Chemistry: Post your doubts here!

[Eugene99]

totally stuck...no idea how to do this help!!

 

[qwertypoiu]

totally stuck...no idea how to do this help!!
View attachment 60038

Is it C?

 

[Eugene99]

Is it C?

no, D

 

[Rizwan Javed]

The product of the reaction with Cold, KMnO4 and Hot KMnO4 is shown.
In reaction with Cold KMnO4, there is addition of two chiral centers which are shown by the Circle in the product with cold as both the carbons of double bond are now attached with 4 differnet groups. So addition of 2 chiral centers.

In Hot KMnO4, the C=C bond is broken, to form one carboxylic acid, and one ketone group, as is shown by the thinner red circle. This beaking of C=C bond has no effect on number of chiral carbons.

Now the thing to note here is that in hot KMnO4, the one OH group present in cholesterol is also oxidised to form a ketone as is encircled by a thicker red circle. As a result of this, the one chiral carbon atom to which this -OH was attached no longer remains as chiral carbon, since it is now joined to 3 different groups only, not four! So there's a reduction of 1 chiral center.

Hence, the answer is D.

Got it?

 

[Eugene99]

View attachment 60041
The product of the reaction with Cold, KMnO4 and Hot KMnO4 is shown.
In reaction with Cold KMnO4, there is addition of two chiral centers which are shown by the Circle in the product with cold as both the carbons of double bond are now attached with 4 differnet groups. So addition of 2 chiral centers.

In Hot KMnO4, the C=C bond is broken, to form one carboxylic acid, and one ketone group, as is shown by the thinner red circle. This beaking of C=C bond has no effect on number of chiral carbons.

Now the thing to note here is that in hot KMnO4, the one OH group present in cholesterol is also oxidised to form a ketone as is encircled by a thicker red circle. As a result of this, the one chiral carbon atom to which this -OH was attached no longer remains as chiral carbon, since it is now joined to 3 different groups only, not four! So there's a reduction of 1 chiral center.

Hence, the answer is D.

Got it?

wow!! U did it so cool! I get it...thanks a lot!

 

[Aishayasin]

answer is c

 

[Eugene99]

View attachment 60044answer is c

equation for the decomposition of CaNO3 is:
Ca(NO3)2 >>> 2CaO + 4NO2 +O2
solid residue is CaO
Since there are 4.1 g of calcium nitrate:
Mr of Ca(NO3)2=164
no. of mol of 4.1 g of calcium nitrate = 4.1/164
=0.025 mol
1 mol of Ca(NO3)2 produce 1 mol of CaO so
0.025 mol will produce 0.025 mol

no. of mol of CaO x Mr of CaO = mass in g
0.025 x (40+16)
=1.4 g

 

[qwertypoiu]

equation for the decomposition of CaNO3 is:
CaNO3 >>> CaO + NO2
solid residue is CaO
Since there are 4.1 g of calcium nitrate:
Mr of CaNO3=102
no. of mol of 4.1 g of calcium nitrate = 4.1/102
=0.04 mol
1 mol of CaNO3 produce 1 mol of CaO so
0.04 mol will produce 0.04 mol

no. of mol of CaO x Mr of CaO = mass in g
0.04 x (40+16)
=2.25 g
is the answer D?

Well firstly calcium nitrate is Ca(NO3)2 so your Mr should be a bit higher and number of moles slightly lower. (and thus final mass lower)

And secondly (although the mole ratio is still 1:1) the thermal decomposition of Ca(NO3)2 also produces oxygen:
2Ca(NO3)2 -----> 2CaO + 4NO2 + O2

 

[Eugene99]

Well firstly calcium nitrate is Ca(NO3)2 so your Mr should be a bit higher and number of moles slightly lower. (and thus final mass lower)

And secondly (although the mole ratio is still 1:1) the thermal decomposition of Ca(NO3)2 also produces oxygen:
2Ca(NO3)2 -----> 2CaO + 4NO2 + O2

oh my god! so wrong
thank u for pointing out, I'll edit

 

[holoholo]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w14_qp_51.pdf

Question 1 (a) complete explanation please

 

[Aishayasin]

equation for the decomposition of CaNO3 is:
Ca(NO3)2 >>> 2CaO + 4NO2 +O2
solid residue is CaO
Since there are 4.1 g of calcium nitrate:
Mr of Ca(NO3)2=164
no. of mol of 4.1 g of calcium nitrate = 4.1/164
=0.025 mol
1 mol of Ca(NO3)2 produce 1 mol of CaO so
0.025 mol will produce 0.025 mol

no. of mol of CaO x Mr of CaO = mass in g
0.025 x (40+16)
=1.4 g

no the ANSWER IS C

 

[Eugene99]

no the ANSWER IS C

yes I edited, it's C now

 

[Aishayasin]

equation for the decomposition of CaNO3 is:
Ca(NO3)2 >>> 2CaO + 4NO2 +O2
solid residue is CaO
Since there are 4.1 g of calcium nitrate:
Mr of Ca(NO3)2=164
no. of mol of 4.1 g of calcium nitrate = 4.1/164
=0.025 mol
1 mol of Ca(NO3)2 produce 1 mol of CaO so
0.025 mol will produce 0.025 mol

no. of mol of CaO x Mr of CaO = mass in g
0.025 x (40+16)
=1.4 g

answer is c and the mistake in yours is Ca(NO3)2

 

[nadeen64]

hiii, does anyone have any clue on what is coming in chemistry paper 31 this may/june?

 

[cute97]

hiii, does anyone have any clue on what is coming in chemistry paper 31 this may/june?

Titration
Enthalpy with a metal carbonate
Ketones and Aldehydes identification for Q3

 

[princess Anu]

Can acidic amino acids also from zwitter ions?Or basic amino acids?, for that matter
for example this one ??
Nh3-C(H)(COOH)-Cooh

 

[Shah34]

Q8 anyone
Ans is A

 

[qwertypoiu]

Can acidic amino acids also from zwitter ions?Or basic amino acids?, for that matter
for example this one ??
Nh3-C(H)(COOH)-Cooh

Sorry I'm not too sure what you mean can you rephrase your question?
Amino acids can form zwitterions, but they can also act as acids or base by either donating or accepting a proton.

 

[qwertypoiu]

Q8 anyone
Ans is A

Ive made an attempt at this question here before. Read this too it might help.
As for temperature, of course you should know that lower temp = lower rate.

 

[Rizwan Javed]

Ive made an attempt at this question here before. Read this too it might help.
As for temperature, of course you should know that lower temp = lower rate.

I also think that this is the only possible explanation to this question. When the temperature is less, a higher conc. of OH ions would still increase the rate of reaction. So to keep the reaction rate minimum, we must decrease the conc. of OH ions too along with the temperature. Meaning, we must reduce all the factors which are directly proportional to reaction rate.