P1 MCQ's preparation thread for chemistry ONLY!!!!

[bogus]

need help with nov 2012 question 4 h4rriet

 

[A*(a*)]

need help with nov 2012 question 4 h4rriet

can you tell me how to tag please?

 

[!>@/^~^\@<!]

T

Q21. B) Citric acid (from top to bottom): The first carbon is bonded to 2 hydrogens, so it's not chiral. The second carbon is bonded to the top and bottom chains which are the same, so it's not chiral. The third carbon is bonded to 2 hydrogens, so it's not chiral.
Isocitric acid (from top to bottom): The first C is bonded to 2 hydrogens, so it's not chiral. The second carbon is bonded to the top chain, an H, a CO2H, and the bottom chain, so it is chiral. The third carbon is bonded to the upper part of the molecule, an OH, an H, and a CO2H, so it is chiral.
Conclusion: Citric acid has 0 chiral centres and isocitric acid has 2 chiral centres

Q22. D) P is an alkene and a primary alcohol. Alkenes are oxidized by hot concentrated KMnO4 to carboxylic acids with two carboxylic acid groups. Primary alcohols are oxidized to a carboxylic acid. This means that the products of this oxidation are CH3CH2CO2H (from primary alcohol oxidation) and HO2CCH2CO2H (from alkene oxidation)

Q23. B) Write down some equations for the complete combustion of simple alkanes so as to determine how the ratio of carbon atoms to moles of oxygen changes as there are more carbon atoms in the alkane.
CH4 + 2O2 → CO2 + 2H2O
2C2H6 + 7O2 → 4CO2 + 6H2O
C3H8 + 5O2 → 3CO2 + 4H2O
The ratio of carbon atoms to moles of oxygen in each equation respectively is:
1:2, 2:7, 3:5
The number of carbon atoms is directly proportional to the number of moles of oxygen, so the answer has to be B

Q24. B) When an alkene reacts with cold liquid bromine, dibromoalkene is formed. Since the compound above is made up of two alkenes, then four bromine atoms are added on to the molecule. This reaction is an electrophilic addition reaction. Each bromine atom adds on to a carbon engaged in the double bond. The answer is 1,3,4,6 - tetrabromocyclohexane


Q25. D) Reaction D is a nucleophilic addition (CN- is the nucleophile)

Q26. A) When a halogenoalkanes reacts with KCN in ethanol, a nucleophilic substitution reaction takes place, in which the CN- ions substitute the halogen atoms (in this case the bromine atoms from 1,4-dibromobutane)


Q27. C) The weakest bond is the C-Cl bond. The C-H bond is very strong because it is an ionic bond. The C-F bond is strong because fluorine is very electronegative. The C-Cl bond however, is covalent and Cl is not as electronegative as Fluorine, which means that it is a relatively weak bond (relative to the other bonds in the molecule). This is why in the reaction, the most susceptible atom to leave the molecule is chlorine, and so the radical in option "C" is formed

Q28. D) When an alcohol reacts with sodium an alkoxide ion is formed (O-Na+). When a carboxylic acid reacts with sodium, since it's an acid, a salt is formed (COONa). The only compound that forms one mole of hydrogen is the last one.
CH3CH(OH)CO2H + 2Na → CH2CH(ONa)COONa + H2

Q29. C) An alcohol and concentrated sulphuric acid under reflux will produce an alkene which can be purified by dilute sodium hydroxide (base hydrolysis of esters).

Q30. D) When propanone reacts with hydrogen cyanide, nucleophilic addition takes place and a hydroxynitrile is formed.
CH3COCH3 + HCN → CH3C(OH)CH3CN
2-hydroxybutanenitrile is hydrolysed under acidic conditions to Butanoleic acid
CH3C(OH)CH3CN + H2O + H+ → (CH3)2C(OH)CO2H + NH4

Q31. C) Silicon tetrachloride doesn't have co-ordinate bonding because it follows the octet rule, sharing all of its valence electrons with the chlorine atoms.
Both silicon and chlorine are non-metals, so, it has covalent bonding.
There are instantaneous dipole-induced dipole forces between the molecules (Van der Waals forces)

Q32. A) The right-hand side of the structure is polar and since water is a dipole, it is attracted to water.
The alkyl chain is non-polar and attracted to other alkyl chains by Van der Waals forces. Since oil is of a similar character to this alkyl chain, the alkyl chain is soluble in oil droplets.
In alkanes, each carbon atom forms a tetrahedral structure (due to sp3 hybridisation), so the C-C-C bond angles are tetrahedral

Q33. A) The reaction is endothermic, which means that diamond has more energy than graphite. The enthalpy change of atomisation is the enthalpy change when one mole of gaseous atoms is formed from one mole of the element in the standard state. Since diamond has more energy than graphite, it requires a smaller enthalpy change of atomisation.
Since the enthalpy change of atomisation is smaller in diamonds, it means that the C-C bonds in diamond are weaker than in graphite because it requires less energy to change into the gaseous state.
Since diamond has more energy than graphite, there is a higher energy requirement to break the C-C bond to form new C=O bonds (in carbon dioxide) in combustion.


Q34. C) The electronegativity difference decreases between the elements (3.05, 2.13, 1.43, 0.65)
All of the compounds fulfill the octet rule and are isolelectronic.
The compounds become increasingly covalent (starting from ionic)

Q35. B) Sulphur dioxide is a reducing agent which prevents oxidation.
Since it's an anti-oxidant, it prevents alcohols from oxidizing to carboxylic acids (prevents sour-tasting acids).
It does smell and is toxic in large quantities.

Q36. C) Iodide ions are strong reducing agents and so they reduce the sulphuric acid, first to sulphur dioxide, then to sulphur and then hydrogen sulphide. Barely any hydrogen iodide is formed because it's displaced by the sulphuric acid.
Iodide ions are reducing agents, and become oxidised to iodine.
The majority of the products of the reaction are sulphur compounds (as explained above)

Q37. B) A chiral centre is an atom bonded to four different groups.
An optical isomer (geometric isomer) occurs when there's a chiral centre.
Chiral carbon atoms DO NOT need to have structural isomers.


Q38. B) Step X is a nucleophilic substitution because the reagent is hot aqueous sodium hydroxide (OH- being the nucleophile).
A chloroalkane cannot be formed by reacting sodium chloride with alcohol. It can be done with phosphorus (III) chloride or phosphorus (V) chloride.

Q39. C) Only an aldehyde forms a brick-red precipitate with Fehling's solution. Aldehydes are formed by the oxidation (in acidified dichromate) of primary alcohols. The two primary alcohols are CH3CH2CH2OH and CH3OH

Q40. B) It only has one chiral carbon (The one in the middle).
It has a carboxylic acid group, so it can be esterified by ethanol. It has an OH group, so it can be esterified by ethanoic acid.
The molecule contains tertiary and primary alcohols, not secondary.

P.S :- Others please post work solutions for all P1 yearly like this..

Thanks a lot!!!

 

[KWIKIW]

16. Yes it does, differently with cold dilute NaOH and hot concentrated NaOH. It's an equation we have to memorise. With cold dilute NaOH you get HClO and with hot concentrated NaOH you get HClO3. You also got NaCl and H2O in both.
32. Because the alkyl chain is hydrophobic, and oil too is hydrophobic.
20. The trans isomers you drew are the same! And there can be more than 1 cis and trans isomer depending on the groups attached. I'll draw all the cis and trans isomers and upload them in a sec. For the last one: change the positions of the Cl and H atoms to get different structural isomers.

Thanks.

 

[bogus]

can you tell me how to tag please?

you press @ and then write down the name like A*(a*)

 

[haha101]

can someone tell mei which variants from 11 12 13 are similar and which are different ?

 

[A n g e l a ♥]

Year/variant what?

v 1 year 2009

 

[ahmed abdulla]

1__Two springs P and Q both obey Hooke’s law. They have spring constants 2k and k respectively.
The springs are stretched, separately, by a force that is gradually increased from zero up to a
certain maximum value, the same for each spring. The work done in stretching spring P is WP,
and the work done in stretching spring Q is WQ.
How is WP related to WQ?
A WP =
 1/4 WQ B WP =1/2
 WQ C WP = 2WQ D WP = 4WQ
i get C ... but the answer is B


2___A wave of amplitude a has an intensity of 3.0Wm–2.
What is the intensity of a wave of the same frequency that has an amplitude 2a
A_12 .. i used I>A^2 and still not coming

 

[ahmed abdulla]

?

 

[Mairaxo]

1__Two springs P and Q both obey Hooke’s law. They have spring constants 2k and k respectively.
The springs are stretched, separately, by a force that is gradually increased from zero up to a
certain maximum value, the same for each spring. The work done in stretching spring P is WP,
and the work done in stretching spring Q is WQ.
How is WP related to WQ?
A WP =
 1/4 WQ B WP =1/2
 WQ C WP = 2WQ D WP = 4WQ
i get C ... but the answer is B


2___A wave of amplitude a has an intensity of 3.0Wm–2.
What is the intensity of a wave of the same frequency that has an amplitude 2a
A_12 .. i used I>A^2 and still not coming

Q1- P is 2k and Q is k. For the same force applied, P will have extension of 2X and Q will have extension X (as k and extension are directly proportional when F is constant) so work done=Force*distance. they have same force so for P work done=F*2X and Q=F*X. hence work done by P will be twice of that of Q. 2WP=WQ and substituting you get WP=1/2WQ
Q2- Use the formula I1/I2 = A1^2/A2^2
then u get 3/I=a^2/(2a)^2
3/I=a^2/4a^2
cross multiplying u get 12a^2=a^2*I
I=12a^2/a^2
a^2 gets cancelled and u get 12

 

[ahmed abdulla]

Q1- P is 2k and Q is k. For the same force applied, P will have extension of 2X and Q will have extension X (as k and extension are directly proportional when F is constant) so work done=Force*distance. they have same force so for P work done=F*2X and Q=F*X. hence work done by P will be twice of that of Q. 2WP=WQ and substituting you get WP=1/2WQ
Q2- Use the formula I1/I2 = A1^2/A2^2
then u get 3/I=a^2/(2a)^2
3/I=a^2/4a^2
cross multiplying u get 12a^2=a^2*I
I=12a^2/a^2
a^2 gets cancelled and u get 12

thanks ... but some one expained it before u

 

[ahmed abdulla]

Q1- P is 2k and Q is k. For the same force applied, P will have extension of 2X and Q will have extension X (as k and extension are directly proportional when F is constant) so work done=Force*distance. they have same force so for P work done=F*2X and Q=F*X. hence work done by P will be twice of that of Q. 2WP=WQ and substituting you get WP=1/2WQ
Q2- Use the formula I1/I2 = A1^2/A2^2
then u get 3/I=a^2/(2a)^2
3/I=a^2/4a^2
cross multiplying u get 12a^2=a^2*I
I=12a^2/a^2
a^2 gets cancelled and u get 12

help me with this .... drawing on it will be good ...i never understand from where the "O " of CHO comes ... i think i should leave it in my exam if this question comes

 

[Mairaxo]

help me with this .... drawing on it will be good ...i never understand from where the "O " of CHO comes ... i think i should leave it in my exam if this question comes

check this example tell me if i should elaborate

 

[ahmed abdulla]

check this example tell me if i should elaborate

thAt the PERFECT way ... thanks ... now i will see it i can do others

 

[Mairaxo]

thAt the PERFECT way ... thanks ... now i will see it i can do others

your welcome! let me know if u need help in any other

 

[itallion stallion]

can someone plz help me with this papers(http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w09_qp_11.pdf) question no 14,21,31.

 

[Alice123]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s03_qp_1.pdf
no 26, 27, 35, 38
your help will be highly appreciated....
Thanks

 

[h4rriet]

need help with nov 2012 question 4 h4rriet

Variant what?

 

[Rajvinder]

Guys I'm sitting for chemistry p1 on the 11th and haven't really done much prep.Doing prep after my bio p5 paper.I was thinking off finishing rereading on Wednesday and then just spam questions. How many papers should I try to do per day and from what year? Isit too late to try aiming for 35/40?

 

[h4rriet]

1__Two springs P and Q both obey Hooke’s law. They have spring constants 2k and k respectively.
The springs are stretched, separately, by a force that is gradually increased from zero up to a
certain maximum value, the same for each spring. The work done in stretching spring P is WP,
and the work done in stretching spring Q is WQ.
How is WP related to WQ?
A WP =
 1/4 WQ B WP =1/2
 WQ C WP = 2WQ D WP = 4WQ
i get C ... but the answer is B


2___A wave of amplitude a has an intensity of 3.0Wm–2.
What is the intensity of a wave of the same frequency that has an amplitude 2a
A_12 .. i used I>A^2 and still not coming

1. Use WD=1/2Fe an find e using F/k. WP will be F/4k and WQ F/2k. F/2k multiplied by 1/2 will give F/4k.
2. (A^2)/3 = (2A)^2/x.