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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
(Q7) E=mc(Temperature change)
assuming 1g ~ 1cm3
E = 100 x 4.18 x (78-20)
E= 24244 J
Now since propanone is limiting u take its moles into consideration here
Change in enthalpy = -(E/0.025) = -969.76 KJmol-1
(Q8)
Down the group, thermal stability of group II carbonates increases and solubility increases.
As solubility increases, the value of the pH decreases
Ba and Sr is both stronger than Ca and more soluble so not C and D.
Ca(OH)2 is lime water so not B. Hence, answer is A.
(Q21)
Oh. this is a bit hard question. U need to think a lot. First, in my opinion, draw the displayed formulae for each compound provided. Eliminate B as it contains only 13 carbon atoms but Q contains 14 carbon atoms. Take (A) into consideration first. This compound only has the corner branch of Q twisted down so it is not an isomer of Q. Take (D) into consideration. That has the same thing as (A). Displayed formulae miight help u with this. Hence (C) is the answer.
(Q24)
SN1 Nucleophillic substitution reactions is undergone by tertiary halagenoalkanes. (3 carbons bonded to carbon containing halogen). (C) only clearly shows this fact. Hence (C) is the answer.
(Q9)
Eliminate (C) and (D) both cos they are not empirical formulae.
Number of Carbon atoms - 6
Number of Hydrogen atoms - 12
Number of Nitrogen atoms - 4
Divide everything by 2 to get empirical formula.
It comes to C3H6N2 (B)
(Q38)
3 gives only 1 product. This is because when the OH goes away from the methyl group a C=C bond is formed with the cyclo compound with the methyl. Hence a double bond cannot be formed with the carbon-carbon bond joined to the methyl group. Hope u get the idea. It's hard to explain. Look at the molecule closely. 2 is obviousy correct as either side the C=C bond goes to, it is the same formula. 1 is wrong cos when u put C=C bond to each side, when counting to get structural formula u see it is different.
Thanks