Chemistry MCQ thread...

[Farhad221]

SALAM
I have problem with this question plz help me fast thank u

 

[Liv2Stdy]

SALAM
IView attachment 50808 have problem with this question plz help me fast thank u

Hi, Here you go, hope you understand what I've done (If not I'll explain ) and I've also included the way i found the oxidation numbers (Don't think you need that...but just in case!lol). Btw, answer is B!

 

[Farhad221]

I need help with these questions.

 

[Farhad221]

and this one

 

[Farhad221]

Guys i am starting a revision group for upcoming may june 2o15 exams more details herehttps://www.xtremepapers.com/community/threads/revision-group-on-facebook-or-viber-for-science-subjects-economics-business-studies.37031/

 

[manya]

can someone tell me y is the answer C and not D

 

[Xaptor16]

can someone tell me y is the answer C and not D

look at the starting compound and the product, the product has 2 more carbon atoms than the beginning compound right? this is because the beginning compound has 2 -OH groups which are both replaced by CN and then converted to COOH. if we used NaOH the ending product would have had the same number of carbon atoms as the beginning compound. i hope you get it

 

[manya]

look at the starting compound and the product, the product has 2 more carbon atoms than the beginning compound right? this is because the beginning compound has 2 -OH groups which are both replaced by CN and then converted to COOH. if we used NaOH the ending product would have had the same number of carbon atoms as the beginning compound. i hope you get it

ahhh! thankyou so much i just noticed that the no. of carbon atoms was more in the product

 

[Xaptor16]

ahhh! thankyou so much i just noticed that the no. of carbon atoms was more in the product

anytime

 

[ashcull14]

 

[M.Fazy]

View attachment 52193

Hi there,

The Answer is B.

We'll simply work by ratio here, as we know that first four Alkanes are mostly gases so moler ratio for a general equation should work on volume ratio here.

For 1st :

CH4 + 2O2 = CO2 + 2H2O

Moler Ratio : 1 = 1 + 2
Volume : (1x10) = (1x10) + (1x20) ........................ (For oxygen, 70 cm3 is given, reqired 20, so we carry own!)
10+20 = 30cm3 which is the product.

For 4th : Because we're pretty sure it's B as it's the only one with 30cm3 on ch4 and we're doing 4 just to verify.

C4H10 + 6.5O2 = 4CO2 + 5H2O

Moler ratio : 1 = 4 + 5
Volume : (1x10) = (1x40) + (1x50) ........................ (For oxygen, 70 cm3 is given, reqired 20, so we carry own!)
40+50 = 90cm3 which is the product.

Similarly, you should do this for the rest of the 2 so that you understand the process.

gl hf ,
l8r.

 

[ashcull14]

Hi there,

The Answer is B.

We'll simply work by ratio here, as we know that first four Alkanes are mostly gases so moler ratio for a general equation should work on volume ratio here.

For 1st :

CH4 + 2O2 = CO2 + 2H2O

Moler Ratio : 1 = 1 + 2
Volume : (1x10) = (1x10) + (1x20) ........................ (For oxygen, 70 cm3 is given, reqired 20, so we carry own!)
10+20 = 30cm3 which is the product.

For 4th : Because we're pretty sure it's B as it's the only one with 30cm3 on ch4 and we're doing 4 just to verify.

C4H10 + 6.5O2 = 4CO2 + 5H2O

Moler ratio : 1 = 4 + 5
Volume : (1x10) = (1x40) + (1x50) ........................ (For oxygen, 70 cm3 is given, reqired 20, so we carry own!)
40+50 = 90cm3 which is the product.

Similarly, you should do this for the rest of the 2 so that you understand the process.

gl hf ,
l8r.

got it ...thnk u so much

 

[The Sarcastic Retard]

View attachment 45163

did u get the answer?

 

[ashcull14]

did u get the answer?

cn u xplain????????? im still a bit cnfused

 

[The Sarcastic Retard]

cn u xplain????????? im still a bit cnfused[/QU
D is the correct answer bcoz
We knw ideal gas eqn is pV =nRT
pV=mRT/(M)
We know rho =m/V
Hence p=rhoRT//(M)
Hence pV = nRtOTE]p

cn u xplain????????? im still a bit cnfused

 

[Lostsoul]

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[ForeverAnonymous]

Hi there, can someone please explain this MCQ?

Butanedioic acid occurs in amber, algae, lichens, sugar cane and beets. It may be synthesised in two steps from 1,2-dibromoethane.
BrCH2CH2Br ----> X ----> HO2CCH2CH2CO2H
Which reagents could be used for this synthesis?
step 1 step 2
A HCN(g) HCl(aq)
B HCO2Na(aq) HCl(aq)
C KCN(aq/alcoholic) H2SO4(aq)
D NaOH(aq) K2Cr2O7 /H2SO4(aq)
Thanks in advance! (Btw the answer is C)

 

[ashcull14]

Hi there, can someone please explain this MCQ?

Butanedioic acid occurs in amber, algae, lichens, sugar cane and beets. It may be synthesised in two steps from 1,2-dibromoethane.
BrCH2CH2Br ----> X ----> HO2CCH2CH2CO2H
Which reagents could be used for this synthesis?
step 1 step 2
A HCN(g) HCl(aq)
B HCO2Na(aq) HCl(aq)
C KCN(aq/alcoholic) H2SO4(aq)
D NaOH(aq) K2Cr2O7 /H2SO4(aq)
Thanks in advance! (Btw the answer is C)

the compound will first be nitrated to remove Br and then through acidic hydrolysis ul get a carboxyl group which will remove the nitrogen on adjacent sides and replace it with COOH.....since it isn't a carbonyl compound so HCN cant be used fr nitification ....therefore BrCH2CH2Br + KCN = NCCH2CH2CN + H2SO4 = HO2CCH2CH2CO2H so C is the correct option

 

[insharah]

http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_s15_qp_12.pdf

question 15 please help!!

 

[Thelastmoment]

http://papers.gceguide.com/A Levels/Chemistry (9701)/9701_s15_qp_12.pdf

question 15 please help!!

Q15- Electronegativity increases across the period and so does first ionisation energy ( since atomic size decreases as nuclear charge increases)
So if you check the data booklet Mg has a higher first ionisation energy than Al so answer is B.