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Chemistry: Post your doubts here!

cHeStEr

cHeStEr

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Casablanca said:
Okay, here goes:

CO2(g) + H2(g) -> CO(g) + H2O(g)

CO2 is being broken, so instead of -283, we take +283
H2O is being formed, so we use -286
BUT it is being formed in gaseous state, not liquid state, so we also take +44.

283 + 44 - 286 = +41.

So the answer is C. :)
Click to expand...
Thnx a tonne..i dont undrstnd this typ of prblmes...:notworthy:
 
Casablanca

Casablanca

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Which equation applies to an ideal gas? (p = pressure, V = volume, M = molar mass, P = density, c = concentration, R = gas constant)

1. p = PRT/M
2. pV = MRT
3. pV = cRT/M

A. 1, 2, 3
B. 1, 2
C. 2, 3
D. 1

How?
 
gary221

gary221

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Casablanca said:
Which equation applies to an ideal gas? (p = pressure, V = volume, M = molar mass, P = density, c = concentration, R = gas constant)

1. p = PRT/M
2. pV = MRT
3. pV = cRT/M

A. 1, 2, 3
B. 1, 2
C. 2, 3
D. 1

How?
Click to expand...
the ans is D...1 only as Mr = Mass/No of moles... Rearranging the equation , v gt p = P * R* T * n / Mass.
we know density = mass/ volume..so d/m = 1/volume...
Rearranging it once agn, v gt ...pv = nRT
 
Goku

Goku

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KurayamiKimmi said:
PCl5-->PCl3+Cl2
bonds broken in PCl5= P-Cl bonds broken= 33o*5=1650
bonds formed in PCl3=330*3 =990
bonds formed in Cl2=240
Bonds broken-bonds formed= 1650 -(990+240)
therefore enthalpy change=+420
Click to expand...
God bless,thankyou!
You made it seem so easy.. o_O
 
nerdybookworm

nerdybookworm

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nerdybookworm said:
q 10 is C
cz juss leme tell yu my internet prob 4 min ok w8
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q16 is also C ...first write the equation for anhydrous calcium nitrate...
thn find th RMM..ie 200.1....
now the no of oles..ie 4.10/200.1=0.0204..tht is moles of CaNo3
thn for Cao..4.10/164=0.025
thn 0.025x56=1.4g
 
nerdybookworm

nerdybookworm

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Ashique said:
For question 2, 2 double bonds in he compound were removed, thus 4 mole of hydrogen was required (since 2 hydrogens are added to the chain once you break a double bond). If you notice, CH2 was changed to CH3 thus one more mole os Hydrogen was required. So in total, 5 mole of hydrogen was required.

Question 4, look for a major change in the ionisation energy. The first 6 value lie close to each other, (or say, one value ot numerically close to the previous value), but the 7th value 13200. Which is BIG increase in ionisation energy. But the first six values lie close to one another. What does this tell about the atom and the ionisation energy? This indicates that 6 electrons were easily removed. And a lot of energy was required to remove the 7th electron. This indicates that the 7th electron lies in a different shell than the first 6. And we can also figure out than the quantum shell (or the outer most shell) has 6 electrons in it. Hence it's in group six. Te is the only element in group VI, hence answe is C.

Question 8- you have to know how to calculate the oxidation number, of a compound in a chemical equations. Go through your book, and I' m sure you'll find out how to do it, (if you don't know that is). The greatets oxidation number change happened in A, which is 20. In B, C and D the oxidation number changes are 6, 4 and 2 respectively.

Question 10- (Since no volume was given, we assume the volume to be 1 dm3, hence the concentrations will have the same numerical value as their mol.)
Now, First calculate the total number of moles at the begining- 0.20+0.15= 0.35 mol.
At equilibrium the number of moles of the products(ie HI) was 0.26. So what is the TOTAL volume of the reactant? 0.35-0.26= 0.09.
Now 0.09 was the TOTAL volume of the reactant. Now look at the denominator, addition of which two concentrations willl give us 0.09? Hence answer is C.

Question 16- Addition of hot alkali will make chlorine go through a disproportionation reaction. Which will give chlorine of two oxidation states -1 and +5, hence answer is D.

Question 20- you simply have to draw out all possible isomers of the compund. It should be 7.

Question 27- You have to find the carbon which has lost the chlorine, since homolytic fission using ultraviolet light, will only give chlorine and the corresponding compound as free radicals. Since C is the only one which lost the chlorine, answer should be C.

Question 28- Reaction of the first three compounds with sodium would give- sodium ethoxide for A and B, with the loss of producing moles of hydrogen, when you balance the equation). Reaction of C with sodium would give sodium ethanote again giving off two moles of hydrogen. It is only compound D which gives of only 1 mole of H2.

Question 30- nucleophilic addiotn of HCN to propanone would give you CH3CH(OH)(CN)CH3. When you reflux this this H2SO4, You'll lose the nitrile and get a carboxylic group , CH3CH(OH)(CO2H)CH3. This compound is represented by D, so thats your answer.
Click to expand...



http://www.xtremepapers.com/papers/...Level/Chemistry (9701)/9701_w11_qp_11.pdf...q 9
 
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