Chemistry Predictions/Thoughts (1 Viewer)

[Unovan]

marking 2022 hsc rn, that 6 marker titration was chef's kiss it felt so good to finish off

ester question was so free as well didn't even ask for a method just the basics

just did a few headings and collected the marks easy af

 

[SadCeliac]

ester question was so free as well didn't even ask for a method just the basics

just did a few headings and collected the marks easy af

the synthesis question??

 

[Unovan]

the synthesis question??

Yeah but idk what you'd use other than esterification

 

[synthesisFR]

no titrations are FREEEEE
the questions that r not free is this BS bc i have no clue how to do them (the 2021 ammonia question where they ask how much must be added or sum shi. other that that the whole paper was perfect)

 

[synthesisFR]

do we get marks for not writing things on the line
e.g i wanna save space so if they have a chemical eqn in the question with its pka sometimes i write the ka next to the question but not in the working lines. is that fine?

 

[SadCeliac]

do we get marks for not writing things on the line
e.g i wanna save space so if they have a chemical eqn in the question with its pka sometimes i write the ka next to the question but not in the working lines. is that fine?

probably

 

[rockydo4u]

How good of a score is 85 in a peak trial paper?

 

[SadCeliac]

How good of a score is 85 in a peak trial paper?

probs b6 idk

 

[synthesisFR]

Hey guys so im marking my paper



can someone help teach me how to do this, its from the 2021 hsc paper. Idk like when i read the solutions it makes a lot of sense but when i was doing the calc i didn't know what to do cus i was like u need to add nitrogen gas so u do 4.5 + x but then also in equlibrium this amount then decreases bc of LCP so i didn't know where to start? idkk




Also with this one. like it seems easy u just write out ksp of the ppt and then u have the concentration of sulfate and on data sheet u have ksp so u just solve for concentration of silver then from there times by .25L to get moles then get mass so thats what i did but its. not working out?

IGNORE THE SECOND ONE I FEEL STUPID LMFAOO I WAS FINDING MAS SOF SILVER NOT SILVER NITRATE LMFAOO JOKES

 

[Masaken]

Hey guys so im marking my paper


View attachment 41093
can someone help teach me how to do this, its from the 2021 hsc paper. Idk like when i read the solutions it makes a lot of sense but when i was doing the calc i didn't know what to do cus i was like u need to add nitrogen gas so u do 4.5 + x but then also in equlibrium this amount then decreases bc of LCP so i didn't know where to start? idkk



View attachment 41094
Also with this one. like it seems easy u just write out ksp of the ppt and then u have the concentration of sulfate and on data sheet u have ksp so u just solve for concentration of silver then from there times by .25L to get moles then get mass so thats what i did but its. not working out?

IGNORE THE SECOND ONE I FEEL STUPID LMFAOO I WAS FINDING MAS SOF SILVER NOT SILVER NITRATE LMFAOO JOKES

ok so for q31. you're given a bunch of moles and vessel info, so i started by calculating [N2]initial, [H2]initial, [NH3]initial because we have all of that. then we chuck all of that info into an ice table. but one thing to note is that the q is asking you to increase the amount of N2 by a certain amount to increase ammonia by 0.050 moles. so for the initial concentration, you found [N2] = 0.45M. let the number of moles you add be X (don't stress too much that it's moles not concentration, the reaction vessel is a nice clean number anyway you'll see)

Concentration (mol/L)	N2	3H2	2NH3
Initial	0.45 + X	0.1	0.58
Change	-0.0025	- 0.0075	+0.0050 moles (this bit is important, remember to divide 0.050 by 10 cos of rxn vessel)
Equilibrium	0.4475 + X	0.0925	0.585

you know that Keq = 748, and you can make the expression from the given equation. sub in everything and then rearrange to find X. then remember to multiply by 10 to convert it back to moles.
0.4475 + X = (0.585)^2/[(0.0925)^3 x 748]
X = 0.130575... mol/L
n(N2) added to vessel = 0.130575... x 10.0L (n = cV) = 1.30575... = 1.3 mol added (2sf)

 

[carrotsss]

Hey guys so im marking my paper


View attachment 41093
can someone help teach me how to do this, its from the 2021 hsc paper. Idk like when i read the solutions it makes a lot of sense but when i was doing the calc i didn't know what to do cus i was like u need to add nitrogen gas so u do 4.5 + x but then also in equlibrium this amount then decreases bc of LCP so i didn't know where to start? idkk

just finished this paper im not sure how they did it but basically my logic was that we know what the initial concentrations of H2 and NH3 are, and we know the change for H2 and NH3 so basically we know everything for them, whereas for N2 we know the change and what we’re trying to find the initial (or final). Note that to find the final or initial, we don’t need to use the knowledge we have of the 4.5 since that can come in later once we’re finding the difference. From there, I just let the initial for N2 be x + 0.0025 (this just makes it easier to work with) and therefore it’s final conc is x, which you can sub into the Keq formula and solve for x very easy, hence finding the final concentration and then you work backwards from there to find the addition which that indicates. dunno if that makes sense but it worked for me

 

[carrotsss]

ok so for q31. you're given a bunch of moles and vessel info, so i started by calculating [N2]initial, [H2]initial, [NH3]initial because we have all of that. then we chuck all of that info into an ice table. but one thing to note is that the q is asking you to increase the amount of N2 by a certain amount to increase ammonia by 0.050 moles. so for the initial concentration, you found [N2] = 0.45M. let the number of moles you add be X (don't stress too much that it's moles not concentration, the reaction vessel is a nice clean number anyway you'll see)

Concentration (mol/L)	N2	3H2	2NH3
Initial	0.45 + X	0.1	0.58
Change	-0.0025	- 0.0075	+0.0050 moles (this bit is important, remember to divide 0.050 by 10 cos of rxn vessel)
Equilibrium	0.4475 + X	0.0925	0.585

you know that Keq = 748, and you can make the expression from the given equation. sub in everything and then rearrange to find X. then remember to multiply by 10 to convert it back to moles.
0.4475 + X = (0.585)^2/[(0.0925)^3 x 748]
X = 0.130575... mol/L
n(N2) added to vessel = 0.130575... x 10.0L (n = cV) = 1.30575... = 1.3 mol added (2sf)

you really had to post one second before me

 

[Masaken]

you really had to post one second before me

life's a sprint, not a marathon

 

[SadCeliac]

life's a sprint, not a marathon

and I'm usain bolt

because I'm fast

and bla-

 

[SadCeliac]

what's a limitation of BL theory of acids and bases

 

[carrotsss]

what's a limitation of BL theory of acids and bases

bf3

 

[SadCeliac]

bf3

???? wdym

not being able to not explain metal oxides as bases??

 

[Masaken]

what's a limitation of BL theory of acids and bases

can't explain acidic behaviour of some non metal oxides such as SO2

 

[Nacseo]

hope they make a very hard back titration question because if i cant do it no one else can.

 

[synthesisFR]

what's a limitation of BL theory of acids and bases

it can't explain stuff like BF3

hope they make a very hard back titration question because if i cant do it no one else can.

sorry to break it to u but idt hard titrations exist