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CH4(g) + 2O2(g) -------> CO2(g) + 2H2O(l)A mixture of 10cm3 of methane and 10cm3 of ethane was sparked with an excess of oxygen. After cooling to room temperature, the residual gas was passed through aqueous potassium hydroxide. All gas volumes were measured at the same temperature and pressure. What volume of gas was absorbed by the alkali?
A 15 cm3
B 20 cm3
C 30 cm3
D 40 cm3
the answer is C. i keep getting 36 cm3. now i know that their reaction would produce carbon dioxide and water. then the gases remaining would be excess O2 and CO2 at room temp and pressure. i think only CO2 would be absorbed by the alkali right? I'm not getting 30, could someone help me please?
Eqn 1 : Sn2+ = Sn4+ + 2e-A solution of Sn2+ ions will reduce an acidified solution of MnO4 – ions to Mn2+ ions. The Sn2+ ions are oxidised to Sn4+ ions in this reaction. How many moles of Mn2+ ions are formed when a solution containing 9.5 g of SnCl 2 (Mr: 190) is added to an excess of acidified KMnO4 solution?
A 0.010
B 0.020
C 0.050
D 0.125
the answer is B. i dont get how?
CH4(g) + 2O2(g) -------> CO2(g) + 2H2O(l)
C2H6(g) + 3.5O2(g) --------> 2CO2(g) + 3H2O(l)
Residual gas is CO2 which is absorbed by KOH(aq)
CO2(g) + OH-(aq) ---------> CO3(-2)(aq) + H2O(l)
Volume of CO2 produced by CH4 and C2H6 is 10 and 20 cm^3 respectively, total volume of CO2 produced = 30 cm^3
thank you so very much!!Eqn 1 : Sn2+ = Sn4+ + 2e-
Eqn 2 : MnO4- + 8H+ + 5e- = Mn2+ + 4H2O
These equations are available in data booklet : http://www.cie.org.uk/images/164870-2016-specimen-data-booklet.pdf
Balance the number of electrons, SHOULD BE EQUAL IN BOTH THE EQN.
Eqn 1 : 5Sn2+ = 5Sn4+ + 10e-
Eqn 2 : 2MnO4- + 16H+ + 10e- = 2Mn2+ + 8H2O
Add Eqn 1 with Eqn 2 = 5Sn2+ + 2MnO4- + 16H+ = 5Sn4+ + 2Mn2+ + 8H2O
Ignore liquids, we are just taking ions into account.
2moles of MnO4- reacts with 5moles of Sn2+
2 Mn---> 5 Sn
0.4 Mn <- 1 Sn
Hence, 9.5/190 * 0.4 = 0.02mol
thank you so very much!!
Did the question ask for hot conc K2Cr2O7 or hot dilute K2Cr2O7 ?Q what will be the result of oxidising this compound with Hot conc K2cR2O7.
Acc to me there should be no change!! any help will be appreciated!
Q6 : From where did 1/5 and 4/5 came?Try to type out the corresponding answers next time.
Q4. The N has a -3 charge. So it has 10 electrons.
Q6. (1/5 x 10) + (4/5 x 11) = 10.8
Q17. Outer shell 7s2 7p6, total 8 outer electrons. Group 0.
Q18. Easiest is to use the data booklet and look at IE of the four options , IE 2 > IE 1 > IE3
Q27. From data booklet. Tc has 43 protons. In the question, the information is that it has 99 nucleons (protons + neutrons).
3 sigma bond ; sp2 hybridization = 120 degrees.How do we figure out the bond angle in this case?
3 sigma bond ; sp2 hybridization = 120 degrees.
You should know
sp = 180
sp2 = 120
sp3 = 109.5
Surely, A isn't impossible if u work smartly.Oh, riiiight. There is just so much to remember!! And I forget after two days, the little details.
Thanks for your help though, if I somehow manage to get an A, it'll be because of you guys helping me out here
I don't know the question
My friend just asked me what this compound resulted in under hot conc k2cr207
Apply the knowledge that C can form 4 bonds, O can 2 and H can 1 it will look like this :How do I know what the damn cyclic compound looks like? :'(
I am so DONE with chemistry. -_-
Q6 : From where did 1/5 and 4/5 came?
Q17,18 : I dont understand.
Q27 : IDK how statement 1 is correct.
Ty.
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