Chemistry: Post your doubts here!

[princess Anu]

& this too

 

[princess Anu]

For enthalpy change of formation, the arrow must point from the individual constituents to the final product. If the arrow is in the opposite direction, the enthalpy change will have the same magnitude but opposite direction.

yea right, but I'm talking about ENTHALPY CHANGE OF COMBUSTION of H2

 

[Mathemagical]

oh.. I get your point but why not B?

The change from B to A involves atomisation, which takes in energy and therefore is endothermic. Thus, A has a higher energy level than B.

 

[Mathemagical]

yea right, but I'm talking about ENTHALPY CHANGE OF COMBUSTION of H2

That's exactly the same. Recall the definition of combustion and you should realise that the arrow should be pointing toward the product of combustion.

 

[nehaoscar]

Iodine: Covalent and van der waals forces
Silicon dioxide: Giant covalent
Sodium chloride: Ionic
Zinc: Metallic

But don't all molecules have van der waals forces?

 

[princess Anu]

That's exactly the same. Recall the definition of combustion and you should realise that the arrow should be pointing toward the product of combustion.

Oh.. so it shud have been +2* enthalpy change of combustion? am I right?

 

[Mathemagical]

Oh.. so it shud have been +2* enthalpy change of combustion? am I right?

Yes, I suppose so

 

[qwertypoiu]

somebody solve me this pleasee

Ratio is 3:2
3Ti+ + 2V5+ -----> 3Ti3+ + 2V(?)+
Left = 3+2*5=13
Right=3*3+2*?=9+2*?
Left=right
13=9+2*?
?=2

 

[qwertypoiu]

But don't all molecules have van der waals forces?

All MOLECULES. Silicon dioxide, sodium chloride, and zinc have a giant lattice structure. They are NOT molecules.

 

[Mathemagical]

Oh.. so it shud have been +2* enthalpy change of combustion? am I right?

Yes, I suppose so

 

[fantastic girl]

http://onlineexamhelp.com/wp-content/uploads/2014/08/9701_s14_qp_11.pdf
Guys , in Q8 - we are choosing ph 7 becaz the products are basic in nature ?
but this reaction isnt equilibrium related to do the shifting :/

 

[qwertypoiu]

Help please??
View attachment 54318

Initial: b, 0, 0
Final: b-x, x/2, x/2
Total moles = b-x+x/2+x/2= b
Mole ratio: (b-x)/b, (x/2)/b, (x/2)/b
Partial pressure = p(b-x)/b, px/(2b) px/(2b)

Kp = px/(2b)*px/(2b) / (p(b-x)/b)^2 = D

Alternatively you can realise that we can use concentration and find out Kp expression without taking pressure or mole ratio to consideration, since you can see Kp will have no units as pressure cancels out

 

[qwertypoiu]

http://onlineexamhelp.com/wp-content/uploads/2014/08/9701_s14_qp_11.pdf
Guys , in Q8 - we are choosing ph 7 becaz the products are basic in nature ?
but this reaction isnt equilibrium related to do the shifting :/

I think it might be because if the pH is 14, there is a very high concentration of hydroxide ions. This would lead to OH- ions having high chances of collision with the aluminum and barium nitrate. It's just a guess I'm not sure though

 

[Xaptor16]

Which molecules, each with a linear carbon chain, can have an optically active isomer?
I - C3H6BrI
II - C3H4BrI
III - C3H6I2
IV - C3H4Br2
A. I and II only
B. I, II and III only
C. II and III only
D. I, II and IV only

answer is B, explain?

 

[Xaptor16]

Carbon monoxide burns readily in oxygen to form carbon dioxide.
What can be deduced from this information?

1 The +4 oxidation state of carbon is more stable than the +2 state.
2 The standard enthalpy change of formation of carbon dioxide is more negative than that of
carbon monoxide.
3 The value of the equilibrium constant for the reaction, 2CO(g) + O2(g) 2CO2(g), is likely to
be high.

answer is A, i need to know why 3 is correct? explanation please!

 

[nehaoscar]

Which molecules, each with a linear carbon chain, can have an optically active isomer?
I - C3H6BrI
II - C3H4BrI
III - C3H6I2
IV - C3H4Br2
A. I and II only
B. I, II and III only
C. II and III only
D. I, II and IV only

answer is B, explain?

For optical isomerism, you need a carbon with a chiral center ( 4 different substituents on the C)
As you can see only 1,2 and 3 have a chrial carbon (marked red)
So it is B

 

[Mathemagical]

Carbon monoxide burns readily in oxygen to form carbon dioxide.
What can be deduced from this information?

1 The +4 oxidation state of carbon is more stable than the +2 state.
2 The standard enthalpy change of formation of carbon dioxide is more negative than that of
carbon monoxide.
3 The value of the equilibrium constant for the reaction, 2CO(g) + O2(g) 2CO2(g), is likely to
be high.

answer is A, i need to know why 3 is correct? explanation please!

It burns readily, so there will be a high concentration of CO2 in the equilibrium mixture. Therefore, applying the formula for equilibrium constant, we can deduce that the value is likely to be high.

 

[nehaoscar]

Help please??
View attachment 54318

What's the answer?

 

[bubbles1997]

Aoa. Please post your guess for the upcoming chemistry practical code 34. Thanks!
https://www.xtremepapers.com/commun...vel-practical-guess-papers.41560/#post-926251

 

[princess Anu]

why not B?