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Chemistry: Post your doubts here!

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a problem:
The compound pentan-1,4-diol has two OH groups per molecule and can be oxidised.
Which statements about pentan-1,4-diol or its oxidation products are correct?
1. When one mole of pentan-1,4-diol reacts with an excess of sodium metal, one mole of hydrogen molecules is produced.
2. At least one of the possible oxidation products of pentan-1,4-diol will react with 2,4-dinitrophenylhydrazine reagent.
3. Dehydration of pentan-1,4-diol could produce a compound with empirical formula C5H8.

which statement(s) is/are correct?
may/june 2014, variant 11 question 39.
all 3 r correct i think
 
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How many different substitution products are possible, in principle, when a mixture of bromine and ethane is allowed to react?
A 3 B 5 C 7 D 9
 
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hello~
can someone please solve the following problem for me?
  1. Use of the Data Booklet is relevant to this question.
    The gas laws can be summarised in the ideal gas equation.

    pV = nRT
    0.96 g of oxygen gas is contained in a glass vessel of volume 7000 cm3 at a temperature of 30 °C.

    What is the pressure in the vessel?
    A 1.1kPa B 2.1kPa C 10.8kPa D 21.6kPa

    thank you x
 
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hello~
can someone please solve the following problem for me?
  1. Use of the Data Booklet is relevant to this question.
    The gas laws can be summarised in the ideal gas equation.

    pV = nRT
    0.96 g of oxygen gas is contained in a glass vessel of volume 7000 cm3 at a temperature of 30 °C.

    What is the pressure in the vessel?
    A 1.1kPa B 2.1kPa C 10.8kPa D 21.6kPa

    thank you x
P×7000×10*-6=0.96/32×8.31×303


Ans:10.8 kPa

So it's C
 
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Perhaps you can list down the other option(s) you thought were also correct, then we can guide you accordingly.

We can try replacing option A (correct answer) with something familiar

CH4 (g) --> C (g) + 4 H (g) ∆H

Bond energy of C-H is ∆H/4
1/4 CH4 (g) --> 1/4C (g) + H (g)

H4 (g) --> C (g) + 4 H (g) ∆H

Analogously
XYn (g) --> X (g) + n Y (g) ∆H

Bond energy of X-Y is ∆H/n
1/n XYn (g) --> 1/n X (g) + Y (g) ∆H
Thankyou soo muchh!! <33
 
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K2O + H2SO4 ----> K2SO4 + H2O
Moles of acid = 15/1000*2 = 0.03mol
Moles of K2O in 25cm3 = 0.03mol, due to ratio.
Moles of K2O in 250cm3, = 0.03*10 = 0.3mol
Mass = (39.1*2+16)*0.3 = 28.26g

this question: October/November 2014 variant 12 question number 15.
the equation: K2O + H2SO4 ---> K2SO4 + H2O i don't think this is correct as the question said that K2O is dissolved in water.
so i think these 2 equations may be used to solve the question:
1) K2O + H2O ---> 2KOH
2)
2KOH + H2SO4 ---> K2SO4 + 2H2O.
but i can't seem to find the answer. However i found this on yahoo answers:

(2 mol/dm^3 H2SO4) x (0.015 dm^3) x (2 mol KOH / 1 mol H2SO4) x (1 mol K2O / 2 mol KOH) x
(250 cm^3 / 25 cm^3) x (94.1960 g K2O/mol) = 28 g K2O


can anyone explain this to me on plain paper and/or step by step?
 
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g

guyssss help me with this
When ethane reacts with Bromine, it forms CH2BrCH3 but the product can react with Br2 again and form disubstituted product which reacts again with Br2 and form trisbustituted product and so on, we can actually form even hexasubstituted product. So it goes like this:
Monosubstituted:
CH2BrCH3
Disubstituted:
CH2BrCH2Br
and
CHBr2CH3

Trisubstituted:
CH3CBr3
and
CHBr2CH2Br
Tetrasubstituted:
CBr3CH2Br
and
CHbr2CHBr2
Pentasubstituted:
CBr3CHBr2
Hexasubstituted:
CBr3CBr3
 
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How does one prove if a reaction is a first order process or not using graph?
What is a first order process?
What are half lives and how do i draw half life lines?
I fear this might come in the practical
Help me out here
 
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Oh this one. See, now when these acids react with calcium, there are several structural formulae. So for ethanoic acid it will be (CH3COO)2Ca. This will be C4H6O4Ca. Now this doesnt match up with the answer. So 1 is incorrect. 2 & 3 are hence correct. This is not AS question. This is a higher level one. But cambridge have given it in such a way that the AS candidate can cross out the first answer to get to the correct answer . Tactics. The other two molecules are complex, so not to worry.
 
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When ethane reacts with Bromine, it forms CH2BrCH3 but the product can react with Br2 again and form disubstituted product which reacts again with Br2 and form trisbustituted product and so on, we can actually form even hexasubstituted product. So it goes like this:
Monosubstituted:
CH2BrCH3
Disubstituted:
CH2BrCH2Br
and
CHBr2CH3

Trisubstituted:
CH3CBr3
and
CHBr2CH2Br
Tetrasubstituted:
CBr3CH2Br
and
CHbr2CHBr2
Pentasubstituted:
CBr3CHBr2
Hexasubstituted:
CBr3CBr3
yesss, thankyouuu
 
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Can anyone help me with these : http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_w15_qp_13.pdf

[Q7] Use of the Data Booklet is relevant for this question. In an experiment, the burning of 1.45g (0.025mol) of propanone was used to heat 100g of water. The initial temperature of the water was 20.0°C and the final temperature of the water was 78.0°C. Which experimental value for the enthalpy change of combustion for propanone can be calculated from these results?
A –1304kJ mol–1 B –970kJ mol–1 C –352kJ mol–1 D –24.2kJ mol–1

[Q9] Please.

Q12 X is a Group II metal. The carbonate of X decomposes when heated in a Bunsen flame to give carbon dioxide and a white solid residue as the only products. This white solid residue is sparingly soluble in water. Even when large amounts of the solid residue are added to water the pH of the saturated solution is less than that of limewater. What could be the identity of X?
A magnesium B calcium C strontium D barium

Q21 please, I dont get how to solve this one.

Q24 This one too.

Q38

I'll be so happy if these questions are solved :)
 
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http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_s15_qp_13.pdf

10 Use of the Data Booklet is relevant to this question. 1.00g of a metallic element reacts completely with 300cm3 of oxygen at 298K and 1 atm pressure to form an oxide which contains O2– ions. The volume of one mole of gas at this temperature and pressure is 24.0dm3 .
What could be the identity of the metal?
A calcium B magnesium C potassium D sodium

Q22.

26 Alkane X has molecular formula C4H10. X reacts with Cl 2(g) in the presence of sunlight to produce only two different monochloroalkanes, C4H9Cl. Both of these monochloroalkanes produce the same alkene Y, and no other organic products, when they are treated with hot ethanolic KOH.
What is produced when Y is treated with hot concentrated acidified KMnO4?
A CO2 and CH3CH2CO2H B CO2 and CH3COCH3 C HCO2H and CH3COCH3 D CH3CO2H only

28 This one is really confusing.
 
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