Chemistry: Post your doubts here!

[Mstudent]

Could you guys help me here?

 

[darks]

Could you guys help me here?

I asked this q. a few pages back.
Metallic9896 answered it as follows...

As zellyman said, P is same in both so in first situation:

P = nRT/V

You don't know n but you do know that n = mass/Mr and you have mass so substitute that in, convert T from C to K, convert V from 0.025 to 0.000025 if you want and R is 8.31 as you know, but keep V in dm^3 anyway cause units will cancel out as you'll be able to see if you're good at maths.

P = mass x R x T/V x Mr
P = (0.1 x 8.31 x 373)/(0.025 x Mr), no need to go any further as the final answer isn't a full answer either. It's enough to see that

P = 0.1 x 8.31 x 373/0.025 x Mr

In second situation:

P = nRT/V

You have n (1), rest as usual.

P = 1 x 8.31 x 273/22.4, again, enough to see that P = 8.31 x 273/22.4

Equate both P and Mr subject of formula:

(8.31 x 273)/(22.4) = (0.1 x 8.31 x 373)/(0.025 x Mr)

0.025 x Mr = 0.1 x 8.31 x 373 x 22.4/8.31 x 273

8.31 cancel out

0.025 x Mr = 0.1 x 373 x 22.4/273

Mr = 0.1 x 373 x 22.4/273 x 0.025

D is the answer.

 

[Mstudent]

I asked this q. a few pages back.
Metallic9896 answered it as follows...

As zellyman said, P is same in both so in first situation:

P = nRT/V

You don't know n but you do know that n = mass/Mr and you have mass so substitute that in, convert T from C to K, convert V from 0.025 to 0.000025 if you want and R is 8.31 as you know, but keep V in dm^3 anyway cause units will cancel out as you'll be able to see if you're good at maths.

P = mass x R x T/V x Mr
P = (0.1 x 8.31 x 373)/(0.025 x Mr), no need to go any further as the final answer isn't a full answer either. It's enough to see that

P = 0.1 x 8.31 x 373/0.025 x Mr

In second situation:

P = nRT/V

You have n (1), rest as usual.

P = 1 x 8.31 x 273/22.4, again, enough to see that P = 8.31 x 273/22.4

Equate both P and Mr subject of formula:

(8.31 x 273)/(22.4) = (0.1 x 8.31 x 373)/(0.025 x Mr)

0.025 x Mr = 0.1 x 8.31 x 373 x 22.4/8.31 x 273

8.31 cancel out

0.025 x Mr = 0.1 x 373 x 22.4/273

Mr = 0.1 x 373 x 22.4/273 x 0.025

D is the answer.

Thank you v. much Darks!

 

[mastermind13]

Around 33..34

What was the hardest question

 

[mastermind13]

How was P1???????

Which topics came the most

 

[selrey]

What do you guys think is the expected grade for Chem p1? Paper was easy but i made some silly mistakes sigh

 

[Mstudent]

Some of the guys had to leave out questions. So you guys had a hard bio paper and we had alhamdulilah rather easy and you guys had an easy chem paper and we had a hard one!

 

[chichibung]

Which paper did you do, I did 11 and I personally found it easy, but my friends don't think so...

 

[qwertypoiu]

this question: October/November 2014 variant 12 question number 15.
the equation: K2O + H2SO4 ---> K2SO4 + H2O i don't think this is correct as the question said that K2O is dissolved in water.
so i think these 2 equations may be used to solve the question:
1) K2O + H2O ---> 2KOH
2) 2KOH + H2SO4 ---> K2SO4 + 2H2O.
but i can't seem to find the answer. However i found this on yahoo answers:

(2 mol/dm^3 H2SO4) x (0.015 dm^3) x (2 mol KOH / 1 mol H2SO4) x (1 mol K2O / 2 mol KOH) x
(250 cm^3 / 25 cm^3) x (94.1960 g K2O/mol) = 28 g K2O

can anyone explain this to me on plain paper and/or step by step?

Hi, good question.

Firstly, you should understand that in most cases, "dissolving" of a substance does not imply a reaction. I do not say this is the case always, as a chemical may be able to react with water to dissolve. However, I find no evidence that K2O would react with water in the manner described by you. Perhaps you are right, and you can refer me to a source that shows that a group 1 oxide dissolves in water in such a manner?

Secondly, it should be noted that even if your reactions were correct, the final answer remains the same. This is because the stoichiometric ratio of H2SO4 to K2O is still 1:1. Thus the method of finding the number of moles of K2O in 25cm^3 of solution using the ratio with number of moles in H2SO4 still yields the same answer.

 

[shahzaib ihsan]

Hi, good question.

Firstly, you should understand that in most cases, "dissolving" of a substance does not imply a reaction. I do not say this is the case always, as a chemical may be able to react with water to dissolve. However, I find no evidence that K2O would react with water in the manner described by you. Perhaps you are right, and you can refer me to a source that shows that a group 1 oxide dissolves in water in such a manner?

Secondly, it should be noted that even if your reactions were correct, the final answer remains the same. This is because the stoichiometric ratio of H2SO4 to K2O is still 1:1. Thus the method of finding the number of moles of K2O in 25cm^3 of solution using the ratio with number of moles in H2SO4 still yields the same answer.

i have seen it on the internet :

&
https://answers.yahoo.com/question/index?qid=20120317205915AAiIWOm

and yes you are right the answers are same both way

 

[Wei Ling]

Can anyone please solve this ? Much appreciated

 

[Slayer10199]

Will CIE accept dot and cross diagrams if I drew them without the circles? But I did put in crosses and dots.

 

[Metanoia]

Can anyone please solve this ? Much appreciated

Do include the year of the paper and answer next time for easier reference

 

[Wei Ling]

Do include the year of the paper and answer next time for easier reference

View attachment 62563

Noted
Wow that was really helpful , Thanks !

 

[Wei Ling]

(GCE AS/A LEVEL) MAY JUNE 2011 PAPER 23 QUESTION 2C

For this question , is it correct to say that Nitrogen has stronger Van Der Waals forces compared to Neon because the size of Nitrogen atom is bigger than Neon atom ( Size of atom decreases across the period ) .

(I assume that there was a typo in the mark scheme and the word "argon" should've been replaced by Neon)

Thanks in advance !

 

[Metallic9896]

(GCE AS/A LEVEL) MAY JUNE 2011 PAPER 23 QUESTION 2C

For this question , is it correct to say that Nitrogen has stronger Van Der Waals forces compared to Neon because the size of Nitrogen atom is bigger than Neon atom ( Size of atom decreases across the period ) .

(I assume that there was a typo in the mark scheme and the word "argon" should've been replaced by Neon)

Thanks in advance !

Yeah youre right about the typo, it seems that way. However, the reason isn't the size. Van der Waals' depend on the number of electrons per unit forming them. Nitrogen exists as a diatomic gas so even though it seems that neon should have greater Van der Waals' based on what I said, because it has more electrons than a nitrogen atom, the opposite is true because in nitrogen the Van der Waals' exist between MOLECULES each containing TWO nitrogen atoms while Van der Waals' in Neon exist simply between atoms of neon. Thus, as a diatomic molecule of nitrogen has more electrons than monoatomic neon, it will also have greater Van der Waals' d thus lesser ideal behavior because ideal requires as less Van dee Waals' as possible which is in neon in this question.

 

[Mstudent]

Will CIE accept dot and cross diagrams if I drew them without the circles? But I did put in crosses and dots.

Yeah they should

 

[Wei Ling]

Yeah youre rught about the typo, it seems that way. However, the reason isn't the size. Van der Waals' depend on the number of electrons per unit forming them. Nitrogen exists as a diatomic gas so even though it seems that neon should have greater Van der Waals' based on what I said, because it has more electrons than a nitrogen atom, the opposite is true because in nitrogen the Van der Waals' exist between MOLECULES each containing TWO nitrogen atoms while Van der Waals' in Neon exist simply between atoms of neon. Thus, as a diatomic molecule of nitrogen has more electrons than monoatomic neon, it will also have greater Van der Waals' d thus lesser ideal behavior because ideal requires as less Van dee Waals' as possible which is in neon in this question.

Thanks a lot !

 

[Wei Ling]

Cambridge International AS/A Level – May/June 2015 paper 21 (QUESTION 1C iii)

My initial answer was (TeCl4 + 4H2O --> Te(OH)4 + 4 HCL)
Can someone please explain to me why the reaction doesn't produce hydroxide as a product? Thanks in advance !

 

[Metanoia]

Cambridge International AS/A Level – May/June 2015 paper 21 (QUESTION 1C iii)

My initial answer was (TeCl4 + 4H2O --> Te(OH)4 + 4 HCL)
Can someone please explain to me why the reaction doesn't produce hydroxide as a product? Thanks in advance !

When the element used in a question is unfamiliar, we try to use known equations as a comparison.
We know that SCl4 + 2H2O --> SO2 + 4HCl, since Te and S are in the same group, we can expect TeCl4 to follow the reaction

Edit: Typed Si instead of S