Chemistry: Post your doubts here!

[abcde]

AoA!
In what range our answers to the practical titrations must fall within? 0.1 cm^3, is it? If some readings are more than, lets say, 0.2 cm^3 apart from the original, must they be recorded or not? Is there enough time to take more than 3 set of readings?
JZK for replies!

W.S!
Yes, the titration experiment requires us to obtain at least two titres that are within 0.1 cm^3 of each other. In some cases, a difference of 0.2 cm^3 is accepted, too but don't give yourself that leverage. You should record all three 3 readings but ensure that two of them are within 0.1 cm^3 of each other. Usually, no. But if you're confidently done with all of your paper correctly, you may perform one more for greater accuracy. That would be a rarity worth applause.

 

[VelaneDeBeaute]

^AoA!
Thankyou abcde!

 

[saleemsamer]

Hello Just wondering any guides to scoring high on P3 for chemistry available?

 

[abcde]

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

 

[OakMoon!]

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

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.

 

[abcde]

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.

I get 23. Thank you.
Regarding 39, is my counting flawed? There are six carbon atoms in each of the compounds. :S

 

[Gémeaux]

Regarding 39, is my counting flawed? There are six carbon atoms in each of the compounds. :S

no your counting is perfectly alryt there are six carbon atoms in each, including 3rd one.
we know that whatever the compound will be made will have to b formed by the combination of C3H7( the radical formed besides H). split each of the options 1,2 and 3 into C3H7. 1 and 2 can be split up easily as in the attached image, but for 3rd one its not possible.if you split at C-3 youre bound to include C-4 written in blue, as it is the branched end.

i noe its all written so random but i tried hope u get it.

 

[OakMoon!]

Yeah! Sorry. Dillusioned. You can't have a 4 carbon free radical from a propane molecule. And for the formation of 3 there has to be a 4 carbon free radical as shown below:

 

[abcde]

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?

 

[OakMoon!]

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?

You shouldn't look for the number of Hydrogen atoms increased. Instead, look for the double bonds removed. There were 9 double bonds initially that have been reduced to 4. So 5 double bonds have been removed. To hydrogenate one double bond we need one mole of hydrogen or 2 atoms. So 5 moles are required for 5 double bonds.

 

[mus123]

i want organic worksheets asap

 

[XPFMember]

Assalamoalaikum wr wb!

Check this out: Chemistry Application Booklet: Mistakes and Corrections!

 

[Pals_1010]

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

 

[XPFMember]

assalamoalaikum!

I'd prefer PCl3....no particular reason...i dont remembe why i chose this, but while doing pastpapers and seeing the mark schemes, i thought i better go for that!

all are correct though!

 

[Pals_1010]

Thank You

 

[Doctor Nemo]

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

The preferred reagent to use is supposedly SOCl2 because the by-products are both gases that can easily be removed from the product.

CH3CH2CH2CH3OH (l) + SOCl2 (l) -----> CH3CH2CH2CH2Cl (l) + HCl(g) + SO2 (g)

With the other reagents the by products are liquids that are more difficult to separate from the desired chloroalkane.

 

[Pals_1010]

Thank you Doctor Nemo

 

[Pals_1010]

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!

 

[Doctor Nemo]

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

 

[Pals_1010]

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

Thanks!!!