Chemistry: Post your doubts here!

[A*****]

can u plz explain why not Mn2+ instead of Mno4- cuz in the question he asked for mn2+

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.

 

[A*****]

and this one also plzz shouldnt X hv lone pairs so one CL will join with lone pairs so it will be XCL4

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.

 

[Mahrukh Iftikhar]

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

 

[Hamnah Zahoor]

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

We will take all of the options as possible compound x and react them in the following way :
X +H2O =Al(OH)3+3CH4
as you already told the answer as D i Started from it...
Al4C3+12H2O=4Al(OH)3+3CH4
Further as the reaction proceeds,
CH4+O2=CO2+H2O

108cm3 of co2 gives (108/24000)=4.5*10-3 moles
as 1 mol of co2 obt from 1 mol of ch4,thus
4.5*10-3 moles of ch4
Going to previous eq.
3 moles of CH4 produced from 1 mole of X
thus, (4.5*10-3/3)=1.5*10-3
Finding its' Mr=(0.216/1.5*10-3)=144


As we see Al4C3 have molar mass =(27*4)+(12*3)=144
Thus ans is D....Hope it helps

 

[KNOWLEDGE IS KEy]

Can anyone give me ans to chap 6 rogernorris end of chap questions?

 

[Metanoia]

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?

"forms orange crystals on reaction with 2,4-dinitrophenylhydrazine" means that the products contain aldehydes and/or ketone functional groups

A. Will produce CH3COOH , CO2 , H2O , no reaction with 2,4-DNPH
B. Will remain as CH3COCH2CH3 (ketone) which reacts with 2,4-DNPH to form orange crystals
C. Will produce CH3CH2CH2COOH (acid), no reaction with 2,4-DNPH
D. will remain as CH3CH(OH)CH2COOH (note that the hydroxyl not be oxidized by warm acidified solution of manganate(VII) ions as it is a tertiary alcohol), no reaction with 2,4-DNPH

 

[Hikaze123]

Can someone please explain Q40 to me? The answer is B

 

[Hamnah Zahoor]

Can someone please explain Q40 to me? The answer is B

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.

 

[Hikaze123]

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.

Thanks for replying!

 

[Metanoia]

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

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.

 

[Hikaze123]

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~

 

[Hikaze123]

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.

K Thx!

 

[Hamnah Zahoor]

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~

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.

 

[Hikaze123]

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.

Thanks for your help again!!!

 

[Unknown202]

 

[Unknown202]

i need a help in this question why is it A?

 

[studyingrobot457]

i need a help in this question why is it A?

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[Unknown202]

 

[Unknown202]

Now ?!

no image

 

[Hamnah Zahoor]

View attachment 62921

When ever two stages are involved in reaction its ethalpy profile diagram will have two humps in its graph .......as it is stated in the question X-Y is positive thus it will be endo and reactants will have low energy than the products....in Y-Z ethalpy change is neg thus its exo and reactants will have high energy than products as shown in option A