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Chemistry: Post your doubts here!

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Ans is B.
Enthalpy change=(Energy given out when bonds are made) - (Energy taken in when bonds are broken)
1800=2X-(994=496)
X=835KJ
SOrry i didnt get this part
How is it 1800?? shudnt it be 180?
n How is 994=496?? its a typing mistake i guess?
can u xplain it again?
1800=2X-(994=496)
X=835KJ
 
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snip...chem.PNG
so is it correct if i do the sum lyk this

n=(120*10^-3)/24
=5*10^-3

(5*10^-3)=0.23/mr
mr=46

(14*1)+(16*2)=46

am i ryt or is there a diff way 2 get it done?
 
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SOrry i didnt get this part
How is it 1800?? shudnt it be 180?
n How is 994=496?? its a typing mistake i guess?
can u xplain it again?
1800=2X-(994=496)
X=835KJ
Sorry it goes:
(N2 +O2)+2(NO)=180
(994+496)-180=2(NO)
=-655KJ
This is energy released when NO bonds formed but in question examiner only asks the energy of NO bond which is given in positive value ie he asks how much u will need for NO bond to break.
Hope u will understand.
 
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During the electrolysis of molten aluminium oxide to produce aluminium, using carbon electrodes,
two consecutive reactions occur at the anode, each producing a different gas.
How does the oxidation number of oxygen change in these reactions?

A decreases by 2, then increases by 2
B increases by 2, then decreases by 2
C increases by 2, then decreases by 4
D no change, then decreases by 2
 
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During the electrolysis of molten aluminium oxide to produce aluminium, using carbon electrodes,
two consecutive reactions occur at the anode, each producing a different gas.
How does the oxidation number of oxygen change in these reactions?

A decreases by 2, then increases by 2
B increases by 2, then decreases by 2
C increases by 2, then decreases by 4
D no change, then decreases by 2

A Because the two consecutive reactions in anode could possibly be:
1. 2C + O2 -> 2CO
2. 2CO + O2 -> 2CO2
The oxidation state of O2 in CO is -2
And oxidation state of O2 in CO2 is -4

It is a pretty confusing question, considering decreases as subtraction and increases as addition, Else it is easy.
 
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Can someone explain this question for me? How to deduce the answer for such questions? (Answer = B)
 

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View attachment 49941
so is it correct if i do the sum lyk this

n=(120*10^-3)/24
=5*10^-3

(5*10^-3)=0.23/mr
mr=46

(14*1)+(16*2)=46

am i ryt or is there a diff way 2 get it done?

Yes their is :)

You can also find the correct ans by elimination method ( checking options by putting the value of x and y in NxOy)

In short Answer is B

NO2

Mr - 46

You have mass 0.23 g
put that in n=mass/mr
you will get moles , multiply it with 24000 and the volume will be 120 cm3 :) Hope you got it.
 
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Can someone explain this question for me? How to deduce the answer for such questions? (Answer = B)

Answer is A .
I seriously need to get a stylus :p
Screenshot_33.png

See..
You know that group no in periodic table represents the no of valence electron. so X+ shows group 1 , X3+ shows group 3 such as
so Just find the difference between two ionization energy and compare
You can see that the first one is 510 but the second one is way too high 3800. If energy shows slight increase it is in the same shell but second ionization energy (3800) shows it is in an another shell...
I.E always start by pulling the valence electrons.. and 510 shows its the valence shell with 1 electron so it is fact that it is in GROUP 1 :)
 
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Can someone please explain these two questions as soon as possible?
 

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Please someone explain me this statement of the bronsted-lowry theory
"A weak acid forms a strong conjugate base and vice versa. Similarly,a weak base generates a strong conjugate acid and viceversa" why this is so??? what is the purpose of equilibrium constant?? What would happen if the equilibrium constant is independent of the temperature???
Answer plzz...:(
 
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Is there any book for Chemistry paper 5 (Planning Analysis And Evaluation) where I can get helps, (like planning techniques, suggestions and paper 5 related contents)??
If there then please tell me the name of the book as well as the writer.
Thanks In Advance
 
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In an experiment, 50.0 cm3 of a 0.10 mol dm–3 solution of a metallic salt reacted exactly with
25.0cm3 of 0.10moldm–3 aqueous sodium sulphite. The half-equation for oxidation of sulphite ion is shown below.
SO3^2− (aq) + H2O(I) → SO4^2− (aq) + 2H+(aq) + 2e–
If the original oxidation number of the metal in the salt was +3, what would be the new oxidation
number of the metal?
 
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