Chemistry: Post your doubts here!

[The Sarcastic Retard]

View attachment 51906

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!

 

[DeViL gURl B)]

in chemistry paper 5, when is the percentage error high?
for smaller measurements or for larger measurements?

 

[shazmina]

in chemistry paper 5, when is the percentage error high?
for smaller measurements or for larger measurements?

For large uncertainties in small measurements

 

[My Name]

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!

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

Ty

 

[princess Anu]

help!

 

[shazmina]

http://onlineexamhelp.com/wp-content/uploads/2012/04/9701_w10_qp_43.pdf
Question 7 ( b ) .... how do we know abt the steepest part of the graph ... confused ... help pls

 

[shazmina]

View attachment 51973 help!

may i know the answer pls?

 

[Xylferion]

View attachment 51906

N2O4 <--> 2NO2 at equilibrium.

Initially, there is 1 mol of N2O4. NO2 is yet to be formed.

When N2O4 dissociates, it gives 2 moles of NO2. 1:2
But it says in the question only 50% dissociates.

so 0.5 remains N2O4, and 0.5 dissociates.
This gives 1 mol of NO2 for every o.5 moles of N2O4. 0.5:1

Now with partial pressures there is something called a mole fraction. If you have the mole fraction, and the total pressure, you can find the partial pressure.

The formula is:
Mole fraction * Total pressure = Partial pressure

In the above reaction, there is 1 +0.5 = 1.5 mols in total.
N2O4 is 0.5/1.5 and NO2 is 1/1.5
This gives you 1/3 and 2/3.

Multiply these values by the total pressure, which is 1 in this case, and you get the partial pressures.

Partial pressure for N2O4 is 1/3 atm and 2/3 atm for NO2.

At equilibrium, there are 2 moles of NO2 for every 1 mole of N2O4.
so Kp = ( 2/3 )^2 / (1/3)^1

This gives you 4/3 atm.

Hope that helped ^_^

 

[Xylferion]

View attachment 51973 help!

Density of ice = 1 gcm^-3
Volume of ice being used = 1 cm^3
Mass of ice being used = 1 gcm^-3 * 1 cm^3
Moles of ice being used = 1 g / 18 g since ice is H2O basically.

This gives you 0.0555 moles of ice being used.

1 mole of a gas occupies a volume of 24 dm^3 at 298 K.
0.0555 moles of a gas would occupy a volume of 0.055 * 24 / 1 = 1.332 dm^3 at 298 K.

We want to find out the volume of steam at 596 K.
So if at 298 K the volume is 1.332 dm^-3.
The volume at 596 K would be 596 K * 1.332 / 298 = 2.664dm^-3

The closest answer to this is 2.67 ( There may be some degree of significance missing ), so C is the answer.

Hope that helps!

 

[My Name]

N2O4 <--> 2NO2 at equilibrium.

Initially, there is 1 mol of N2O4. NO2 is yet to be formed.

When N2O4 dissociates, it gives 2 moles of NO2. 1:2
But it says in the question only 50% dissociates.

so 0.5 remains N2O4, and 0.5 dissociates.
This gives 1 mol of NO2 for every o.5 moles of N2O4. 0.5:1

Now with partial pressures there is something called a mole fraction. If you have the mole fraction, and the total pressure, you can find the partial pressure.

The formula is:
Mole fraction * Total pressure = Partial pressure

In the above reaction, there is 1 +0.5 = 1.5 mols in total.
N2O4 is 0.5/1.5 and NO2 is 1/1.5
This gives you 1/3 and 2/3.

Multiply these values by the total pressure, which is 1 in this case, and you get the partial pressures.

Partial pressure for N2O4 is 1/3 atm and 2/3 atm for NO2.

At equilibrium, there are 2 moles of NO2 for every 1 mole of N2O4.
so Kp = ( 2/3 )^2 / (1/3)^1

This gives you 4/3 atm.

Hope that helped ^_^

Yes that helped!
Ty.

 

[shazmina]

http://maxpapers.com/wp-content/uploads/2012/11/9701_s13_qp_42.pdf
Can anyone please explain me Question 1 part b ( iii ) ...
like do i have to draw 2 tangents for both the graphs at time = 0 and find out the initial rate ie the gradient of these 2 lines or is there some other method for finding it??? i dont rele get the mark scheme .. so plz help

 

[shazmina]

Tha value of Kw is 9.55 * 10^-14 at a certain temperature. Calculate the pH of the water at this temperature ...

CAN someone please explain me how to do this ......

 

[Sarosh Jameel]

plz help !

 

[princess Anu]

plz help !

is 6 C,15 B?

 

[princess Anu]

Isn't this question weird? :/
they are asking for isomers of PENTAN-3- ONE
and I don't think there are any isomers of pentane 3 one because in any isomer I make, the place of carbonyl group no longer stays at 3 as stated by the question

 

[Xylferion]

plz help !

For Question 6, You need to write the reaction out.
Al4C3 + NaOH + H2O --------> NaAlO2 + CxHy

Tackle the carbons first, there's 3 carbons on the left, so there should be 3 on the right.
This gives you:
Al4C3 + NaOH + H2O --------> NaAlO2 + 3CHy

There's 4 Al's on the left, so there should be 4 on the right,
Al4C3 + NaOH + H2O --------> 4NaAlO2 + 3CHy

Now balance sodium,
Al4C3 + 4NaOH + H2O --------> 4NaAlO2 + 3CHy

8 oxygen atoms on the right, so there must be 8 in total on the left.
Al4C3 + 4NaOH + 4H2O --------> 4NaAlO2 + 3CHy

12 hydrogen atoms on the left, so there must be 12 on the right, but since there are 3 Hydrocarbon molecules, we can have it as H4.
Al4C3 + 4NaOH + 4H2O --------> 4NaAlO2 + 3CH4

Giving you CH4 as the hydrocarbon, with your answer being A.

---------------------
K2O is dissolved in 250 cm^3 of water.
25 cm^3 of this is titrated with H2SO4 of conc. 2.00 mol/dm^3 needing 15 cm^3.

First write out a reaction between K2O and H2SO4.

K2O + H2SO4 --------> K2SO4 + H2O

The molar ratio for this is 1:1:1:1

First find out the number of moles of H2SO4 using n = cV

n = 2.00 x 15/1000
n = 0.03 moles

1:1 between K2O and H2SO4 so 0.03 moles of K2O.

They want to know the amount in 250 cm^3 of water, while we only used 25 cm^3.

So multiply 0.03 by 10.
You get 0.3 moles of K2O.

Get the Molar Mass of K2O using the data booklet for that year, which is: ( K = 39.1, O = 16.0 )
39.1 x 2 + 16.0 = 94.2

94.2 x 0.3 = 28.26 ---> 28.3 g

Giving you B as your answer.

Hope that helped

 

[Xylferion]

Isn't this question weird? :/
they are asking for isomers of PENTAN-3- ONE
and I don't think there are any isomers of pentane 3 one because in any isomer I make, the place of carbonyl group no longer stays at 3 as stated by the question

How about these?

 

[princess Anu]

How about these?
View attachment 52028
View attachment 52029

These are no longer Pentanones. You've used 6 carbons

 

[Xylferion]

These are no longer Pentanones. You've used 6 carbons

The main chain is still 5 carbons long, those extra CH3 groups are considered methyl.

The first one is 2-methylpentan-3-one.
The second one is 2,2-dimethylpentan-3-one.

Structural Isomerism splits up into:
1) Positional isomerism
2) Functional group isomerism
3) Chain Isomerism

1 and 3 would not work as you want the ketone to stay on the 3rd carbon, and you cannot rearrange the other carbons or branch them since the functional group in question lies on an odd numbered carbon.

Hope that clears it up a bit

 

[Sarosh Jameel]

is 6 C,15 B?

6 A and 15 B