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Chemistry: Post your doubts here!

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ashcull14

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Saif Qureshi said:
Doubts. Can anyone tell the answer and explain it?
Click to expand...
in Q1 is the ans D?....if it is then for chlorine gas you count the peaks of both atoms and molecules ....that is for an atom =2 (35/37)...and for molecules 3 (Cl 70,72,74)
for Q2 i think its C because for nonmetals the rule is s before p subshell and then you add the figures on both s and p to know the group number ...since u cn see tht there is a huge rise in ionisation energies after a gap of 6 IEs thus the element belongs to grup 6 so s+p shud be equal to 6
 
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shazmina

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ashcull14 said:
Which reactions have a coloured organic product?
1 ethanal + 2,4-dinitrophenylhydrazine reagent
2 ethanol + acidified potassium dichromate(VI)
3 ethene + cold dilute acidified potassium manganate(VII)
help plz
Click to expand...
What abt all????? the others dnt produce an organic product or what???
 
Xylferion

Xylferion

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Saif Qureshi said:
Doubts. Can anyone tell the answer and explain it?
Click to expand...

For Q3, you're already told that you have a sample containing " isotopes " of mass numbers 35 and 37. On the mass spectrum you're going to end up with different peaks for different combinations of the mass numbers.

You can have
- Cl-35 and Cl-37,
- Cl-35 and Cl-35
- Cl-37 and Cl-37

Which are 3 different combinations that result in 3 different peaks. So your answer is B.

-------------

For Q9, you're given 7 ionization energies. From these you need to find out where the biggest jump in energy lies, in this case it's between the 6th and the 7th energy.

Now the large jump from the 6th to the 7th ionization energy, means that the 7th electron is being removed from a shell closer to the nucleus. Which basically means it's being removed from a shell with a lower quantum number.

They're asking you for the configuration of the " outer " shell, so you need only pick the one with 6 electrons in total. That is option C, which is 2s2,2p4.
The 7th electron will be removed from 1s2.

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For Q28 and 29, the answers are both D. There is only 1 correct statement in either case.

Chlorine with a Mr of 72, is an isotope. Isotopes will have more neutrons, but the same number of protons and electrons. With more neutrons both the Nucleon number and isotopic mass will INCREASE.

They're asking you what always stays the same, and that is the radius, since adding more neutrons will have no affect, as they have no charge. With no effect on the overall charge, there will be no greater attraction or repulsion between the nucleus and the electrons.

Hope that helped :)
 
Xylferion

Xylferion

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ashcull14 said:
Which reactions have a coloured organic product?
1 ethanal + 2,4-dinitrophenylhydrazine reagent
2 ethanol + acidified potassium dichromate(VI)
3 ethene + cold dilute acidified potassium manganate(VII)
help plz
Click to expand...

To begin with, you need to be able to identify what each reagent does..

2,4-DNPH is used to test for the presence of the C=O double bond that are present in both aldehydes and ketones. If there is one, it will show either an "orange" or "yellow" or "red" precipitate. Which is also why you're seeing it react with ethanal, which is an aldehyde.

Acidified (Sodium or Potassium) dichromate(VI) is used to distinguish between primary, secondary and tertiary alcohols. It takes an orange solution ( which is due to the chromium ions ) and if there is either a primary or a secondary alcohol present, the solution will turn green. Since tertiary alcohols do not undergo oxidation reactions.

"Cold" dilute acidified potassium manganate ( VII ) is used to oxidize and alkene to form a diol. The diol has no double bond as opposed to the alkene, so as a result cold dilute acidified potassium manganate can be used as a test for the presence of a double bond, just like bromine water is.

The "dark purple" solution due to the manganate ions will turn " colorless "

Now the reason I've highlighted the words precipitate and solution are because these clearly distinguish between forming a colored product as opposed to visibly exhibiting a change in color.

Both the tests with dichromate and manganate are simply changing the color of a solution as opposed to 2,4-DNPH which is essentially creating a product that is colored. The product created is a hydrazone, and like I stated above, it can be either red, orange or yellow.

The other two form products, but the color visible is not because of the products but instead the chromium and manganate ions.

Hope this clears up any confusion, there's a lot of theory behind it, and also the ability to deduce what's responsible for the color. :)
 
Xylferion

Xylferion

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ashcull14 said:
View attachment 51870
Click to expand...

2,4-DNPH is for aldehydes and ketones only, so it will only work for propanal.
Sodium will react with alcohols as a metal, but to react with an aldehyde, it'd have to be bonded to cyanide, so as to undergo nucleophilic addition. So it only really reacts with propan-2-ol.
Tollens reagent is directed towards carbonyl compounds, which are ketones and aldehydes, so it won't prove useful for propan-2-ol.

The only one that works for both is potassium dichromate, since its job is to oxidize both of them. By oxidizing a secondary alcohol you get an aldehyde. You oxidize it further and you end up with a carboxylic acid.

Propan-2-ol is a secondary alcohol, which is CIE already giving you a hint that it will oxidize.

Hope it all made sense ^_^
 
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ashcull14

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Xylferion said:
To begin with, you need to be able to identify what each reagent does..

2,4-DNPH is used to test for the presence of the C=O double bond that are present in both aldehydes and ketones. If there is one, it will show either an "orange" or "yellow" or "red" precipitate. Which is also why you're seeing it react with ethanal, which is an aldehyde.

Acidified (Sodium or Potassium) dichromate(VI) is used to distinguish between primary, secondary and tertiary alcohols. It takes an orange solution ( which is due to the chromium ions ) and if there is either a primary or a secondary alcohol present, the solution will turn green. Since tertiary alcohols do not undergo oxidation reactions.

"Cold" dilute acidified potassium manganate ( VII ) is used to oxidize and alkene to form a diol. The diol has no double bond as opposed to the alkene, so as a result cold dilute acidified potassium manganate can be used as a test for the presence of a double bond, just like bromine water is.

The "dark purple" solution due to the manganate ions will turn " colorless "

Now the reason I've highlighted the words precipitate and solution are because these clearly distinguish between forming a colored product as opposed to visibly exhibiting a change in color.

Both the tests with dichromate and manganate are simply changing the color of a solution as opposed to 2,4-DNPH which is essentially creating a product that is colored. The product created is a hydrazone, and like I stated above, it can be either red, orange or yellow.

The other two form products, but the color visible is not because of the products but instead the chromium and manganate ions.

Hope this clears up any confusion, there's a lot of theory behind it, and also the ability to deduce what's responsible for the color. :)
Click to expand...
VERY WELL EXPLAINED ...May God bless you :)
 
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My Name

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The Sarcastic Retard said:
Assume there are initially 2 moles of N2O4. How many moles of N2O4 and NO2 would there be at equilibrium ?

How many moles of gas are there altogether at equilibrium ?

Can you find the partial pressures now ?

->
Initially, N2O4:NO2 is 2:0
At equilibrium, it'd be 1/3 : 2/3
Thus Kp would be (2/3)^2/ (1/3), which is 4/3!
Click to expand...

Where did you get the 1/3 :2/3 ?

Ty
 
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