Chemistry: Post your doubts here!

[The Famous One]

When I solve recent years past papers that is from 2007 to 2015(summer) summer and witner both I use to score always above 50. Now when I look into old years like this paper I ended up at the score of just 46. I have so many doubts in this paper. If anyone can help me out solving, Q1(c)(d), Q2(c)(d), Q5(c), Q7(b)(c) in Q7(b) arent answer in ms for uses of esters? Please do it as soon as possible. Also, When I was solving papers I found that mostly moles thingy are in summer papers, in winter papers its mostly inorganic part and theory. Am I correct? Should I expect 2015 november paper to be easier with moles thingy??

Thank you,
Regards,
XPC member,
The Sarcastic Retard

imma do you still need them?

 

[Sinzzzz]

can someone please explain question 37 in this paper

 

[The Sarcastic Retard]

imma do you still need them?

No thanks.

 

[The Sarcastic Retard]

can someone please explain question 37 in this paper

The connecting carbon (C) should be sp2 hybridizes.
1 and 2 are sp2 hybridised and 3 is sp3.

 

[The Sarcastic Retard]

 

[The Sarcastic Retard]

imma do you still need them?

Solve these :~
https://www.xtremepapers.com/commun...qs-9701-and-9702-discussion-here.42450/page-4 (my doubts are at last posts of the page 4 and starting of page 5) Thanks.

 

[Metanoia]

help plzView attachment 57400
this is a paper 5 question, that means no application booklet, so how am i supposed to write the equation without it

In acidic solutions, balance the equations in order of AOHE (atoms, oxygen, hydrogen and electrons).

1. Balance the atoms apart from oxygen and hydrogen.
2. Balance the oxygens by adding water molecules.
3. Balance the hydrogens by adding hydrogen ions.
4. Balance the charges by adding electrons.

For example, MnO4- to Mn2+

MnO4- --> Mn2+ .....................( Step 1 not necessary as Mn is already balanced)

MnO4- --> Mn2+ + 4H2O ................. (balancing the O atoms)

MnO4- + 8H+ --> Mn2+ + 4H2O .........(balancing the H atoms)

MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O........... (balancing the charges)

You can try for the HCOO- to CO2.

 

[Metanoia]

Solve these :~
Q : 6, 8, 17, 18, 19, 30, 39 and 4o

https://www.xtremepapers.com/commun...qs-9701-and-9702-discussion-here.42450/page-4 (my doubts are at last posts of the page 4 and starting of page 5) Thanks.

Qn 6.
Original Mr (CaCl2) = 111
Final Mr (CaCl2.2H2O) = 147
% Increased in Mr =(36/111) x 100% = 32%

Qn8.
Moles of Cr2O7 2- = 0.00131 mol
Moles of Fe2+ = 0.00131 x 6 = 0.00786 mol
Mass of Fe = 0.00786 x 56 = 0.44 g
% of Fe = (0.44/1) x 100% = 44%

Qn17.
Describing an element that forms an amphoteric oxide, so answer is Al.

Qn 18.
P4O10 + 6H2O --> 4H3PO4

Qn 19.
CO2 is released when HCl is added to CaCO3.

Qn30.
You can post your isomers, and I'll help to see what's missing.

Qn39.
Statement 1: true, hydrazone is orange
Statement 2: false, ethanoic acid is colorless
Statement 3: false, ethan-1,2- diol is colorless.

Qn40.

 

[manya]

In acidic solutions, balance the equations in order of AOHE (atoms, oxygen, hydrogen and electrons).

1. Balance the atoms apart from oxygen and hydrogen.
2. Balance the oxygens by adding water molecules.
3. Balance the hydrogens by adding hydrogen ions.
4. Balance the charges by adding electrons.

For example, MnO4- to Mn2+

MnO4- --> Mn2+ .....................( Step 1 not necessary as Mn is already balanced)

MnO4- --> Mn2+ + 4H2O ................. (balancing the O atoms)

MnO4- + 8H+ --> Mn2+ + 4H2O .........(balancing the H atoms)

MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O........... (balancing the charges)

You can try for the HCOO- to CO2.

Ooooo thankyu so much
jazakalla khair

 

[SadiaMaryam]

Qn39.
Statement 1: true, hydrazone is orange
Statement 2: false, ethanoic acid is colorless
Statement 3: false, ethan-1,2- diol is colorless.

But for Statement #2, I thought that ethanol + acidified potassium dichromate will turn from orange to green, forming an aldehyde, isn't that right??

 

[awesomaholic101]

23rd question please. If possible please draw out the isomers?
Is there any tip for these type of questions? Any particular way to work it out? I feel like it's only a waste of time coz in the end, i don't get it right anyway.

 

[krishnapatelzz]

why the answer is c and not d :/

 

[awesomaholic101]

View attachment 57442why the answer is c and not d :/

The question asks for the MAJOR product. The carbocation intermediate that is formed for C to be the product is more stable as it is tertiary. That is, the carbon atom with the +ve charge has 3 other carbons attached to it. The more stable the carbocation, the more of that product formed.
For D however, the intermediate carbocation would only be secondary.

 

[Metanoia]

But for Statement #2, I thought that ethanol + acidified potassium dichromate will turn from orange to green, forming an aldehyde, isn't that right??

It could be ethanal or ethanoic acid, depending on the conditions. The question is asking for an organic product that is colored, and neither ethanal or ethanoic acid is colored. The green color is due to the Cr3+ ions.

 

[Metanoia]

View attachment 57441
23rd question please. If possible please draw out the isomers?
Is there any tip for these type of questions? Any particular way to work it out? I feel like it's only a waste of time coz in the end, i don't get it right anyway.

No short cut. At the most, you can focus on the tertiary isomers first to narrow down the options to 2 choices. Then move on to the secondary isomers.

 

[awesomaholic101]

No short cut. At the most, you can focus on the tertiary isomers first to narrow down the options to 2 choices. Then move on to the secondary isomers.

ok ... so more practice atleast should help?

 

[nehaoscar]

When you say equilibrium shifts to the left, which equilibrium equation are you referring to? And how do you conclude it would shift to the left?

Since it's added to the Fe cell the equation would be FeSO4 == Fe^2+ + SO4^2-
So as Na2SO4 is added the concetration of SO4^2- ions increases so the equilibrium shifts to the left??

For weak acid,

[H+] = square root (Ka*[acid])

[H+] = square root (8.9 x 10^-4 *(0.1)) = 9.43 x 10^-3

pH = - log [H+] = - log (9.43 x 10^-3) = 2.02

For the second step you have done (8.9 x 10^-4 *(0.1)) instead the mark scheme has done (8.9 x 10^-4 *(0.2)) so the pH works out to be 1.87 instead of 2.02

So I'm confused how they got 0.2 as concentration :/

 

[SadiaMaryam]

It could be ethanal or ethanoic acid, depending on the conditions. The question is asking for an organic product that is colored, and neither ethanal or ethanoic acid is colored. The green color is due to the Cr3+ ions.

 

[The Sarcastic Retard]

View attachment 57441
23rd question please. If possible please draw out the isomers?
Is there any tip for these type of questions? Any particular way to work it out? I feel like it's only a waste of time coz in the end, i don't get it right anyway.

Practice. Once you will, it will be easy for you.

 

[awesomaholic101]

Practice. Once you will, it will be easy for you.

hopefully, and thanx