Can anyone tel the differnce b/w OH of alcohal and OH of alkali? PLEASE..!
Plx...!
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Can anyone tel the differnce b/w OH of alcohal and OH of alkali? PLEASE..!
OH is a bond at the alcohol moleculeCan anyone tel the differnce b/w OH of alcohal and OH of alkali? PLEASE..!
Q8)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
Assuming the volume of the mixture does not change after addition of CuSO4.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
help dudeAssuming 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.
Firstly, find charge.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.
Thanks a lot!!!!!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 man..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.
Thank you very muchQ2) 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
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?
can u explain question 14 in more detailed pls ?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.
what is that troubles u.. any part of this which u cant understand well one.can u explain question 14 in more detailed pls ?
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