Chemistry: Post your doubts here!

[ahmed abdulla]

ΔHf = ΔHp - ΔHr


You don't need the Hess Law for this.

Since enthalpy of formation is formation of 1 mol from constituent elements (which is a two-step reaction in this case)

So, ΔHf for 2ICl = +14, so ICl = +7

Add that to the second step you have your enthalpy change (In that, -88+7 = -81 thus C)

Wait a few mins I'll upload a handout on Hess's Law. It's very very easy.

thanks aloot bro ... now at the end of the AS year .. i dont think i need this for paper 1 ! it takes time

 

[ahmed abdulla]

ΔHf = ΔHp - ΔHr


You don't need the Hess Law for this.

Since enthalpy of formation is formation of 1 mol from constituent elements (which is a two-step reaction in this case)

So, ΔHf for 2ICl = +14, so ICl = +7

Add that to the second step you have your enthalpy change (In that, -88+7 = -81 thus C)

Wait a few mins I'll upload a handout on Hess's Law. It's very very easy.

as per your equation .. p-r .... shouldnt it be -88-(+7) =-95 ... or did u add them ?

 

[AbbbbY]

thanks aloot bro ... now at the end of the AS year .. i dont think i need this for paper 1 ! it takes time

For P1, simple use ΔHf = ΔHp - ΔHr and ΔHc = ΔHr - ΔHp (I remember it from Crap. Crap = Cr-p. And thus, F is the other way round).

 

[AbbbbY]

as per your equation .. p-r .... shouldnt it be -88-(+7) =-95 ... or did u add them ?

Enthalpy of formation is given for both the reactions. If the enthalpy of formation of individual products was given we'd have done that.
Basically, they've already done the p-r and have given the value (+14 and -88 respectively)

 

[AbbbbY]

HELLLPPPPP!

S12_qp11 Question 1
Why is the answer BF3 and NOT CH3-? :S

And same paper, question 14. Thanks a ton!

1- BF3 =
Each F has 7 outer electrons
Each B has 3 outer electrons
So, B forms 3 bonds and hence has 6 electrons (incomplete octet!)

CH3-
Each C has 4 outer electrons
Each H has one outer electron
CH3 has an overall negative charge.

So, C forms 3 bonds and will have 6 electrons.
One electron is left but there is a negative charge so it is gaining an electron. [Well that's what my teacher told me back when I asked him!]


14- 4Al + Ba(NO3)2 -> 2(Al2O3) + Ba + N2

1 mol : 1 mol
0.783/(137+62(2)) = 0.003 mol

1 mol occupies 24dm3
0.003 mol will occupy 0.072dm3
Hence,B

 

[Rahma Abdelrahman]

12. Going down periodic table, electronegativity decreases. Going right, electronegativity increases. So go right from Be, it increases, and go down one it decreases. These cancel so it's electronegativity is similar to Al.
Answer B

13. Did you consider the cis and trans isomers?

29. Two -OH group's add at the double bond with cold,dilute KMnO4. The carbon atoms attached to the -OH groups both become chiral, so +2.
With hot, conc. KMnO4, double breaks, and ketone group and carboxylic acid group form. The carbons still not chiral. The -OH group on the very left of the original molecule is oxidised to ketone, so that carbon becomes non-chiral, so -1.
Answer D.

30. B is acid hydrolysis of a nitrile. You would get ethanoic acid. Question says which would not give. A is base hydrolysis of nitrile, where you get the sodium salt sodium ethanoate.
Answer A.

12--> I got ur point, but I assumed its Mg as its closest to Al in the periodic table..
(Its 26, not 13) --> Oh.. I guess not thanks for reminding me
29--> Got it..
30--> Yeah..
Thanks ..
Sorry but what about Q35 ?

 

[omarjaved619]

1- BF3 =
Each F has 7 outer electrons
Each B has 3 outer electrons
So, B forms 3 bonds and hence has 6 electrons (incomplete octet!)

CH3-
Each C has 4 outer electrons
Each H has one outer electron
CH3 has an overall negative charge.

So, C forms 3 bonds and will have 6 electrons.
One electron is left but there is a negative charge so it is gaining an electron. [Well that's what my teacher told me back when I asked him!]

14- 4Al + Ba(NO3)2 -> 2(Al2O3) + Ba + N2

1 mol : 1 mol
0.783/(137+62(2)) = 0.003 mol

1 mol occupies 24dm3
0.003 mol will occupy 0.072dm3
Hence,B

I understood the first question. But I dont really get how you got 137 and 62 for the second question. I understand you took mass/molar mass. Which substance's moles are you tryna find?

 

[AbbbbY]

I understood the first question. But I dont really get how you got 137 and 62 for the second question. I understand you took mass/molar mass. Which substance's moles are you tryna find?

I need the mols of Ba(NO3)2 to deduce how many mols of N2 will be produced.

I didn't have a periodic table but I remember Ba and NO3 values having used them ample times so I simply plugged 137 and 62x2. You could also do 137+2(14+16+16+16) and it'd give you the same thing. (137+62+62 is the Mr of Ba(NO3)2)

Did you get it?

 

[omarjaved619]

I need the mols of Ba(NO3)2 to deduce how many mols of N2 will be produced.

I didn't have a periodic table but I remember Ba and NO3 values having used them ample times so I simply plugged 173 and 62x2. You could also do 173+2(14+16+16+16) and it'd give you the same thing. (137+62+62 is the Mr of Ba(NO3)2)

Did you get it?

I think you're mixing up 173 and 137 lol.
My Periodic Table shows that the Mr of Ba is 39.

 

[AbbbbY]

I think you're mixing up 173 and 137 lol.
My Periodic Table shows that the Mr of Ba is 39.

My bad. Typo.

What sort of period table do you have?
http://www.ptable.com/Images/periodic table.png

 

[omarjaved619]

I think you're mixing up 173 and 137 lol.
My Periodic Table shows that the Mr of Ba is 39.

Gawwwwd I was looking at something else. -__-
Haahaha I get it now. Thanks mate.
Mind if I ask you some more questions?

 

[AbbbbY]

Gawwwwd I was looking at something else. -__-
Haahaha I get it now. Thanks mate.
Mind if I ask you some more questions?

Inorganic and Physical, fire away!

 

[omarjaved619]

Inorganic and Physical, fire away!

s12_qp12
4, 6,

I've got a lot more but sadly all of them are from Inorganic Will have to ask my lousy teacher lol.

 

[AbbbbY]

s12_qp12
4, 6,

I've got a lot more but sadly all of them are from Inorganic Will have to ask my lousy teacher lol.

Oh jeez. Contradiction in 4. I'll text my teacher and ask what the method was.

6- P = nRT/V
=> ((0.96/32)(8.31)(30+273))/7000 * 10^-6
=10791.1 Pa
Hence, C

I do know inorganic.
Just haven't done organic.

 

[omarjaved619]

Oh jeez. Contradiction in 4. I'll text my teacher and ask what the method was.

6- P = nRT/V
=> ((0.96/32)(8.31)(30+273))/7000 * 10^-6
=10791.1 Pa
Hence, C

I do know inorganic.
Just haven't done organic.

Thanks man! Appreciate it.

 

[AbbbbY]

Thanks man! Appreciate it.

For the first question,

we know CH3- is complete

So, CH3+ has to be incomplete as it's losing 2 electrons. It didn't make sense when I first did it because I thought it was CH3- that was listed.

 

[omarjaved619]

For the first question,

we know CH3- is complete

So, CH3+ has to be incomplete as it's losing 2 electrons. It didn't make sense when I first did it because I thought it was CH3- that was listed.

2 electrons? What is actually losing the electrons, the C or the H?

 

[AbbbbY]

2 electrons? What is actually losing the electrons, the C or the H?

The lonepair on Carbon is lost (from CH3-)

 

[omarjaved619]

So, CH3+ has to be incomplete as it's losing 2 electrons

How do you know its losing 2 electrons?

 

[AbbbbY]

So, CH3+ has to be incomplete as it's losing 2 electrons

How do you know its losing 2 electrons?

Because it has a positive charge on top!