Chemistry: Post your doubts here!

[desterman]

Try Znotes, they'd probably have it all. As for chemistry chemguide is pretty good!

Do you have others? Also can you get chemguide as a pdf file?

 

[Panchod Ijaz]

which ions are present in a solution of ethanol in an excess of concentrated sulphuric acid?
1) CH3CH2O^-1
2) CH3CH2^+OH2
3) HSO4^-1

 

[Mistar-Kigi]

which ions are present in a solution of ethanol in an excess of concentrated sulphuric acid?
1) CH3CH2O^-1
2) CH3CH2^+OH2
3) HSO4^-1

Not sure ummm but 2&3?

 

[Metanoia]

which ions are present in a solution of ethanol in an excess of concentrated sulphuric acid?
1) CH3CH2O^-1
2) CH3CH2^+OH2
3) HSO4^-1

1) is definitely out as it requires the OH group to act as an acid and release a H+ . which is not possible in the presence of H2SO4

 

[Emmris]

Can anyone explain to me how to arrive to the answer .... https://drive.google.com/file/d/1BkV12DgTZwm7fl42yz1lTP0m7SrHlKRq/view?usp=drivesdk

 

[MShaheerUddin]

Can anyone tell me why in a(ii) they multipled experiment 1's initial rate with 27 to get K?
where that 27 came from?
explain?

 

[MShaheerUddin]

Can anyone explain to me how to arrive to the answer .... https://drive.google.com/file/d/1BkV12DgTZwm7fl42yz1lTP0m7SrHlKRq/view?usp=drivesdk

The two equations are:

CH4 + 2O2 --> CO2 + 2H2O
CH4 + 3/2O2 --> CO + 2H2O

multiply equation 1 by 99 and add to equation 2

99CH4 + 198O2 --> 99CO2 + 198H2O
CH4 + 3/2O2 --> CO + 2H2O
---------------------------------------------------------------------------------------- add
100CH4 + 199(1/2)O2 --> 99CO2 + CO + 200H2O

So, burning 100 moles needs 199(1/2) moles of oxygen


Therefore burning y moles needs 2y - y/200

or expressed as a decimal fraction:

2y - (0.01/2)y

i.e. answer A

 

[Mistar-Kigi]

View attachment 64527
Can anyone tell me why in a(ii) they multipled experiment 1's initial rate with 27 to get K?
where that 27 came from?
explain?

Reaction is first order wrt O2 and 2nd order wrt NO. So concentration of NO is increased 3 times as well as [O2]. You get (3^2) x 3 considering the orders which'd give you 27 times the rate in reaction 1.


Edit : An alternative way would be to calculate the rate constant using any experiment of your choice and then using it along with concentrations in the 4th experiment to get your desired rate.

 

[MShaheerUddin]

Reaction is first order wrt O2 and 2nd order wrt NO. So concentration of NO is increased 3 times as well as [O2]. You get (3^2) x 3 considering the orders which'd give you 27 times the rate in reaction 1.


Edit : An alternative way would be to calculate the rate constant using any experiment of your choice and then using it along with concentrations in the 4th experiment to get your desired rate.

bro where did this 3 came from?

 

[Mistar-Kigi]

bro where did this 3 came from?

O2 is first order and it's concentration is tripled [3]^1 will be 3

 

[MShaheerUddin]

Please help me with this one too.. ?
What is the procedure of building these sort of equations and then telling which is the slow one?

 

[Panchod Ijaz]

When making sparkler fireworks, a mixture of barium nitrate powder with aluminium powder, water and glue is coated onto wires and allowed to dry. At this stage, the following exothermic reaction may occur. 16Al + 3Ba(NO3)2 + 36H2O → 3Ba(OH)2 + 16Al(OH)3 + 6NH3 Which conditions would be best to reduce the rate of this reaction during the drying process, and would also keep the aluminium and barium nitrate unchanged? temperature/K pH A 298 7 B 298 14 C 398 7 D 398 14 https://papers.gceguide.com/A Levels/Chemistry (9701)/9701_s14_qp_11.pdf plz help

 

[Panchod Ijaz]

https://papers.gceguide.com/A Levels/Chemistry (9701)/9701_w14_qp_11.pdf q 26 plz help

 

[Hamnah Zahoor]

https://papers.gceguide.com/A Levels/Chemistry (9701)/9701_w14_qp_11.pdf q 26 plz help

 

[Hamnah Zahoor]

When making sparkler fireworks, a mixture of barium nitrate powder with aluminium powder, water and glue is coated onto wires and allowed to dry. At this stage, the following exothermic reaction may occur. 16Al + 3Ba(NO3)2 + 36H2O → 3Ba(OH)2 + 16Al(OH)3 + 6NH3 Which conditions would be best to reduce the rate of this reaction during the drying process, and would also keep the aluminium and barium nitrate unchanged? temperature/K pH A 298 7 B 298 14 C 398 7 D 398 14 https://papers.gceguide.com/A Levels/Chemistry (9701)/9701_s14_qp_11.pdf plz help

to reduce the rate of reaction lower temperature will be used i.e 298 K and to keep the amount of barium nitrate and aluminum constant a lower or neutral pH will be used for if the pH is increased more hydroxides on the right side will be produced resulting in increased amount of aluminium reacting with the nitrates thus decreasing their amounts. to keep the reactants amount constant lower pH will be maintained.

 

[Fadli Alim]

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[Emmris]

The two equations are:

CH4 + 2O2 --> CO2 + 2H2O
CH4 + 3/2O2 --> CO + 2H2O

multiply equation 1 by 99 and add to equation 2

99CH4 + 198O2 --> 99CO2 + 198H2O
CH4 + 3/2O2 --> CO + 2H2O
---------------------------------------------------------------------------------------- add
100CH4 + 199(1/2)O2 --> 99CO2 + CO + 200H2O

So, burning 100 moles needs 199(1/2) moles of oxygen


Therefore burning y moles needs 2y - y/200

or expressed as a decimal fraction:

2y - (0.01/2)y

i.e. answer A

Thanks again bro

 

[Panchod Ijaz]

https://papers.gceguide.com/A Levels/Chemistry (9701)/9701_s11_qp_22.pdf QUESTION 1 PART C

 

[Ebrahim12]

https://papers.gceguide.com/A Levels/Chemistry (9701)/9701_s11_qp_22.pdf QUESTION 1 PART C

From part b we we know that:
0.04 mol of NaOH reacted with ethanoic acid
mol ratio of NaOH to ethanoic acid in the reaction is 1:1

therefore, there was 0.04 mol of ethanoic acid in the equilibrium state
before the equilibrium state there was 0.1 mol of ethanoic acid
so the other 0.06 mol of ethanoic acid underwent the forward reaction

mole ratio of acid to alcohol is 1:1
0.06 mol of alcohol reacted with the 0.06 mol of acid
this leaves 0.04 mol of alcohol

mole ratio of acid to both products is 1:1
0.06 mol of both products was formed from the 0.06 mol of acid

therefore, table is:
0.04 0.04 0.06 0.06

 

[jinggggg]

how can we determine element b and c are in group 2 and 1,but not group 12 and 11? should we refer to data booklet for checking the standard 1st ionisation energies to get the elements and correct group?