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can u plz explain why not Mn2+ instead of Mno4- cuz in the question he asked for mn2+
and this one also plzz shouldnt X hv lone pairs so one CL will join with lone pairs so it will be XCL4
You must write Mn2+ instead of MnO4- in the working. Here it is not affecting the answer coz both have the same number of moles.can u plz explain why not Mn2+ instead of Mno4- cuz in the question he asked for mn2+
If Cl joins with the lone pair so it will have 9 electrons in its outer shell(7 its own and 2 of X). Element X will have 5 electrons, so 3 chlorine atoms will join with it so that all of them have an octet configuration.and this one also plzz shouldnt X hv lone pairs so one CL will join with lone pairs so it will be XCL4
A 0.216g sample of an aluminium compound X reacts with an excess of water to produce a single hydrocarbon gas. This gas burns completely in O2 to form H2O and CO2 only. The volume of CO2 at room temperature and pressure is 108 cm3.
What is the formula of X?
A Al2C3 B Al3C2 C Al3C4 D Al4C3
Can someone help me with this Q, the ans is D
28 Compound X will decolourise a warm acidified solution of manganate(VII) ions and forms orange
crystals on reaction with 2,4-dinitrophenylhydrazine.
What is X?
A CH3CH=CHCH2OH
B CH3COCH2CH3
C CH3CH2CH2CHO
D CH3CH(OH)CH2CO2H
And why?
Can someone please explain Q40 to me? The answer is B
Thanks for replying!I don't know weather i am 100% right or not.....But in statement 1 for the formation of bromoethane it requires distillation at the end of the process and C2H5Br is collected in ice cooled beaker, and in statement 2 formation of ethanal requires gentle heat and distillation.Thus b is correct.
In statement 3 formation of 1,2-dibromoethane, from bromine and ethene does not require distillation...Hope it helps.
I don't know weather i am 100% right or not.....But in statement 1 for the formation of bromoethane it requires distillation at the end of the process and C2H5Br is collected in ice cooled beaker, and in statement 2 formation of ethanal requires gentle heat and distillation.Thus b is correct.
In statement 3 formation of 1,2-dibromoethane, from bromine and ethene does not require distillation...Hope it helps.
Can someone please explain Q40 to me? The answer is B
K Thx!For statement 3, we need to be aware that ethene is a gas at room conditions, and will not be contained in the distillation flask for the reaction to take place.
9701_w12_qp_12
23 The cracking of a single hydrocarbon molecule, CnH2n+2, produces two hydrocarbon molecules
only. Each hydrocarbon product contains the same number of carbon atoms in one molecule.
Each hydrocarbon product has non-cyclic structural isomers.
What is the value of n?
A 4 B 6 C 8 D 9
The answer is C. Can someone please explain this? I tried to understand but i still couldn't get it. I managed to eliminate D though.
24 But-2-ene-1,4-diol is converted in two steps through an intermediate X into ketobutanedioic acid.
HOCH2CH=CHCH2OH ->step 1->X->step 2(hot acidified KMnO4)-> HO2CCOCH2CO2H
but-2-ene-1,4-diol ketobutanedioic acid
What could be the reagent for step 1 and the intermediate X?
reagent for step 1 X
A cold acidified KMnO4 HOCH2CH2CH(OH)CH2OH
B hot acidified KMnO4 OHCCH(OH)CH2CHO
C steam and concentrated H2SO4 HOCH2CH(OH)CH2CH2OH
D warm acidified K2Cr2O7 HO2CCH=CHCO2H
The answer is C. Please explain~
Thanks for your help again!!!For Question 23
Minimum number of carbon required for isomerism is 4..less than this isomerism can not be done thus, each product having equal carbon will have 4 carbons leading to N=8 so when 8/2=4 carbon each
for question 24
i don't understand it much but from seeing the option C...addition of steam and Conc.H2SO4 leads to removal of the double bond and water being added at the point
the -OH attached is secondary alchohal for it being attached to carbon that has two carbons attached.further on addition of hot acidified KMnO4 it oxidises the secondary alchohal group to ketone and the primary alcohals on each side to carboxylic acid.....You will understand it much better when you draw displayed formulas of the compond given.
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