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Chemistry: Post your doubts here!

DivinoDD

DivinoDD

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

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Namehere

Namehere

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Jinosupreme said:
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
Click to expand...

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.
 
J

Jinosupreme

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Namehere said:
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.
Click to expand...

Love your explanation! Thanks :D
 
sandwich147

sandwich147

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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 HELPcapture-20130913-191417-horz.jpgcapture-20130913-191417-horz.jpg
 
J

Jaspreet Dhaliwal

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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???
 
K

KZW

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sandwich147 said:
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
Click to expand...


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

Namehere

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Jaspreet Dhaliwal said:
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???
Click to expand...

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
 
adithyaXX

adithyaXX

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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

adithyaXX

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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 ..
 
Namehere

Namehere

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SIstudy said:
can sum1 explain me the shielding effect of inner electrons?
Click to expand...

Shielding effect plays a role in ionisation energies and therefore in reactivity. Shielding usually decreases the effective nuclear force, which makes the outermost electrons experience less of an attraction meaning the electron can be removed with more ease. Shielding effect increases as you go down the group as you have increasing number of energy levels, whereas going across the period, shielding effects usually remain rather constant.
In exam questions they usually ask for a general trend, e.g. Why is there a general increase in the first ionisation energy as you go across the period? or they ask you to compare e.g. Why is the first ionisation energy of phosphorus greater than that of sulfur? or Why is the first ionisation energy of of magnesium greater than that of aluminium?
When answering those questions you should include: distance from the nucleus, effective nuclear charge (causes the atomic radius to decrease, meaning ionisation energies are higher since the electrons are nearer to the nucleus) , increasing nuclear charge, shielding effects and interelectronic repulsion (from paired electrons).
If you know the answers to those questions you should be ok for that section.

Hope it helps.
 
Adiizz

Adiizz

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Asalamualaikum!
I need help for qu 2 in novemeber 2003 paper 1
A garden fertiliser is said to have a phosphorous content of 30.0% 'P2O5 is soluble in water'
What is the percentage by mass of phosphorous in the fertiliser?
A. 6.55%
B. 13.1%
C. 26.2%
D. 30.0%
 
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