Chemistry: Post your doubts here!

[anastasia grey113]

can someone help PLEASE? :/

okay first see what youve been provided with
the volume, pressure, and temperature
what can you use here? pV=nRT
this will give you the number of moles of CxHy
rearrange the equation to make n=pV/RT
put in the values
n = 100x103 x 25x10-6/8.31 x 310
the temperature is converted to Kelvins by adding 273 to 37
and volume is converted to cubic metres
the answer is 9.7x10-4

now its been told that when CO2 is absorbed, the volume changes from 150cm3 to 50cm3.
so to find out how much CO2 was produced, we subtract final volume from the initial one which gives us
V= 150 - 50 = 100cm3 <----- volume of CO2

now we have the volume of the gas and we know that one mole of any gas occupies 24dm3 or 24000cm3 of volume at rtp.
So we can use No. of moles = Volume/24000 to find the no. of moles of CO2
putting in the values leaves us with
Moles = 100/24000 = 1.67x10-3 moles of CO2

Now that we have the moles of both the compounds we can simply use the ratio method to find the moles of CO2 per unit CxHy i.e. X.

If 9.7x10-4 moles of P produce 1.67x10-3 moles of CO2, then 1 mole should produce 1.67x10-3/9.7x10-4 moles.
The answer you'll get will be 4.

part b)ii)
Now we have to find out how many moles of O2 have been used up.
As it's said, the O2 was provided in excess (200cm3).
But by the end the gas left was 50cm3.
So subtraction leaves us with 150cm3 of O2 used up.
We have the volume and the same molar law of volume applies to this gas as well
so we will be finding the no. of moles of O2 in the same way as CO2
the no. of moles will be 6

 

[Hamnah Zahoor]

okay first see what youve been provided with
the volume, pressure, and temperature
what can you use here? pV=nRT
this will give you the number of moles of CxHy
rearrange the equation to make n=pV/RT
put in the values
n = 100x103 x 25x10-6/8.31 x 310
the temperature is converted to Kelvins by adding 273 to 37
and volume is converted to cubic metres
the answer is 9.7x10-4

now its been told that when CO2 is absorbed, the volume changes from 150cm3 to 50cm3.
so to find out how much CO2 was produced, we subtract final volume from the initial one which gives us
V= 150 - 50 = 100cm3 <----- volume of CO2

now we have the volume of the gas and we know that one mole of any gas occupies 24dm3 or 24000cm3 of volume at rtp.
So we can use No. of moles = Volume/24000 to find the no. of moles of CO2
putting in the values leaves us with
Moles = 100/24000 = 1.67x10-3 moles of CO2

Now that we have the moles of both the compounds we can simply use the ratio method to find the moles of CO2 per unit CxHy i.e. X.

If 9.7x10-4 moles of P produce 1.67x10-3 moles of CO2, then 1 mole should produce 1.67x10-3/9.7x10-4 moles.
The answer you'll get will be 4.

Should we not solve it in a more simpler way You know as the mark for this question is only '1' thus, by using the simpler molar ratio we can calculate the value of X

CxHy + (x+y/2)O2 -------------> CO2 + H2O
25cm3 : 50cm3------------------>100cm3
1 : 2 -------------------> 4
thus from this molar gas ratio we can see that X=4

In the second part we can equate
(x+y/4)=2
and substituting the value of X=4
to find y=8
thus, (4+8/4)=(4+2)=6

Now we have to find out how many moles of O2 have been used up.
As it's said, the O2 was provided in excess (200cm3).
But by the end the gas left was 50cm3.
So subtraction leaves us with 150cm3 of O2 used up.

The 150 cm3 is NOT oxygen it is a mixture of CO2 and the excess oxygen. later CO2 was removed by reacting it with NaOH and 50 cm3 excess O2 was left.
it is mentioned in the question statement...

 

[anastasia grey113]

Should we not solve it in a more simpler way You know as the mark for this question is only '1' thus, by using the simpler molar ratio we can calculate the value of X

CxHy + (x+y/2)O2 -------------> CO2 + H2O
25cm3 : 50cm3------------------>100cm3
1 : 2 -------------------> 4
thus from this molar gas ratio we can see that X=4

In the second part we can equate
(x+y/4)=2
and substituting the value of X=4
to find y=8
thus, (4+8/4)=(4+2)=6



The 150 cm3 is NOT oxygen it is a mixture of CO2 and the excess oxygen. later CO2 was removed by reacting it with NaOH and 50 cm3 excess O2 was left.
it is mentioned in the question statement...

lol thats actually a better method lol
but 150cm3 of O2 reacted. Not 50cm3. So y did you use 50 in the above ratios?
well v were given 200 initially n were left with 50 in the end so this shows that 150 of O2 WAS USED UP right? thaz wut i said

 

[Hamnah Zahoor]

lol thats actually a better method lol
but 150cm3 of O2 reacted. Not 50cm3. So y did you use 50 in the above ratios?
well v were given 200 initially n were left with 50 in the end so this shows that 150 of O2 WAS USED UP right? thaz wut i said

It is the question statement.....

The sample was completely burned in 200cm3 of oxygen (an excess).

The final volume, measured under the same conditions as the gaseous sample (so that the water produced is liquid and its volume can be ignored), was 150cm3 .

.....This shows that the total volume of oxygen that reacted was 50cm3.... From (200-150)=50cm3
Treating the remaining gaseous mixture with concentrated alkali, to absorb carbon dioxide, decreased the volume to 50cm3 .
when CO2 was removed volume left was 50 cm3 which is the unreacted oxygen(amount in excess)

This is from the examination report:

This question was not well answered. Many answers used the volume of 150 cm3 to calculate the value of x and did not appreciate that this volume contained both CO2 and excess oxygen. There were a number of answers that included attempts to use the general gas equation.

 

[anastasia grey113]

It is the question statement.....

The sample was completely burned in 200cm3 of oxygen (an excess).

The final volume, measured under the same conditions as the gaseous sample (so that the water produced is liquid and its volume can be ignored), was 150cm3 .

.....This shows that the total volume of oxygen that reacted was 50cm3.... From (200-150)=50cm3
Treating the remaining gaseous mixture with concentrated alkali, to absorb carbon dioxide, decreased the volume to 50cm3 .
when CO2 was removed volume left was 50 cm3 which is the unreacted oxygen(amount in excess)

This is from the examination report
This question was not well answered. Many answers used the volume of 150 cm3 to calculate the value of x and did not appreciate that this volume contained both CO2 and excess oxygen. There were a number of answers that included attempts to use the general gas equation.

exactly
it included 100cm3 of CO2 as well and 50cm3 of O2
but after absorbing CO2 only O2 was left which was 50cm3 in the end
but after using 50cm3 the answer is not 6
it is 8
the correct answer for x+y/4 is 6

 

[Hamnah Zahoor]

it included 100cm3 of CO2 as well
but after absorbing CO2 only O2 was left which was 50cm3 in the end
but after using 50cm3 the answer is not 6
it is 8
the correct answer for x+y/4 is 6

Well, the answer is 6
i.e (x+Y/4)=6
8 is the value of y only.

 

[anastasia grey113]

Well, the answer is 6
i.e (x+Y/4)=6
8 is the value of y only.

yes exactly
and u get this only if the ratio is 1:6 of CxHy:O2 which is only if the volumes used r 25:150

 

[Hamnah Zahoor]

yes exactly
and u get this only if the ratio is 1:6 of CxHy:O2 which is only if the volumes used r 25:150

I still don't get it why are you using O2=150 cm3 when only 50cm3 was used.

 

[anastasia grey113]

I still don't get it why are you using O2=150 cm3 when only 50cm3 was used.

okay let me explain
at first
200cm3 of O2 was added right?
then after the reaction, the volume of the mixture of gases was measured and it was 150cm3
now this mixture still contains CO2...this volume was measured BEFORE its elimination
so there was still some CO2 in the mixture of 150cm3 as it was no removed yet
so CO2 + O2 = 150cm3...not just O2
then they 'absorbed' the CO2 and were left with 50cm3 OF O2 ONLYYYY RIGHT?
so at first, 200 was added and at the end only 50 was left
so final - initial of O2 ONLY gives 150cm3

if u still don't get it you can inbox me v can talk about this
n lol thx for suggesting the shorter method..i totally forgot abt it lol n fell for the pV=nRT trap

 

[Hamnah Zahoor]

Got it .... thanks ...Just got a little confused.

 

[anastasia grey113]

oh okay happy to help

Got it .... thanks ...Just got a little confused.

 

[anastasia grey113]

View attachment 62948
Question 2 b(iii)

if you have done part b)ii) correctly, you must've gotten a value of 0.51V.
0.51V = Eelectrode for a half cell with a different [Ag+]
so to calculate the [Ag+] we should be using the equation given above
i.e.
Eelectrode = E°electrode + 0.06log[Ag+]
the E°electrode is available in the data booklet (the value of E under standard conditions)
for the above silver's reduction, it is 0.80 V.
so after putting in the values, the equation will be something like this

0.51 = 0.80 + 0.06log[Ag+]
solve it like this

0.51 - 0.80 = 0.06log[Ag+]
-0.29/0.06 = log[Ag+]
-4.83 = log[Ag+]
10^4.83 = [Ag+]
so [Ag] = 1.47x10-5

 

[anastasia grey113]

can someone help,

19.0 cm3 of 1View attachment 6320810-1 moldm-3 sodium hydroxide are mixed with 50.0 cm3 of 1View attachment 6320910-1 moldm-3 Na2H2PO4 solution. The resulting mixture can act as a buffer. Ka of H2PO4 is 6.2 View attachment 6320710-8 moldm-3 at 25 View attachment 63210.Calculate the (a) ratio of [H2PO4-]: [HPO42-] and (b) the pH

Ans: (a)= 45:28 and (b) pH=7

I did manage to solve it and my answer is very close. However, the ratio I've calculated is not to the nearest whole numbers, rather it's in decimal places.
For your answer, the value is 45/28 = 1.61 however I got 1.63.
Here's my method.

a)
Start with finding the no. of moles of each compound you've been given: NaOH and Na2H2PO4.
NaOH= 19/1000 x 0.1 = 1.9x10-3 moles
Na2H2PO4 = 50/1000 x 0.1 = 5x103 moles

Now write the IONIC equation which shows how the solution is acting as a buffer.
It should be as follows.
OH- + H2PO4- ----------> HPO4 2- + H2O
1 : 1 1 : 1
So as you can see, all 4 compounds react in a 1 to 1 ratio.

This goes on to show that if 1.9x10-3 moles of OH- are added, only 1.9x10-3 of the buffer solution will react to make HPO4 2-
The rest will stay as HPO4-
So now we know no. of moles of HPO4- (acid as it lost a proton) = 1.9x10-3
And we calculate those of H2PO4- by subtracting the moles of HPO4 2- from total inital moles of the buffer (5x10-3)
Moles of H2PO4- (salt) = 5x10-3 - 1.9x10 -3 = 3.1x10-3

Calculate ratios using both the values i.e. 3.1/1.9 = 1.63

b)
To calculate pH we use the following formula
pKa = pH + log[salt/acid]
Calculate pKa using pKa = -log[Ka]
Ka is given above as 6.2x10-8
pKa = -log[6.2x10-8] = 7.2

We already calculated salt/acid in part (a)
So put all the values in the equation and it'll be as follows
7.2 = pH + log[1.63]
the answer will be 7
i hope it helped u can inbox me if u still don't get it

 

[dumbledore.]

Ans :21/9

 

[Hinafatima]

can anyone kindly help me to solve q#4 october november 2016 component 12?

 

[Hinafatima]

can anyone kindly help me to solve q#4 october november 2016 component 12?

it's okay, got it !!! thanks anyway

 

[anastasia grey113]

Ans :21/9

Look here's the thing
The ratio we already have for tertiary : primary is 21 : 1.
However, as you can see the number of atoms of hydrogen which can be substituted by Cl to make primary halogenoalkane are 9
while in tertiary there is only 1.
Thus, 9 kinds of primary products can be formed but only one kind of tertiary
thus the ratio of products formed tertiary : primary would be 1 : 9
thus the chances of primary product being formed are 9 times greater than tertiary and so products made will be 9 times higher in concentration IF RATE OF REACTION IS NOT CONSIDERED
however, as the reaction RATE is slower, the ratio will be somewhat different

this can be calculated by multiplication
21 x 1 : 1 x 9
thus the ratio would be 21 : 9

 

[hamadh]

Sulfur can be oxidised in two ways.
S(s) + O2(g) → SO2(g) ∆H o = –296.5 kJ mol–1 2S(s) + 3O2(g) → 2SO3(g) ∆H o = –791.4 kJ mol–1

Sulfur trioxide can be made from sulfur dioxide and oxygen.

2SO2(g) + O2(g) → 2SO3(g)

What is the standard enthalpy change for this reaction? A –1384.4 kJ mol–1 B –989.8 kJ mol–1 C –494.9 kJ mol–1 D –198.4 kJ mol–1
please somebody explain me this twist in the hess's law and the point of doing it. i would be grateful.

 

[Hamnah Zahoor]

Sulfur can be oxidised in two ways.
S(s) + O2(g) → SO2(g) ∆H o = –296.5 kJ mol–1 2S(s) + 3O2(g) → 2SO3(g) ∆H o = –791.4 kJ mol–1

Sulfur trioxide can be made from sulfur dioxide and oxygen.

2SO2(g) + O2(g) → 2SO3(g)

What is the standard enthalpy change for this reaction? A –1384.4 kJ mol–1 B –989.8 kJ mol–1 C –494.9 kJ mol–1 D –198.4 kJ mol–1
please somebody explain me this twist in the hess's law and the point of doing it. i would be grateful.

First of all reverse the first reaction and multiply it by 2 as we do in writing equations for contact process this will also change the sign of enthalpy change required for this reaction i.e 2(296.5)=+593 make an overall equation of the both equations given to get 2SO2(g) + O2(g) → 2SO3(g)
then
-791.4+593= -198.4 thus D is correct option.

 

[anastasia grey113]

Sulfur can be oxidised in two ways.
S(s) + O2(g) → SO2(g) ∆H o = –296.5 kJ mol–1 2S(s) + 3O2(g) → 2SO3(g) ∆H o = –791.4 kJ mol–1

Sulfur trioxide can be made from sulfur dioxide and oxygen.

2SO2(g) + O2(g) → 2SO3(g)

What is the standard enthalpy change for this reaction? A –1384.4 kJ mol–1 B –989.8 kJ mol–1 C –494.9 kJ mol–1 D –198.4 kJ mol–1
please somebody explain me this twist in the hess's law and the point of doing it. i would be grateful.

The reaction
S(s) + O2(g) → SO2(g) ∆H o = –296.5 kJ mol–1 is basically the Δ H f ∘ of SO2 while
S(s) + 1.5O2(g) → 2 SO3(g) ∆H o = –791.4 kJ mol–1 is of (2 x Δ H f ∘ SO3).
You can see in the 2nd reaction, I have multiplied the enthalpy by 2 to find enthalpy for formation of only two moles of SO2.
because as you can see, in the given reaction the product is made from 2 moles of SO2.

This is the Hess's law diagram.
The arrows both moving upwards are showing the two reactions stated.
As you can see, the two moles of SO3 can be made by two routes. Either directly, or by doing the 1st reaction followed by oxidation of SO2.
The indirect route is shown with the green arrow.
So the equation for the reaction 2SO2 + O2 -> 2SO3 will be as follows.
2 x Δ H f ∘ of SO2 + ΔHr⊖ = 2 x Δ H f ∘ SO3