Chemistry: Post your doubts here!

[Xeshan16]

Can anyone tel the differnce b/w OH of alcohal and OH of alkali? PLEASE..!

Plx...!

 

[memorypriest]

Can anyone tel the differnce b/w OH of alcohal and OH of alkali? PLEASE..!

OH is a bond at the alcohol molecule

OH- is an ion from a alkali (eg. NaOH)

 

[Beaconite007]

http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_w02_qp_1.pdf Q.8 Q.11 Q.24 Q.38 PLZ HELP ME ANY ONE

 

[NokiaN95638]

Plz help me with dis questions:

http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_w10_qp_12.pdf

Q.no: 7,8,13,38

 

[beeloooo]

link for the chemistry topical past paper questions ?

 

[memorypriest]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w02_qp_1.pdf Q.8 Q.11 Q.24 Q.38 PLZ HELP ME ANY ONE

Q8)
A 2 g sample of hydrogen at temperature T and of volume V exerts a pressure p.
Deuterium, 2/1 H, is an isotope of hydrogen.
Which of the following would also exert a pressure p at the same temperature T ?
A 2g of deuterium of volume V
B 4g of deuterium of volume V/2
C a mixture of 1 g of hydrogen and 2 g of deuterium of total volume V
D a mixture of 2 g of hydrogen and 1 g of deuterium of total volume 2V

First find the amount of hydrogen from 2g of hydrogen.
amount of hydrogen = 2/1 = 2mol

Hence, 2mol of gas occupy volume V at temperate T exerts a pressure p.

Option A: 2g of deuterium (1 mol of gas) with volume V. (WRONG)
Option B: 4g of deuterium (2 mol of gas) with volume V/2. (WRONG)
Option C: mixture of 1 g of hydrogen (1 mol of gas) and 2g of deuterium (1mol gas) with volume V (CORRECT)
Option D: 2g of hydrogen (2mol), 1g of deuterium (0.5mol) total volume 2V. (WRONG) 2.5mol of gas would occupy 1.25V instead.

Q11) D
Equilibrium mixture will contain...
HI (b-x)
H2 (x/2)
I2 (x/2)

Kp = [(x/2)(x/2)] / (b-x)^2
= x^2 / 4(b-x)^2

Q24) B
Step 1 : Nucleophilic substitution (Sub Br away with OH)
Step 2 : Oxidation (From secondary alcohol oxidised to Ketone - Lose 2 H)
Choose a secondary halogenoalkane

Q38)
Catalytic cracking is not in my syllabus. Sorry.

 

[Jspake]

Please help me solve the enthalpy question below.. I am really confused!

Copper(II) sulfate is soluble in water. A student dissolved 25g of copper(II) sulfate in 100cm³ of water in a polystyrene beaker stirring all the time. The temperature of water fell by 2.9 °C.
(i) Calculate the enthalpy change of solution of copper(II) sulfate. (Specific heat capacity of water = 4.18 J/g/°C ; relative molecular mass of copper(II) sulfate = 249.7 g/mol).
(ii) Suggest one source of error in this experiment and explain how the error affects the results

Thanks

 

[JulyMei]

http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_s10_qp_53.pdf



Anyone know how to draw this???

Thanks in advance..

 

[memorypriest]

Please help me solve the enthalpy question below.. I am really confused!

Copper(II) sulfate is soluble in water. A student dissolved 25g of copper(II) sulfate in 100cm³ of water in a polystyrene beaker stirring all the time. The temperature of water fell by 2.9 °C.
(i) Calculate the enthalpy change of solution of copper(II) sulfate. (Specific heat capacity of water = 4.18 J/g/°C ; relative molecular mass of copper(II) sulfate = 249.7 g/mol).
(ii) Suggest one source of error in this experiment and explain how the error affects the results

Thanks

Assuming the volume of the mixture does not change after addition of CuSO4.

Heat absorbed = mc DeltaT
= 100 x 4.18 x 2.9
= 1212.2J

Amount of CuSO4 = 25/159.6
= 0.156641604 mol

Enthalpy change of solution of CuSO4 = Heat absorbed / Amount of CuSO4
= +1212.2 / 0.156641604
=+7738.6848 Jmol-1
= +7.74 kJmol-1 (3s.f)

ii) heat gained from surrounding. This will decrease the value of Delta T and subsequently decrease the value of enthalpy change of solution.

 

[snowbrood]

what is the difference between R-MgBr and R"MgBr my book says that R-MgBr reacts with R2CO then H+ and forms R3COOH but when the same R2CO then H+ reacts with R"MgBr it forms R-OH

 

[snowbrood]

Assuming the volume of the mixture does not change after addition of CuSO4.

Heat absorbed = mc DeltaT
= 100 x 4.18 x 2.9
= 1212.2J

Amount of CuSO4 = 25/159.6
= 0.156641604 mol

Enthalpy change of solution of CuSO4 = Heat absorbed / Amount of CuSO4
= 1212.2 / 0.156641604
=7738.6848 Jmol-1
= 7.74 kJmol-1 (3s.f)

ii) heat gained from surround. This will decrease the value of Delta T and subsequently decrease the value of enthalpy change of solution.

help dude

 

[Shuayb]

Could somebody do this one for me:
Calculate the volume of O2 evolved at s.t.p when 2A(ampere) is passed thru dilute H2SO4 for 1830 seconds using inert electrodes.

 

[memorypriest]

snowbrood Grinard reagent is not in my syllabus, can't help.

Could somebody do this one for me:
Calculate the volume of O2 evolved at s.t.p when 2A(ampere) is passed thru dilute H2SO4 for 1830 seconds using inert electrodes.

Firstly, find charge.
Q = It = 2 x 1830 = 3660 C

Secondly, find amount of electrons
Amount of electrons = Q/96500 = 3660/96500 = 0.0379274611 mol

Thirdly, construct the Oxidation half equation
2H2O --> O2 + 4H+ + 4e

Next, find amount of O2 generated
Ratio of O2 : electron is 1 : 4
Amount of O2 = 0.0379274611 / 4 = 0.0094818653mol

Find volume of O2. (1mol of gas occupy 22.4dm3 under standard temperate pressure s.t.p)
Vol of O2 = 0.0094718653 x 22.4 = 0.212 dm3 (3 s.f)

 

[Shuayb]

snowbrood Grinard reagent is not in my syllabus, can't help.


Firstly, find charge.
Q = It = 2 x 1830 = 3660 C

Secondly, find amount of electrons
Amount of electrons = Q/96500 = 3660/96500 = 0.0379274611 mol

Thirdly, construct the Oxidation half equation
2H2O --> O2 + 4H+ + 4e

Next, find amount of O2 generated
Ratio of O2 : electron is 1 : 4
Amount of O2 = 0.0379274611 / 4 = 0.0094818653mol

Find volume of O2. (1mol of gas occupy 22.4dm3 under standard temperate pressure s.t.p)
Vol of O2 = 0.0094718653 x 22.4 = 0.212 dm3 (3 s.f)

Thanks a lot!!!!!

 

[Jspake]

Assuming the volume of the mixture does not change after addition of CuSO4.

Heat absorbed = mc DeltaT
= 100 x 4.18 x 2.9
= 1212.2J

Amount of CuSO4 = 25/159.6
= 0.156641604 mol

Enthalpy change of solution of CuSO4 = Heat absorbed / Amount of CuSO4
= +1212.2 / 0.156641604
=+7738.6848 Jmol-1
= +7.74 kJmol-1 (3s.f)

ii) heat gained from surrounding. This will decrease the value of Delta T and subsequently decrease the value of enthalpy change of solution.

Thanks a lot man..

 

[cowarrior]

Q2) B
1 of the side chain - 1 double bond gets saturated
2 of the side chain - 2 double bond gets saturated x 2 = total of 4 double bond saturated
1+4 = 5

Q6) C
Ideal gas equation work best for gases at High temperature and Low pressure

Q8) A

Q11) A
There is higher % of dissociation of H+ in NH3, thus more acidic.

Q13) C
Pick 1 acidic oxide and 1 alkaline oxide

Q32) C

Q33) A

Thank you very much

 

[izzahzainab]

http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_w06_qp_1.pdf

Can anyone explain me question 32? an 24 (Oxygen is a flammable gas I thought :/)
and Q14?

 

[snowbrood]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_1.pdf

Can anyone explain me question 32? an 24 (Oxygen is a flammable gas I thought :/)
and Q14?

well u got to know that inflammable is the inability to burn. oxygen dont burn itself it just accelerates combustion. flammable means a reaction in which oxygen combines with the substance that is flammable how can oxygen combine with itself hope that makes sense...

for Q14
well balance equation for decomposition for any group 2 nitrate decomposition 2X(NO3)2(s) = 2XO(s) + 4N02(g )+O2(g) (where X is a metal from group 2)
masses of these gases combine account for the total mass lost i.e 3.29. the remaining mass is that of XO 1.71.
5g of nitrate decompose into 1.71 XO so 1g of nitrate will decompose into 0.342=(1.71/5).
now 2X+2(16)/2X+4(14)+12(16) =0.342

now put in the value of calcium i.e 40 in X and it would satisfy the equation.

 

[iKhaled]

well u got to know that inflammable is the inability to burn. oxygen dont burn itself it just accelerates combustion. flammable means a reaction in which oxygen combines with the substance that is flammable how can oxygen combine with itself hope that makes sense...

for Q14
well balance equation for decomposition for any group 2 nitrate decomposition 2X(NO3)2(s) = 2XO(s) + 4N02(g )+O2(g) (where X is a metal from group 2)
masses of these gases combine account for the total mass lost i.e 3.29. the remaining mass is that of XO 1.71.
5g of nitrate decompose into 1.71 XO so 1g of nitrate will decompose into 0.342=(1.71/5).
now 2X+2(16)/2X+4(14)+12(16) =0.342

now put in the value of calcium i.e 40 in X and it would satisfy the equation.

can u explain question 14 in more detailed pls ?

 

[snowbrood]

can u explain question 14 in more detailed pls ?

what is that troubles u.. any part of this which u cant understand well one.
Step 1
u have to first balance a equation for anhydrous nitrate which has a metal ion from group 2 .
X here is the metal from group 2 as this is anhydrous so no water vapor will be released (2XNO3)2(s) = 2XO(s) + 4N02(g )+O2(g) (where X is a metal from group 2). note that this nitrate equation applies for every group 2 metal nitrate that is anhydrous
Step 2.
u know that substance loses mass after reaction only if there is a product including a gas. in this case that is nitrate and oxygen.
step 3
question says that 5g of this nitrate loses 3.29g(this includes the total mass of both these gases) thus u come to conclude that 5g of that substance will give 1.71(5-3.29) as u know that the total mass of reactants and products is the same.
step 4
1g of nitrate will give (1.71/5)=0.342g of that oxide ok
step5
now make a equation
2X+2(16)/2X+4(14)+12(16)=0.342.
now plug in X calcium magnesium and the other two compounds in the equation and see which satisfies the equation.
if i put calcium 2(40)+2(16)/2(40)+4(14)+12(16) u will see it will equal to 0.342 which means calcium is the right option.
if u put magnesium 2(24)+2(16)/2(24)+4(14)+12(16) this will equal to 0.27 see