Chemistry: Post your doubts here!

[tdk]

Could anybody answer these plz : What is the equation representing the enthalpy change of formation of Pb? and what is the end product of complete combustion of Pb : is it red lead, Pb3O4 or PbO? What are the oxidation states of Pb and O in Pb3O4?

 

[messi10]

you know that ΔH = ΔHp - ΔHr

so ΔH = (4NO + 6H20 ) - ( 4NH3)
ΔH = (4(90.3) + 6(-241.8) ) - (4(-46.1)
ΔH = -905.2 therefore the answer is C !

knowing the definition of the enthalpy change of formation will help u a lot in this question. i hope i have helped u though!

I know, but my sir said that to get sign you must do "Reactant-Product", not ""product-reactant"..

This is confusing me, please let me know this thing.

 

[iKhaled]

I know, but my sir said that to get sign you must do "Reactant-Product", not ""product-reactant"..

This is confusing me, please let me know this thing.

i dont know about ur sir but i always studied it that "the enthaply change (ΔH) is equal to the enthalpy change of product - the enthalpy change of reactant" that's how i had been taught bro!

 

[tdk]

Q8)
A 2 g sample of hydrogen at temperature T and of volume V exerts a pressure p.
Deuterium, 2/1 H, is an isotope of hydrogen.
Which of the following would also exert a pressure p at the same temperature T ?
A 2g of deuterium of volume V
B 4g of deuterium of volume V/2
C a mixture of 1 g of hydrogen and 2 g of deuterium of total volume V
D a mixture of 2 g of hydrogen and 1 g of deuterium of total volume 2V

First find the amount of hydrogen from 2g of hydrogen.
amount of hydrogen = 2/1 = 2mol

Hence, 2mol of gas occupy volume V at temperate T exerts a pressure p.

Option A: 2g of deuterium (1 mol of gas) with volume V. (WRONG)
Option B: 4g of deuterium (2 mol of gas) with volume V/2. (WRONG)
Option C: mixture of 1 g of hydrogen (1 mol of gas) and 2g of deuterium (1mol gas) with volume V (CORRECT)
Option D: 2g of hydrogen (2mol), 1g of deuterium (0.5mol) total volume 2V. (WRONG) 2.5mol of gas would occupy 1.25V instead.

Q11) D
Equilibrium mixture will contain...
HI (b-x)
H2 (x/2)
I2 (x/2)

Kp = [(x/2)(x/2)] / (b-x)^2
= x^2 / 4(b-x)^2

Q24) B
Step 1 : Nucleophilic substitution (Sub Br away with OH)
Step 2 : Oxidation (From secondary alcohol oxidised to Ketone - Lose 2 H)
Choose a secondary halogenoalkane

Q38)
Catalytic cracking is not in my syllabus. Sorry.

Q38) In catalytic cracking a large molecule of alkane is decomposed to form smaller molecules of alkanes or alkenes. So all the three options are possible. Answer: A

I would like to add some more explanation to Q24)
Options A & C are primary halogenoalkanes which will produce primary alcohols in Step 1 and a mixture of aldehyde and carboxylic acid in Step 2.
Option D is a tertiary halogenoalkane which will produce a tertiary alcohol in Step 1 which will not be oxidized in Step 2.
So B is correct because it will produce only a ketone with formula C4H8O

 

[messi10]

need explanation to this question
isn't it supposed to be -176 because..

ΔHf = (2(38) + 2(-214) ) / 2 really confused

In this question.. Enthalpy change of combustion is given, but "enthalpy change of formation" is asked, so switch sign of every value.

 

[messi10]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s11_qp_12.pdf

need help with Q # 19, 23 and 24.

Edit:
This one as well, anyone?
http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_w11_qp_13.pdf
in this paper I need help with Q # 1

Thank You!

 

[cowarrior]

HELP GUYS>>>>!!!!!!

http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_w11_qp_12.pdf

in question 1 , the answer is A. Why not B??
I dont know how to solve questions 4 10 26 29

PLEASE HElppp.....!!!!

 

[tdk]

i just figured out question 32 well 1moldm^-3 sulphuric acid is dilute dilute sulphuric acid u will consider an ion as high if there are more moles of that ions formed by the dilute acid than the concentrated one . u see from this equation that 2 moles of H+ ions are formed which are greater than in the concentrated acid where only one is formed. no HS04 ion is formed and the number of moles of SO4 ion is same in both the reactions
H2SO4(l) + aq ==> 2H+(aq) + SO42-(aq)

Q14) When one mole of an anhydrous Group II metal nitrate, M(NO3)2 is strongly heated one mole of the metal oxide, MO is formed. So in total (N2O5) is lost, the relative mass of which is 108.
So 108 g is lost from one mole of the nitrate. Therefore 3.29 g is lost from 3.29/108 = 0.030462962 mole of the nitrate.
So Mr of the nitrate = mass/mol = 5.00/0.030462962 = 164.1337 . So the metal is Ca. Answer : B

Q32) Step 1 is irreversible and H2SO4 is an strong acid, so equal no. of H+ and HSO4- ions are formed.
Step 2 is reversible and HSO4- is a weak acid, so it does not ionises completely and the equilibrium remains to the left. So the conc. of HSO4- and that of (SO4)2- are not equal and also the conc. of (SO4)2- is also not as high as that of H+. Answer : D

 

[tdk]

lol, thanks buddy! Actually, I got numb and then had to post these questions here was too fed up to use my mind on chemistry

Q14) When one mole of an anhydrous Group II metal nitrate, M(NO3)2 is strongly heated one mole of the metal oxide, MO is formed. So in total (N2O5) is lost, the relative mass of which is 108.

So 108 g is lost from one mole of the nitrate. Therefore 3.29 g is lost from 3.29/108 = 0.030462962 mole of the nitrate.

So Mr of the nitrate = mass/mol = 5.00/0.030462962 = 164.1337 . So the metal is Ca. Answer : B

Q32) Step 1 is irreversible and H2SO4 is an strong acid, so equal no. of H+ and HSO4- ions are formed.
Step 2 is reversible and HSO4- is a weak acid, so it does not ionises completely and the equilibrium remains to the left. So the conc. of HSO4- and that of (SO4)2- are not equal and also the conc. of (SO4)2- is also not as high as that of H+. Answer : D

 

[tdk]

Need help with this MCQ..

How you determine the sign?? Ans is D.. How you determine the sign... How you solved it... please write the calculation.

The amount of energy needed to break a specific covalent bond is called the bond dissociation energy or bond dissociation enthalpy or bond enthalpy or bond energy.
The values of bond energies are always positive because they refer to bonds being broken.

In the compound X2H6 there are six X-H bonds and one X-X bond. To break the six X-H bonds we have to supply 6*(+395) = +2370 Joules of energy.

The given value of delta ΔH (-2775kJ/mol) is negative because it was a bond forming process. The same amount of energy would be required to break all the bonds (6 X-H and one X-X bond) in the molecule X2H6 but the value would be positive (+2775) because bond breaking is an endothermic process. So the energy required to break the X-X bond would be (+2775)-(+2370) = +405kJ/mol. Answer : D

 

[messi10]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_13.pdf

Q # 18 and 28 please.

 

[NouranAyman]

people please anyone tell me what should I do??? Im not getting so high i solved all papers almost and exam is tmrw, im revising syllabus but idk why do i make many mistakes pleaseeeeeeeeeeeeeeee any tipssssssss im dying of stress

 

[messi10]

people please anyone tell me what should I do??? Im not getting so high i solved all papers almost and exam is tmrw, im revising syllabus but idk why do i make many mistakes pleaseeeeeeeeeeeeeeee any tipssssssss im dying of stress

These require a lot of practice.. Same here :'(.... Try to help others here, Allah will give you the reward. InShaAllah.

 

[NouranAyman]

i don mind helping others But I need an advice myself, what you people will do for tomorrow?? people who have exam tomorroww

 

[Pals_1010]

http://olevel.sourceforge.net/papers/9701/9701_s12_qp_12.pdf

Help with number 28 please?

 

[Iadmireblue]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_13.pdf

Q # 18 and 28 please.

For 18.The double bond in the centre break and give 2 extra OH groups so the total wll be 3
and with Hot mno4,the double bond will break to give you a ketone and carboxylic acd.and also the other OH group will get oxidised to carboxylic,so your'e left with 3 rings
For 28.You know that free radical substitution is homolytic,so the answer should be C or D
draw the isomers of c3h 7cl
There are 6 ways for drawing 1 chloropropane and 2 for 2chloropropane so in the simplest form it becomes 3:1 which is D
Good luck for tomorrow

 

[hassankhan]

PLzz help me with which book should i take for chemistry because i m really confused...... Plz anyone???

 

[magnesium]

CAN anyone share link of A LEVEL CHEMISTRY BY EN RAMSDEN?

 

[Mayedah]

CAN anyone share link of A LEVEL CHEMISTRY BY EN RAMSDEN?

http://books.google.ae/books/about/...Fourth_Editi.html?id=ZCkKH4bUcPUC&redir_esc=y

 

[Mayedah]

if u have ethanal COHCOH
wht different isomers u can make out of the formula C2H2O2 such tht the isomer reacts with Na metal to form hydrogen ?