Prelim Chem Thread (2 Viewers)

[Nailgun]

true but knowing the syllabus fully is the key, at least thats waht worked for me

you're in yr11 lol

 

[strawberrye]

Eyeseeyou, the main problem is not that you are asking so many questions, which arguably can assist in your understanding of chemistry, the problem rather is you expect every single of these questions to be answered by us. That is highly unrealistic-and there is one skill that every state ranker-or at least definitely those who do well in any subject need to develop-and that is, unsurprisingly, critical thinking skills. You need to learn to think for yourself and find ways of answers to the questions-i.e. via google and all the other resources you have available to you-otherwise how else will you develop research skills for science subjects or indeed any subjects?-because you are likely to get research assessments, and with us helping you so much with the 'why' questions, you will never be able to research competently for these assessments in the future.

To other users-please don't answer all the questions because you are just fuelling lack of critical thinking-a lot of these-if not every single question-can be discovered by some simple googling. Given you (eyeseeyou) have personally admitted they are not going to be examined, then technically spamming threads with literally 50, 60 questions is going to mislead other users into thinking these are important and going to be examined. With us helping you to balance every single equation and answering every small question, then you are going to lack the basic skills to excel in chemistry. And if you are so curious, I highly encourage you to actually ask your teachers these questions or investigate it yourself rather than burdening everyone else with excessive amounts of questions. I am sure other people would greatly appreciate you being more selective in your questions.

If you don't want other people to mistake you being a troll-be more selective in your questions-when you ask the 'WHY' questions, you need to ask the right ones for it to be beneficial. And small additional tip: if you are struggling to balance basic equations at this stage in year 11, then you are in some really serious trouble (because it is a very basic assumed skill by now-or else you really are trolling all of us)-either way-a piece of advice for you to think about for future post and I wish you all the best for your senior studies.

 

[Zoinked]

Well put ^

 

[InteGrand]

how do you solve this. I don't get the question.

View attachment 33287

The answer is just one-millionth of 2000 g, which is 2 mg.

 

[strawberrye]

Can I've some help with this question?

Ethane (C2H6) burns in oxygen to form carbon dioxide (CO2) and water. How many moles of oxygen are needed to react with one mole of ethane? How many moles of: i) CO2 ii) H20 are produced?

Write balanced equation, and you basically will get your answer, i.e. if it is 2 moles of ethane reacting with 3 moles of oxygen, then you divide both by 2, to get 1 mole of ethane reacting with 1.5 moles of oxygen, and the other parts of the question is similar approach.

 

[writeful]

The molar heat of solution of sodium hydroxide is -44.5 kJ.mol^-1. Paula has 200ml of water. Assuming that all the heat produced by the dissolution of the sodium hydroxide heats the water, what mass of sodium hydroxide does she need to add to increase the temperature by 6 degrees centigrade?

^Can I've some help with this question please? c:

molar heat of solution = -44.5kJ/mol = -44500J/mol
change in t = 6 degrees
mass(water) = 200g
specific heat capacity c = 4.18 J/degrees/g

=> Finding the enthalpy absorbed by the water

Q=mc delta t
= 200g x (4.18 J/degrees/g) x 6 degrees
Q = 5016J

=> Then using molar heat of solution=-Q/n, and re-arrange to make n the subject

n= -5016J/(-44500J/mol) = 0.1127....mol

=> converting this into mass

m(NaOH) = nM
= 0.1127...mol x (22.99 + 16.00 + 1.008)g/mol

=> Thus m(NaOH) = 4.51g (to 3 sig.fig.)

I hope I didn't make a mistake.

 

[cottonboy99]

Hi guys,

Can I please have some help in regards to setting up the following practicals:

a). Separate CuSO4 from Solution A
b). Electrolysis of Cu from Solution B

Also can I please have some help in suggestions for writing up the scientific method on one of these.

Thank you all.

 

[Zoinked]

What is solution A and what is solution B ?

 

[cottonboy99]

I don't know re: Solution A & B, They could be anything as I will be assigned them on the day.
I just need to know the basics of setting them up. Once they are set up correctly, I will be given theoretical sample which will be weighed and then to use data to calculate the molarity of Solution A and Solution B, before finally choosing one of these pracs and writing it up using the scientific method.

 

[Zoinked]

There are multiple methods of separation, etc filtration, centrifuge, distillation. I can't really help you unless I know what method you will be using.

Check this website out to find out about a few different types of separation. http://amrita.olabs.edu.in/?sub=73&brch=2&sim=96&cnt=1

 

[Snowflek]

Desperate help please.
A 100ml solution of 0.85mol/L lead nitrate is mixed with a 65ml solution of 0.5mol/L aluminium sulfate. Calculate the mass of lead sulfate you would expect to form
Pb(NO3)2 + Al2(sO4)3 ---> PbSO4 + Al(NO3)3
3 1 3 2
Thats my mole ratio balanced with the numbers underneath it (correct me if im wrong)
So first i tried finding out the moles of Pb(NO3)2 which is (correct me if im wrong):
0.85=n/0.1
0.1 x 0.85 = 0.085
Then i tried finding out the moles of Al2(SO4)3 which is (correct me if im wrong):
0.5=n/0.065
0.5 x 0.065= 0.0325
Now this is where im stuck, please help D:

 

[HeroicPandas]

Desperate help please.
A 100ml solution of 0.85mol/L lead nitrate is mixed with a 65ml solution of 0.5mol/L aluminium sulfate. Calculate the mass of lead sulfate you would expect to form
Pb(NO3)2 + Al2(sO4)3 ---> PbSO4 + Al(NO3)3
3 1 3 2
Thats my mole ratio balanced with the numbers underneath it (correct me if im wrong)
So first i tried finding out the moles of Pb(NO3)2 which is (correct me if im wrong):
0.85=n/0.1
0.1 x 0.85 = 0.085
Then i tried finding out the moles of Al2(SO4)3 which is (correct me if im wrong):
0.5=n/0.065
0.5 x 0.065= 0.0325
Now this is where im stuck, please help D:

Hint: find the limiting reagent.

 

[Snowflek]

limiting reagent is 0.085? since it requires 3 moles of that to make 1 mole of also4

 

[HeroicPandas]

limiting reagent is 0.085? since it requires 3 moles of that to make 1 mole of also4

Yeh, the limiting reagent is Pb(NO3)2. Now what do you do find the mass of PbSO4?

 

[Snowflek]

Actually one moment

 

[HeroicPandas]

0.085 x ((207.2 x6) + 32.07 + (16 x 4))?? Is that correct?

Yep

Nice work!

wait, don't times the molar mass of lead by 6

 

[Snowflek]

to find pbs04 first find the moles for it which is
0.0325( correct me if im wrong)
Then 0.0325 x molecular weight of pbso4 right?

 

[HeroicPandas]

to find pbs04 first find the moles for it which is
0.0325( correct me if im wrong)
Then 0.0325 x molecular weight of pbso4 right?

Why is the moles of PbSO4 = 0.0325?

The rest is fine, well the idea not the numbers.

 

[Snowflek]

because we have the c=n/v we have c and v where c is 0.5 and v is 0.065 and if you times them together you get 0.0325

 

[Snowflek]

Acutlally sorry no wait sorry mb