Chemistry: Post your doubts here!

[cute97]

Ammonia is made by the Haber process. The reactants are nitrogen and hydrogen.
N2(g) + 3H2(g) 2NH3(g) ΔH –ve
What will increase the rate of the forward reaction?
A adding argon to the mixture but keeping the total volume constant
B decreasing the temperature
C increasing the total pressure by reducing the total volume at constant temperature
D removing ammonia as it is made but keeping the total volume of the mixture the same
the answer is c why not b

Decreasing the temperature would lead to an increase in the yield of ammonia and not fasten the rate of reaction, They are asking which one increases the rate of the reaction, so an increase in kinetic energy would increase the rate of reaction with a lower yield (percentage) of ammonia. So its C not B

 

[My Name]

Oh, didn't know that, thanks for letting me know

Welcome

 

[Aishayasin]

can anyone explain q 38 answer is c

 

[Aishayasin]

Flask X contains 5 dm3of helium at 12 kPa pressure and flask Y contains 10 dm3 of neon at 6 kPa pressure.
If the flasks are connected at constant temperature, what is the final pressure?
A 8 kPa B 9 kPa C 10 kPa D 11 kPa
answer is A

 

[sfhn_128]

Flask X contains 5 dm3of helium at 12 kPa pressure and flask Y contains 10 dm3 of neon at 6 kPa pressure.
If the flasks are connected at constant temperature, what is the final pressure?
A 8 kPa B 9 kPa C 10 kPa D 11 kPa
answer is A

use the equation:

P1V1 = P2V2

5 x 12 + 6 x 10 = (10 + 5) x p
p =8

 

[Hamody]

Plz I need help in enthalpy changes

 

[Aishayasin]

can anyone explain q 38 answer is c

i am confused can anyone explain me its organic chemistry question

 

[Aishayasin]

ANSWER IS C how

 

[qwertypoiu]

i am confused can anyone explain me its organic chemistry question

Spoiler: image

First compound will give both A and B as products. So it has two products.
Second compound will give A and C as products, but you must realize these are both equivalent. Therefore, number 2 gives only one product.
Third compound will give B and D as products. However, you must realize that D cannot be possible, since it involves a carbon atom making 5 bonds. Therefore, number 3 also only has one possible product.

 

[Rizwan Javed]

ANSWER IS C howView attachment 60031

The Cl circled red will form a tertiary alcohol after hydrolysis since the alpha carbon, to which this Cl is attached, is attached to 3 other carbon atoms.

The Cl circled blue will form a secondary alcohol.

The -OH ringed black is the only primary alcohol present in this molecule.

The rest of the -OH groups , highlighted yellow represent secondary alcohols.

So if you count the number of secondary alcohols before hydrolysis, it is 4. After hydrolysis this number increases to 5 as nucleophilic substitution of blue-ringed Cl forms a secondary alcohol. There's only 1 tertiary alcohol, produced by the substitution of red-ringed Cl.
And there's only one primary alcohol, which is circled black.

Therefore, the answer is C.

 

[Aishayasin]

how

View attachment 60035
The Cl circled red will form a tertiary alcohol after hydrolysis since the alpha carbon, to which this Cl is attached, is attached to 3 other carbon atoms.

The Cl circled blue will form a secondary alcohol.

The -OH ringed black is the only primary alcohol present in this molecule.

The rest of the -OH groups , highlighted yellow represent secondary alcohols.

So if you count the number of secondary alcohols before hydrolysis, it is 4. After hydrolysis this number increases to 5 as nucleophilic substitution of blue-ringed Cl forms a secondary alcohol. There's only 1 tertiary alcohol, produced by the substitution of red-ringed Cl.
And there's only one primary alcohol, which is circled black.

Therefore, the answer is C.

how do i know which is secondary and tertiary alcohol in this diagram could u explain i get the primary one

 

[Rizwan Javed]

how

how do i know which is secondary and tertiary alcohol in this diagram could u explain i get the primary one

The corners of this skeleton formula are represented by C atoms (I have marked them with blue Cs). You can see that C atoms to which -OH groups which are highlighted are attached, that carbon is attached to 2 other carbon atoms. So these are the secondary alcohols.

 

[The Sarcastic Retard]

2(b) If step 2 is slowest overall then why a b and c = 1? Help me.

 

[qwertypoiu]

View attachment 60037
2(b) If step 2 is slowest overall then why a b and c = 1? Help me.

I think this is the same question I answered here.
Have a look and see if it helps

 

[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