Chemistry: Post your doubts here!

[leosco1995]

I need help again
i need help with question no.2,12,16,18,26,and 36
please do try to explain the answers and thank you for any help ^_^
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s07_qp_1.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s07_ms_1.pdf

Q2) Write down all of the data you have.

Mr = 14x + 16y
Mass = 0.23 g
Moles = 0.23 / (14x + 16y) - (1)
Moles = 0.005 (since they said it occupies 120 cm^3) - (2)

Equate (1) with (2):

0.07x + 0.08y = 0.23
7x + 8y = 23

And then try each option and see if they add up to 23. In B, 7(1) + 8(2) = 23 so it's right.

There's probably a few other methods of solving this question, but this one worked for me.

12) NaCl is completely ionic, SiCl4 is completely covalent. AlCl3 has a lot of covalent character and isn't that much ionic and MgCl2 is the opposite of that. Since they asked which one shows just some covalent character, MgCl2 should be right.

16) "The enthalpy of formation of HCl and HI is represented by the equations:


H2 + Cl2 --> 2HCl

and

H2 + I2 --> 2HI

The enthalpy change of these reactions can be found by looking at the bonds broken and formed in terms of their energies (remembering that bond breaking is endothermic and bond formation is exothermic):

ΔH(rxn) = ΔH(reactant bonds) - ΔH(product bonds)

So if the HI bond energy is much smaller than the HCl bond energy (option C) then the HI reaction will be more endothermic...

... whereas if the bond energy of I2 were smaller than the bond energy of Cl2 (option D) the reaction forming HI would be more exothermic.

And as we know that the reaction forming HI is more endothermic, the correct answer is C"

18) "A solid nitrate fertiliser reacts with an alkali to produce a gas which turns damp pH paper blue."

Read the bolded parts. The anion must be nitrate and the cation must be NH4+.

So you have NH4NO3 whose empirical formula is N2H4O3.

26) For it to be chiral, the central C atom MUST be bonded to 4 different groups. One of them will be OH, the minium the rest will have to be is:

- CH3
- CH3CH2
- CH3CH2CH2

In total, that's 6 + the central C atom which is 7.

36)
1 is right, the more down you go group 7, the more readily dissociation occurs.
2 this only happens with Flourine I think.
3 is wrong because only the halide ions are good reducing agents.

 

[saudha]

Answer is D.
First check the oxidation of Nitrogen atom in NH4(+1)
x+4=1 so x= -3
Oxidation of Nitrogen in Ammonium Ion = -3

Now Oxidation of Nitrogen in NO3(-1)
x-6=-1 so x= 5
Oxidation of Nitrogen in Nitrate Ion is 5

Now Check the Oxidation State of Nitrogen in N20.
2x-2=0 so x=1
So Oxidation of Nitrogen in N20 is 1

Final Step :
Calculate the change : from -3 to +1 ( change of +4)
from 5 to +1 (change of -4)
Therefore answer is D

thxx a lot... u cleared my doubt ....thx for the detailss

 

[saudha]

leosco1995 had answerd this..
Figure out the oxidation numbers in both N atoms in NH4NO3 and compare it with the oxidation number of N in N2O. Like so:

You split up NH4NO3 into the NH4+ and NO3- ions.

In NH4+, the charge on Nitrogen is 1 - 4 = -3.
In NO3-, the charge on Nitrogen is -1 + 6 = +5.

And in N2O, the charge on nitrogen is 2/2 = +1.

And the changes in these are +4 and -4 respectively, meaning D is right.

thxx a lott

 

[leosco1995]

The equation below represents the combination of gaseous atoms of non-metal X and of
hydrogen to form gaseous X2H6 molecules.
2X(g) + 6H(g) → X2H6(g) ΔH = –2775 kJ mol–1
The bond energy of an X–H bond is 395 kJ mol–1.

What is the bond energy of an X–X bond?
A – 405.0 kJ mol–1
B – 202.5 kJ mol–1
C +202.5 kJ mol–1
D +405.0 kJ mol–1

How is the answer D ?

Bond energy is ALWAYS +ve, so the answer can't be A or B. Also, X2H6 has one X-X bond and 6 X-H bonds. Using formula, you have 2775 = x + (6*395) which is 405.

 

[leosco1995]

all these elements belong to group 2. and why cant it be magnesium aluminium is right next to magnesium in periodic table?

Be isn't really a typical group 2 element. Like I said before, Be2+ is smaller and has a higher charge density than the other group 2 elements and its charge density is similar to that of Al's. The rest of the group 2 elements are bigger than Al and don't really have a high charge density.

Let me try to give a better explanation:

Al is a small atom with a 3+ charge. All of these group 2 elements have a 2+ charge. For the atoms to have the same electronegativity, you would want the atom to be smaller to make the net effect the same. Be is the only atom which is smaller, so the effect of charge density on both Al and Be is similar.

 

[hihi]

To you too...
Im hoping if i nail the other exams and even if i get less marks in MCQ's it wont matter much.. but lets hope for the best.

Always think positive

 

[Anoushay]

Hey,can anyone please explain the question number 13 in this paper??? i posted it before too. thanks.
http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_w10_qp_12.pdf

 

[Most_UniQue]

Someone help me with this question! How can i find number of isomers?

 

[Most_UniQue]

Hey,can anyone please explain the question number 13 in this paper??? i posted it before too. thanks.
http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_w10_qp_12.pdf

I gt same problem

 

[saudha]

The esterification reaction
ethanol + ethanoic acid --> ethyl ethanoate + water
is an equilibrium. The forward reaction is exothermic.
How can the value of the equilibrium constant KC be increased?
A by adding a little concentrated sulfuric acid as a catalyst
B by increasing the initial concentration of ethanol
C by lowering the temperature
D by raising the temperature

Ammonia is manufactured on a large scale by the Haber process.
In a particular plant, conditions of 400 °C and 250 atm in the presence of an iron catalyst are
used.
N2(g) + 3H2(g) 2NH3(g) ∆Ho = –92kJmol–1
What could contribute most to increasing the equilibrium yield of ammonia?
A adding more catalyst
B increasing the pressure to 400atm
C increasing the temperature to 1000°C
D using air rather than nitrogen
some one pls explain these question to me!

 

[Most_UniQue]

The esterification reaction
ethanol + ethanoic acid --> ethyl ethanoate + water
is an equilibrium. The forward reaction is exothermic.
How can the value of the equilibrium constant KC be increased?
A by adding a little concentrated sulfuric acid as a catalyst
B by increasing the initial concentration of ethanol
C by lowering the temperature
D by raising the temperature
some one pls explain this question to me!

It shud be either C or D cuz only Temperature has an effect on equilibrium constant. I guess its C cuz if you lower temperature , rate of forward reaction increases since its exothermic and KC also increases. Cuz KC = [H2O][CH3COOC2H5]/ [C2H5OH][CH3COOH]
So since forward reaction is more, [H2O][CH3COOC2H5] will have higher value and [C2H5OH][CH3COOH] lower value.

 

[saudha]

It shud be either C or D cuz only Temperature has an effect on equilibrium constant. I guess its C cuz if you lower temperature , rate of forward reaction increases since its exothermic and KC also increases. Cuz KC = [H2O][CH3COOC2H5]/ [C2H5OH][CH3COOH]
So since forward reaction is more, [H2O][CH3COOC2H5] will have higher value and [C2H5OH][CH3COOH] lower value.

thxx

 

[Most_UniQue]

thxx

My pleasure and mind answering my question which is on previous page?

 

[Most_UniQue]

The esterification reaction
Ammonia is manufactured on a large scale by the Haber process.
In a particular plant, conditions of 400 °C and 250 atm in the presence of an iron catalyst are
used.
N2(g) + 3H2(g) 2NH3(g) ∆Ho = –92kJmol–1
What could contribute most to increasing the equilibrium yield of ammonia?
A adding more catalyst
B increasing the pressure to 400atm
C increasing the temperature to 1000°C
D using air rather than nitrogen
some one pls explain these question to me!

Answer is B . Adding catalyst just increases rate of reaction by lower activation energy. C cant be cuz reaction is exothermic so it favors low temperature otherwise equilibrium will be shifted to the left . D is impossible.
So its B , in haber process the pressure required is 400*C .

 

[Oliveme]

Hydrolyze the ester to get CH3OH and CH3CH2CH2CO2H (so 3 is obviously wrong). Product #2 will be formed, but it doesn't distill off so you only have the alcohol in the distillate. It's also worth noting that if there was another option - CH3CHH2CH2CO2H in the question, that wouldn't be found in the distillate either because all of it would have been converted into the salt.

The answer is D.

thank you

 

[saudha]

Someone help me with this question! How can i find number of isomers?
View attachment 8582

i think C is right and that is bcoz......to find this there is a way..... the formula is 2^(no of double bonds)....in thsi question...
double bonds=3
so 2^3=8 and so =3
C is the ans .... im guessing

 

[Most_UniQue]

i think C is right and that is bcoz......to find this there is a way..... the formula is 2^(no of double bonds)....in thsi question...
double bonds=3
so 2^3=8 and so =3
C is the ans .... im guessing

WOW cool! C is right! Tnx a lot! I didnt knew it was so easy!!

 

[user]

Assalamoalaikum wr wb!

need help with these mcqs

 

[Most_UniQue]

Assalamoalaikum wr wb!

need help with these mcqs

post ur question

 

[saudha]

plss help me with this Q