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Chemistry: Post your doubts here!

Mstudent

Mstudent

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darks said:
For Kc do not consider solid and you should get correct answer!
Good Luck! It was easy for us!
Click to expand...
But the other substances are gases, so could you plz write the steps for me cuz I am VERY confused, and very tense about my exams 3 hours later!
 
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darks

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Mstudent said:
But the other substances are gases, so could you plz write the steps for me cuz I am VERY confused, and very tense about my exams 3 hours later!
Click to expand...
Sure.
I think you have reached moles at equilibrum for NH3 and HCL to be 0.7.
so Kc=[NH3][HCL]
vol. is given.
Kc=[0.7/0.5][0.7/0.5]=C
 
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darks

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Mstudent said:
Could you guys help me here?
Click to expand...
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

Mstudent

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darks said:
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.
Click to expand...
Thank you v. much Darks!
 
selrey

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

Mstudent

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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!
 
qwertypoiu

qwertypoiu

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shahzaib ihsan said:
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?
Click to expand...
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

shahzaib ihsan

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qwertypoiu said:
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.
Click to expand...
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
 
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