chem. questions

[sweetie]

cud sum1 xplain to me Q 7,12,16,20 and 24 of nov05 P1

Thank You

[Saladin]

Q7:

The greater the difference in pH, the greater the enthalpy chage. So tha answer is A.

Q12:

The answer is D. Because MgO takes moles. Al₂O₃ takes mol, and Sulphur takes a lot of moles, because Sulphur can form SO₂, then SO₃.

Q16:

The answer is C. You need to have a base avaiable when you react it with a Carbonate. So it is wither C or D. Now the answer is not D, because MgCO₃ and CaCO₃, are both weak carbonates because both metals have strong positive charges and small radii. So, the answer is C, because Barium Carbonate would not easily decompose under heat.

Q20:

The answer is B. Cis and trans of this: CHCl==CHCl and CH₂==CCl₂. Total, 3.

Q 24:

A. It could be between the 2 middle carbon, between either one of the side carbons.

Hope that explanation is suffice.

[sweetie]

Q7:

The greater the difference in pH, the greater the enthalpy chage. So tha answer is A.

Q12:

The answer is D. Because MgO takes moles. Al₂O₃ takes mol, and Sulphur takes a lot of moles, because Sulphur can form SO₂, then SO₃.

Q16:

The answer is C. You need to have a base avaiable when you react it with a Carbonate. So it is wither C or D. Now the answer is not D, because MgCO₃ and CaCO₃, are both weak carbonates because both metals have strong positive charges and small radii. So, the answer is C, because Barium Carbonate would not easily decompose under heat.

Q20:

The answer is B. Cis and trans of this: CHCl==CHCl and CH₂==CCl₂. Total, 3.

Q 24:

A. It could be between the 2 middle carbon, between either one of the side carbons.

Hope that explanation is suffice.

thanx sooo much!
but cud u xplain que. 24 a lil bit in more detail?
and for Q 12, i guess that 1 mole of Al₂O₃ needs 0.75 moles of O₂, check it plz???
and in 20 , u mean there r 2cis and 1 trans???

[Saladin]

Question 12: One mole of Oxygen is O₂. So O₃ is 1.5 moles of oxygen, what you said there is the ratio.

Q 20: Yes, there are 2 cis-trans isomers. And also one where you have 2 Hs on one carbon and 2 Cls on the other.

Q24: Butan 2 can mean from the right 2 or the left 2. This will mean the double bond is able to be in 3 different positions, in the middle, and other 2 sides.

[sweetie]

got it
thanx

[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.

[sweetie]

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.

but how do we know its not chiral???
like  in butanol one of the carbon atoms cud have H2, OH, CH3CH2, and a CH2 grps around it... so they r diff.
so why is butanol not chiral???

[sweetie]

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

Thank You sooo much
+rep

but in Q21, why is D wrong?
and in que. 19 comp.II also has a double bond , so is it chiral or no?

[tmisterr]

but how do we know its not chiral???
like  in butanol one of the carbon atoms cud have H2, OH, CH3CH2, and a CH2 grps around it... so they r diff.
so why is butanol not chiral???

how can H2 happen? its not possible, each hydrogen has to be bonded to a carbon or an oxygen, you cannot have two hydrogen atoms bonded to each other. again the CH2 will only have three bonds from it, it needs to have 4 bonds so this cannot happen. Butanol is not chiral, the primary, secondary or tertiary forms of it are all not chiral.

[tmisterr]

Thank You sooo much
+rep

but in Q21, why is D wrong?
and in que. 19 comp.II also has a double bond , so is it chiral or no?

thanks for the +rep

metal ion cannot catalyse the reaction.this reaction depends on light!! maybe increasing light intensity can increase rate of reaction but not use of a metal catalyst.

and in q19 it is most definitely chiral if both the Br and I are attached to the carbon which is not involved in the double bond as this carbon will be attached to 4 different groups, H, Br, I and CHCH2