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Hello Just wondering any guides to scoring high on P3 for chemistry available?
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I need help with MCQ 23: http://www.xtremepapers.com/CIE/International A And AS Level/9701 - Chemistry/9701_w09_qp_11.pdf
Why are the other options incorrect?
Also need help with Q39 from: http://www.xtremepapers.com/CIE/International A And AS Level/9701 - Chemistry/9701_w08_qp_1.pdf
I get 23. Thank you.Once ethene has been polymerised, it acts like an alkane i.e. an organic compound without a C=C bond. Hence, the formation of alcohol using KMnO4 is not possible. A and D are eliminated. Secondly, hydration of alkane doesn't result result in the formation of an alcohol. We are only left with C which is quite obvious in the first place. This question deals more with other concepts than the concept of polymerisation, since each option has already been polymerised and ethene has already lost its C=C bond. Hope you get it.
Are you sure you are talking about 39? In this question the 3rd option gives us a compound with 7 Carbons that can't be produced from two Propane free radicals. 1 and 2 can be formed. So the answer is B.
no your counting is perfectly alryt there are six carbon atoms in each, including 3rd one.Regarding 39, is my counting flawed? There are six carbon atoms in each of the compounds. :S
http://www.xtremepapers.com/CIE/International A And AS Level/9701 - Chemistry/9701_s03_qp_1.pdf
MCQ 2 of the above paper deals with hydrogenation. Upon counting, I find that the soft margarine formed has 5 more hydrogen atoms that the reactant, glyceryl trieleostearate. Since hydrogen gas is used for hydrogenation, wouldn't the number of moles of hydrogen required be 2.5 moles rather than 5 moles?
The preferred reagent to use is supposedly SOCl2 because the by-products are both gases that can easily be removed from the product.Hey, can someone explain this to me please?
''An alcohol can be converted into a chloroalkane by reaction with either PCL3 , PCL5 or SOCL2. Which is the best method? Justify your answer.''
Thank you
Another one please?
The oxidation number of a vanadium complex can be determined as follows.
0.013 mol of the ion is dissolved in water and made up to 100cm3. 10 cm3 of this solution was made to react with 20.8 cm3 of 0.025 mol/dm3 of KMNo4 solution. All the Vanadium ions were oxidised to the +5 oxidation state.
Use these values to determine the original oxidation state of the Vanadium ion.
(Answer is supposedly +3)
Thank you!
1/10 of the vanadium ions were used so this is
.0013 moles Vn where n is the oxidation number of the vanadium in the complex.
.0208 X .025 = number of moles KMn O4 that reacted this is equal to .00052 moles
Mn goes from the +7 to the +2 state in the reaction and so each ion gains 5 electrons
.00052 moles of MnO4 X 5 e- = .0026 moles of electrons are given up by .0013 moles of Vn
Write the half reaction for vanadium in the complex and balance for charge.
0.0013 Vn -----> .0013V+5 + .0026 e-
Considering just the charge balance
.0013n = .0013 X +5 + .0026 -
.0013n = .0065 -.0026
.0013n = .0039
n = +3
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