Chemistry: Post your doubts here!

[danial 234]

5) the ideal gas law states that the gas molecules experience no forces of repulsion or attraction. HCl is opposite to this cz it is a large molecule n it has hydrogen bonding. so it is most likely to deviate. the remainig options only experience van der waals which is nt significant

 

[Jinosupreme]

Can anyone help me?
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s06_qp_1.pdf

Q18, 20, 33, 34, 35, 36

18C
20C
33D
34C
35D
36B

Really need help thanks :00

 

[danial 234]

18)) SO2 is acidic n will react with basic CaO. n will be neutralized.
33) statement 1 is ryt. statement 2 is wrong cz there is no triple bond btween Nitrogen. the third is also wrong cz the state of hydrazine doesnt affect.
34) MgCl2 conducts electricity in liwuid state also. while the remaining disassociate only in H20 n conduct electricity

 

[danial 234]

36) NH3 reduces Cl2. nd it is also a base. but the valency of H remains same.(+1)
35) Sr is just below the Ca .so hs almost same size. but both oxides r ionic n cn dissolve/ppt in H20. Hydroxyapatite doesnt contain metallic bond

 

[Namehere]

Can anyone help me?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s06_qp_1.pdf

Q18, 20, 33, 34, 35, 36

18C
20C
33D
34C
35D
36B

Really need help thanks :00

Q20- C -> A Nucleophilic addition reaction usually occurs when there is a C double bond, in this case at the second stage of the reaction the aldehyde (contains double bond) is joined with an oxygen, due to their difference in electronegativity a dipole is set up, leaving the carbon atom with a partial positive charge, allowing nucleophiles (negatively charged species, in this case ethanol due to the dipole in the O-H end) to be attracted to it.

Hope it helps.

 

[KZW]

Can anyone help me?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s06_qp_1.pdf

Q18, 20, 33, 34, 35, 36

18C
20C
33D
34C
35D
36B

Really need help thanks :00

18 ; CaO + SO2 -> CaSO3. You will need to learn it.

20 ; The aldehyde in the second reaction contains a partially positive carbon due to the oxygen bonded to it. The dipole set up makes it vulnerable to nucleophiles (The OH from ethanol).

33 ; Having high activation energy is fairly obvious. You probably got tripped up by 2. Hydrazine does not have a triple bonded nitrogen

34 ; All aqueous solutions will conduct. However, MgCl2 is an ionic solid. It conducts in a liquid state, so it should be excluded, hence only 2 and 3 are right.

35; Calcium hydroxide is less soluble than Strontium hydroxide due to the higher charge density, so 2 is eliminated. 3 should be eliminated, as it is fairly obvious why, and thus D is the only possible answer.

36; 2 is correct, as ammonia is a proton acceptor in this reaction. 1 is correct;8NH3 -> N2 + 6NH4. NH3 reduces itself to NH4, hence answer is B.

Hope it helps

 

[DivinoDD]

Those who need help in Geometric Isomerism Hope this help you, guys!

 

[Jinosupreme]

Hey, need help in
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w07_qp_1.pdf

Thanks!

Q9, 35, 37
9B
35B
37B

 

[A star]

can any one share a complete LABELED diagram of standard electrode potential?

 

[Namehere]

Hey, need help in
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_1.pdf

Thanks!

Q9, 35, 37
9B
35B
37B

Q9 - B ---> Here you have to make two redox reactions (the one they give to you plus the half-equation for the metal ion in the salt) and balance them. The metalic salt reaction would be something like: 2X^3+ + 4e^- --> 2X^x (equation (1) )
The balance: in the equation of the sulphite ion you lose two electrons, so in the metalic salt reaction you should gain two electrons, however, the question says that you need 50cm^3 of a metallic salt to react exactly with 25cm^3 of aqueous sodium sulphite, so you need double the number of moles for the metalic salt reaction, hence the 2 in front of the X and the 4 electrons (2x2). From a mathematical point of view, you could say that conc is constant, therefore say that
n1/vol1 = n2/vol2 (n1 & vol1 stand for the equation 1 above and the 2 for the equation they gave to us). Since vol1 =50 and vol2= 25, in other word vol1=2 and vol2 = 1... cross multiply and cancel out vol... 2n1 = n2. The charge on the ion salt would then be determined using the equation 1: 6^+ + 4^- = answer = 2^+

Q35 - B ---> The reaction happening here is the following: Mg + Ba(NO3)2 --> Mg(NO3)2 + Ba. Since we are in a firework, the temperatures would be pretty high, so the Ba will react with the oxygen in the area to form BaO. Mg(NO3)2... Mg is second in Group 2 so the thermal stability of Mg(NO3)2 is rather low, therefore it will decompose into 2Mg(NO3)2 ---> 2MgO + 4NO2 + O2. Since the question asks for the solid products the only solid products are BaO and Mgo.

Q37 - B ---> 1) 2CaO and 2NH4Cl ----> CaCl2 + Ca(OH)2 + 2NH3 ; in this case produces ND3. For the Second one I´m not sure how it happens so that it produces ND3, so won´t say anything in case I confuse someone.

Hope this helps.

 

[Jinosupreme]

Q9 - B ---> Here you have to make two redox reactions (the one they give to you plus the half-equation for the metal ion in the salt) and balance them. The metalic salt reaction would be something like: 2X^3+ + 4e^- --> 2X^x (equation (1) )
The balance: in the equation of the sulphite ion you lose two electrons, so in the metalic salt reaction you should gain two electrons, however, the question says that you need 50cm^3 of a metallic salt to react exactly with 25cm^3 of aqueous sodium sulphite, so you need double the number of moles for the metalic salt reaction, hence the 2 in front of the X and the 4 electrons (2x2). From a mathematical point of view, you could say that conc is constant, therefore say that
n1/vol1 = n2/vol2 (n1 & vol1 stand for the equation 1 above and the 2 for the equation they gave to us). Since vol1 =50 and vol2= 25, in other word vol1=2 and vol2 = 1... cross multiply and cancel out vol... 2n1 = n2. The charge on the ion salt would then be determined using the equation 1: 6^+ + 4^- = answer = 2^+

Q35 - B ---> The reaction happening here is the following: Mg + Ba(NO3)2 --> Mg(NO3)2 + Ba. Since we are in a firework, the temperatures would be pretty high, so the Ba will react with the oxygen in the area to form BaO. Mg(NO3)2... Mg is second in Group 2 so the thermal stability of Mg(NO3)2 is rather low, therefore it will decompose into 2Mg(NO3)2 ---> 2MgO + 4NO2 + O2. Since the question asks for the solid products the only solid products are BaO and Mgo.

Q37 - B ---> 1) 2CaO and 2NH4Cl ----> CaCl2 + Ca(OH)2 + 2NH3 ; in this case produces ND3. For the Second one I´m not sure how it happens so that it produces ND3, so won´t say anything in case I confuse someone.

Hope this helps.

Love your explanation! Thanks

 

[sandwich147]

Hi guys, need help with June 2013 paper 41, question no. 2 (ii), (iii) & (iv). I've checked the marking scheme but can't understand it. NEED URGENT HELP

 

[Jaspreet Dhaliwal]

Can someone plz xplain this question:
A student mixed 25cm3 of 0.10mol dm–3 sodium hydroxide solution with 25cm3 of 0.10mol dm–3 hydrochloric acid and noted a temperature rise of 2.5°C.
What is the enthalpy change of the reaction per mole of NaOH?
A –209 kJ mol–1
B –104.5 kJ mol–1
C –209 J mol–1
D –522.5 J mol–1
Help please???

 

[KZW]

Hi guys, need help with June 2013 paper 41, question no. 2 (ii), (iii) & (iv). I've checked the marking scheme but can't understand it. NEED URGENT HELPView attachment 31767View attachment 31767

ii) 1st order reaction means that the length of half lives are constant. So for "construction lines" just draw lines from y= 0.1 mol/dm3 , y= 0.05 mol/dm3 , y= 0.025 mol/dm3 to the curve, and their respective x values. The length of time between each subsequent half life value should be constant, therefore proving it is a first order reaction.

iii) I was really confused until I had a look at the full question. You need to do part i) plotting the values with 0.2mol/dm3 HCl.
Doing the same thing I explained for part ii, you'll find that the half lifes are double that of when the experiment was carried out with 1mol/dm3 HCl.
Therefore the order of reaction with respect to HCl is 1.

iv) From the first 2 parts, we can deduce that rate = k[HCl][CH3CO2CH2CH3]
To find k, we just have to find the initial rate via drawing a tangent and finding the gradient of that tangent (you can use either of the 2 graphs), then pop in the numbers for the concentration of HCl and CH3CO2CH2CH3 depending on which graph you used.

k= [HCl][CH3CO2CH2CH3]

Ask if you still don't understand, and I'll go through the full working with you.

 

[Namehere]

Can someone plz xplain this question:
A student mixed 25cm3 of 0.10mol dm–3 sodium hydroxide solution with 25cm3 of 0.10mol dm–3 hydrochloric acid and noted a temperature rise of 2.5°C.
What is the enthalpy change of the reaction per mole of NaOH?
A –209 kJ mol–1
B –104.5 kJ mol–1
C –209 J mol–1
D –522.5 J mol–1
Help please???

Here you have to use Q = mcΔt
Therefore Q=(25+25)(4.18)(2.5)
Q = -522.5 J mol-1 (Q is negative because heat is given off, shown by the rise in temperature - the reaction is exothermic.) D is the answer

 

[DivinoDD]

can any one share a complete LABELED diagram of standard electrode potential?

Sure, check this http://www.docbrown.info/page07/addhoc07/cell2.gif and also http://chemwiki.ucdavis.edu/@api/deki/files/6476/=Galvanic.jpg

 

[adithyaXX]

Can some one tell me which concepts to read in order for A level chem (acc to mark weightage )... im puzzled and i have my a levels this oct ... please guys i want to be a doctor ..

 

[adithyaXX]

Can some one tell me which concepts to read in order for A level chem (acc to mark weightage )... im puzzled and i have my a levels this oct ... please guys i want to be a doctor ..

 

[Lashan Ali]

can anybody give me a link of all the topically compiled pastpapers for chemistry paper 4.

 

[SIstudy]

can sum1 explain me the shielding effect of inner electrons?