Author Topic: HSC Chemistry Question Thread (Read 1293648 times) Tweet Share

meditatingwalrus · « **Reply #3300 on:** February 08, 2018, 06:24:10 pm »

Quote from: arii on February 07, 2018, 05:57:13 pm

You applied the ratio thing in the wrong part.

Finding the mass of CO_2(g):
n(O₂) = 74.4/24.79 = 3.00121... (So you were correct until here)
n(CO₂) = 2/3 * n(O₂) = 2/3 * 3.00121... = 2.000806... moles
m(CO₂) = (2.000806)(12.01+2*16.00) = 88.0555014...grams.

Therefore, your closest answer would be (C).

oh wow im an idiot

thank you so much - you're such a legend!!

RuiAce · « **Reply #3301 on:** February 08, 2018, 06:33:43 pm »

Quote from: meditatingwalrus on February 08, 2018, 06:20:51 pm

hey! im really struggling with this question - i keep getting a humongous answer? can someone help me out if possible? it means a lot,

1.   Calculate the pH of the solution when 25.0ml of 0.750mol L-1 hydrochloric acid solution is added to 10ml of 0.500mol L-1 barium hydroxide solution.

(A)    0.456
(B)   0.602
(C)   1.862
(D)   2.058

thanks so much

$\text{Using }\boxed{n=CV}\text{ we have}\\ \begin{align*}n_{HCl}&= 0.025\times 0.750 =0.01875\text{ mol}\\ n_{Ba(OH)_2}&= 0.005\text{ mol}\end{align*}$
Be careful to make sure you converted all the millilitre measurements to litres.
$\text{The chemical equation of interest is}\\ \boxed{\ce{2 HCl_{(aq)} + Ba(OH)2_{(aq)} -> BaCl_2_{(aq)} + 2 H_2O(l)}}\\ \text{We see that 2 mol of HCl reacts with 1 mol of Ba(OH)}_2$
$\text{So in our scenario, all }0.005\text{ mol of Ba(OH)}_2\\ \text{will react with 0.010 mol of HCl, and hence}\\ \text{HCl is the excess reagent.}$
__________________________________
$\text{Therefore once the reaction is complete, we have }0.01875 - 0.010 =0.00875\text{ mol of HCl left.}$
$\text{Using }\boxed{C = \frac{n}{V}}\text{, noting the combined volume is now 35 mL,}\\ [HCl] = \frac{0.00875}{0.035} = 0.25\text{ mol L}^{-1}$
$\text{Of course, since HCl fully ionises, this would imply that }[H_3O^+] = 0.25\text{ mol L}^{-1}\\ \text{and therefore }pH = -\log_{10}0.25 \approx 0.602$

meditatingwalrus · « **Reply #3302 on:** February 08, 2018, 06:56:49 pm »

Quote from: RuiAce on February 08, 2018, 06:33:43 pm

$\text{Using }\boxed{n=CV}\text{ we have}\\ \begin{align*}n_{HCl}&= 0.025\times 0.750 =0.01875\text{ mol}\\ n_{Ba(OH)_2}&= 0.005\text{ mol}\end{align*}$
Be careful to make sure you converted all the millilitre measurements to litres.
$\text{The chemical equation of interest is}\\ \boxed{\ce{2 HCl_{(aq)} + Ba(OH)2_{(aq)} -> BaCl_2_{(aq)} + 2 H_2O(l)}}\\ \text{We see that 2 mol of HCl reacts with 1 mol of Ba(OH)}_2$
$\text{So in our scenario, all }0.005\text{ mol of Ba(OH)}_2\\ \text{will react with 0.010 mol of HCl, and hence}\\ \text{HCl is the excess reagent.}$
__________________________________
$\text{Therefore once the reaction is complete, we have }0.01875 - 0.010 =0.00875\text{ mol of HCl left.}$
$\text{Using }\boxed{C = \frac{n}{V}}\text{, noting the combined volume is now 35 mL,}\\ [HCl] = \frac{0.00875}{0.035} = 0.25\text{ mol L}^{-1}$
$\text{Of course, since HCl fully ionises, this would imply that }[H_3O^+] = 0.25\text{ mol L}^{-1}\\ \text{and therefore }pH = -\log_{10}0.25 \approx 0.602$

YOU . ARE. AMAZING. that was actually the clearest explanation ever dude
thank you so much!

slinkybench · « **Reply #3303 on:** February 10, 2018, 09:30:38 pm »

Hi there,
I'm a preliminary chemistry student and don't understand the Pauli's Exclusion Principle and Hund's Rule.
Can someone please explain?

Thanks so much!

clovvy · « **Reply #3304 on:** February 10, 2018, 09:50:26 pm »

Quote from: slinkybench on February 10, 2018, 09:30:38 pm

Hi there,
I'm a preliminary chemistry student and don't understand the Pauli's Exclusion Principle and Hund's Rule.
Can someone please explain?
Thanks so much!

Personally I never did the new syllabus and your year will be the very first ones to do it... I never studied it in depth but I liked it way more than the ones that I did.... Anyway enough about myself and I'll try my best to explain

Paul's expulsion rule stated that in any atom no two electrons can have all four quantum numbers the same.
Hund's rule describes the electronic configuration of degenerate orbitals in the ground state. The electronic configuration will have the maximum number of unpaired electrons. The ground state will have the maximum orbital angular momentum....

Since I never studied the new prelim, this is the best way I could explain so if it doesn't make sense than ask your teacher

sweetiepi · « **Reply #3305 on:** February 10, 2018, 09:57:35 pm »

Quote from: slinkybench on February 10, 2018, 09:30:38 pm

Hi there,
I'm a preliminary chemistry student and don't understand the Pauli's Exclusion Principle and Hund's Rule.
Can someone please explain?
Thanks so much!

Hey! Sorry if this is fairly brief, as I only barely touched this in uni chem, so bear with me.

Pauli's exclusion rule is where no two electrons in an atom can have the same quantum numbers. As an orbital can contain a maximum of only two electrons, the two electrons must have opposing spins- an up-spin and a down-spin.

(I.e. a p-shell orbital is degenerate- there are actually 3 orbitals of the same energy level that carry two electrons each

)

Hund's rule is where each orbital is occupied by only one electron before filling up with a second electron. (I.e. 2px1 2py1 2pz1 for 3 electrons).

Edit: was beaten but clovvy is also right

arii · « **Reply #3306 on:** February 12, 2018, 11:00:01 pm »

I just want to confirm something to ensure I don't do it wrong...

For a first hand investigation that I'm doing, one of the questions say: "How do you know the information you have collected is relevant?"

Is the word "relevant" the same meaning as the word "valid"? If so, do I simply say everything not tested was kept constant (i.e. just talk about controlled variables)?

clovvy · « **Reply #3307 on:** February 14, 2018, 01:26:31 pm »

Quote from: arii on February 12, 2018, 11:00:01 pm

I just want to confirm something to ensure I don't do it wrong...

For a first hand investigation that I'm doing, one of the questions say: "How do you know the information you have collected is relevant?"

Is the word "relevant" the same meaning as the word "valid"? If so, do I simply say everything not tested was kept constant (i.e. just talk about controlled variables)?

I may be wrong but as far as my understanding goes, the word 'relevant' implies the close connection of the data you gather to the prac. for example, if you want to use n=m/M then speed is irrelevant as it is NOT connected to what you are trying to find? I think validity and relevance are different because with validity you are testing consistency of your results, if say your results always vary then the prac is unreliable therefore it is invalid. Do that make any sense?

arii · « **Reply #3308 on:** February 14, 2018, 08:09:42 pm »

Quote from: clovvy on February 14, 2018, 01:26:31 pm

I may be wrong but as far as my understanding goes, the word 'relevant' implies the close connection of the data you gather to the prac. for example, if you want to use n=m/M then speed is irrelevant as it is NOT connected to what you are trying to find? I think validity and relevance are different because with validity you are testing consistency of your results, if say your results always vary then the prac is unreliable therefore it is invalid. Do that make any sense?

I think validity is whether or not the experiment actually tests the what you intended it to test. Whilst, the consistency of results is reliability. (There's 3 parts schools normally teach: Reliability, Accuracy and Validity).

cocopops201 · « **Reply #3309 on:** February 14, 2018, 09:48:09 pm »

Hello!
I'm looking for chem notes regarding the Ostwald process.
More specifically the reaction conditions required at each step vis-a-vis Le Chatelier's principle.
It would be greatly appreciated!
Thank you for anyone who replies.

Just tryna graduate

arii · « **Reply #3310 on:** February 15, 2018, 12:28:35 am »

Quote from: cocopops201 on February 14, 2018, 09:48:09 pm

Hello!
I'm looking for chem notes regarding the Ostwald process.
More specifically the reaction conditions required at each step vis-a-vis Le Chatelier's principle.
It would be greatly appreciated!
Thank you for anyone who replies.

Just tryna graduate

The internet is a wonderful source of information.

If you just want stuff on the Ostwald process, check out these links:
http://www.worldofchemicals.com/449/chemistry-articles/manufacturing-of-nitric-acid-by-ostwald-process.html
http://www.bbc.co.uk/bitesize/standard/chemistry/plasticsandothermaterials/fertilisers/revision/7/

The first link is a bit more wordy but there is some good info whilst I find the 2nd one to be more condensed and summarised. You should also utilise YouTube if you don't want to read or scan reputable textbooks (i.e. Jacaranda) for what you want.

beeangkah · « **Reply #3311 on:** February 17, 2018, 04:45:28 pm »

Hii, just need help with this question:

"What is the pH of 0.1mol/L of carbonic acid if only 8% ionises?"

So far I've found mole ratio but I'm confused with how to continue with the degree of ionisation part...

H₂CO₃_(aq) --> 2H⁺_(aq) + CO₃^2-_(aq)

Therefore the calculations involve 2[H⁺]...

Thanks!

Also, the answers say 2.1

RuiAce · « **Reply #3312 on:** February 17, 2018, 04:48:37 pm »

Quote from: beeangkah on February 17, 2018, 04:45:28 pm

Hii, just need help with this question:

"What is the pH of 0.1mol/L of carbonic acid if only 8% ionises?"

So far I've found mole ratio but I'm confused with how to continue with the degree of ionisation part...

H₂CO₃_(aq) --> 2H⁺_(aq) + CO₃^2-_(aq)

Therefore the calculations involve 2[H⁺]...

Thanks!

Also, the answers say 2.1

$\text{If the degree of ionisation is only }0.08\\ \text{then since the concentration of carbonic acid is }0.1\\ \text{the concentration of }H_3O^+\text{ will be }0.08\times 0.1 = 0.008$
$\text{Hence }pH = -\log_{10} 0.008 \approx 2.1$

beeangkah · « **Reply #3313 on:** February 17, 2018, 05:02:13 pm »

Quote from: RuiAce on February 17, 2018, 04:48:37 pm

$\text{If the degree of ionisation is only }0.08\\ \text{then since the concentration of carbonic acid is }0.1\\ \text{the concentration of }H_3O^+\text{ will be }0.08\times 0.1 = 0.008$
$\text{Hence }pH = -\log_{10} 0.008 \approx 2.1$

Oh I see, thanks!

So the mole ratio is irrelevant? Or is that only for fully ionised equations?

kaustubh.patel · « **Reply #3314 on:** February 17, 2018, 05:07:24 pm »

Ummmm just to be sure does that mean carbonic acid is a monoprotic acid cause in biology I learnt that as CO2 dissolves in blood it dissolutes into bicarbonate ion and a proton.
And i suppose the mole ratio is still necessary to look at cause it shows only one proton is donated to make hydronium.

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