Qualification > Sciences

chem. questions

<< < (2/3) > >>

sweetie:
cudnt get these que's
nov07
Q10, 19, 21, 30,33 and 35

may07
Q5, 26,28, and 40

tmisterr:
m/j 07
q5 c, cyanide has a triple bond so 3 bonding pairs. Nitrogen shares 3 electrons leaving 2 forming one lone pair. Carbon has a charge of -1 in cyanide so it has an extra electron. it shares 3 electrons and leaves two again forming one lone pair

q26 c, it cannot react with acidified manganate(VII) ions so it has to be a tertiary alcohol. the simplest tertiary alcohol is butanol but its is not chiral, next is pentanol again not chiral and has no chiral isomers neither does hexanol, but a 7 carbon alcohol (2-methylhexan-2-ol)  is chiral since the central carbon is bonded to four different groups, CH3, C2H5, and C3H7 ( try drawing it to visualize it better)

q28 c, it decolourises 2,4-DNPH so it must be a ketone or an aldehyde but it also de-colourises manganate(VII) ions so it must be able to be oxidised. ketones cannot be oxidised further, but aldehydes can so C

q40 D, 1 only, heating an easter with an acid will hydrolysithe ester bond and therefore break it. alcohol or sodium hydrogen-carbonate will have no effect the ester bond.

tmisterr:
O/N 07
q10 use mole ratios, 0.8 mole of oxygen was made from (0.8*2)=1.6 mole of nitrogen dioxide so at equilibrium, there is 4-1.6=2.4 mole of nitrogen dioxide. since 1.6 mole of oxygen reacts, then there is 1.6 mole of NO. as you know co-efficient of the compounds in the equations is the power of the concentrations in the Kc equation so answer is D

q19 is B. I will have a chiral centre if the Br and I atom are on the same carbon, so this eliminates C. now check III and IV since II is common is all so no need to check it. III is chiral when it is 1,2-di-iodopropane. IV cannot be chiral, the double bond eliminates two carbons and there is no way it can be arranged to make the third chiral.

q21 reaction between methane and chlorine is a free radical substitution and is not an equilibrium, this automatically rules out B and C, the rate of reaction increases if concentration if reagents increase (methyl free radicals) so A.

q30 is B, when the ester breaks it will form butanoic acid which has the molecular formular C4H8O2 which in its simplest form is C2H4O (Divide all by 2)

q35 is B, when a metal nitrite is heated you get the oxide and NO2 and O2 so BaO will be formed. Mg is oxidised so MgO is formed, no way in which Magnesium nitrite can form.

q33 is C, really gotta go but will explain real soon

sundars:
I think the ansewr to Q33 of October November 2007 is B not C

sundars:
Sorry it is C not B.Read the other question.

Navigation

[0] Message Index

[#] Next page

[*] Previous page