Chemistry: Post your doubts here!

[DESTROYER1198]

because the answer we get is the mole of hydrogen molecules , to find the moles of H atoms we times it by two as each hydrogen molecule has two H atoms.

if we take H moecule as Y and H atom as X then Y=2X
so X=Y/2 isnt it?
so why we have to multiply instead of dividing??

 

[nehaoscar]

Practicals!
Anyone have any points and tips for all the reactions and reagents and stuff needed to know for practicals?
And any tips on finishing the practicals fast cause they take time..
I like the titrations but then heating and all takes too much time
And how do you know which gas to test for when? Like it takes too much time to test for each and every gas
and by the time you do the test, the gas surely must have escaped so sometimes the result is wrong...
And tips please and perhaps a revision guide note if anyone has??
Also any predictions on this year's practical for variant 33?

 

[The Sarcastic Retard]

if we take H moecule as Y and H atom as X then Y=2X
so X=Y/2 isnt it?
so why we have to multiply instead of dividing??

1 mole of H2 = ? mol of H atoms
Metanoia's style.

 

[The Sarcastic Retard]

i have another moles question that would be nice if you helped me with s12 qp21 Q5) B)ii & iii:
why in ii when we wanted to find the moles of hydrogen atom we multiplied by 2??

ii) 1 mol = 24dm^3
--> ? = 160 * 10^-3 dm^3
---> n of H2 = 6.67 * 10^-3 mol <----------- This is the moles of molecule. So we know 1 mol of H2 molecule = 2 mol of H atoms. So 2 * n of H2 = 0.0133 mol of H atoms.
iii)If there is 1 OH- then one H+ will be produced and in X; for 1 mole of X, 2 moles of H atoms are produced. So there are two OH- in each molecule of X.

 

[Farhad221]

Practicals!
Anyone have any points and tips for all the reactions and reagents and stuff needed to know for practicals?
And any tips on finishing the practicals fast cause they take time..
I like the titrations but then heating and all takes too much time
And how do you know which gas to test for when? Like it takes too much time to test for each and every gas
and by the time you do the test, the gas surely must have escaped so sometimes the result is wrong...
And tips please and perhaps a revision guide note if anyone has??
Also any predictions on this year's practical for variant 33?

 

[The Sarcastic Retard]

Practicals!
Anyone have any points and tips for all the reactions and reagents and stuff needed to know for practicals?
And any tips on finishing the practicals fast cause they take time..
I like the titrations but then heating and all takes too much time
And how do you know which gas to test for when? Like it takes too much time to test for each and every gas
and by the time you do the test, the gas surely must have escaped so sometimes the result is wrong...
And tips please and perhaps a revision guide note if anyone has??
Also any predictions on this year's practical for variant 33?

http://labskills.co.uk/a-level-labskills-trial/biology/index.html Register yourself here. Its very helpful.
Personal tip, Do the titration practical first as you good at it. Leave the calculations, go to heating stuffs question or the ones you are left and find difficult. Then when u get free time like 2 or 3 minutes heating stuffs and thing move on to calculations. Moles stuffs are very easy... so u'll finish it in very short time. If u master Enthalpy changes, your heating practical gets easier in calculations. And like we use to do in our IGCSE or O levels remember the tests given in last two pages of our every practical paper which will save time instead of recording the observation u know it directly, it doesnt happen in many cases but its good for you only...

 

[DESTROYER1198]

ii) 1 mol = 24dm^3
--> ? = 160 * 10^-3 dm^3
---> n of H2 = 6.67 * 10^-3 mol <----------- This is the moles of molecule. So we know 1 mol of H2 molecule = 2 mol of H atoms. So 2 * n of H2 = 0.0133 mol of H atoms.
iii) It will show that 1 mole of the compond produces one mole of H2 hence it must contain two OH groups

Thanks alot man

 

[The Sarcastic Retard]

Thanks alot man

Look at my edited post for part iii if not understood properly.

 

[nehaoscar]

Thanks a lot!

 

[nehaoscar]

http://labskills.co.uk/a-level-labskills-trial/biology/index.html Register yourself here. Its very helpful.
Personal tip, Do the titration practical first as you good at it. Leave the calculations, go to heating stuffs question or the ones you are left and find difficult. Then when u get free time like 2 or 3 minutes heating stuffs and thing move on to calculations. Moles stuffs are very easy... so u'll finish it in very short time. If u master Enthalpy changes, your heating practical gets easier in calculations. And like we use to do in our IGCSE or O levels remember the tests given in last two pages of our every practical paper which will save time instead of recording the observation u know it directly, it doesnt happen in many cases but its good for you only...

Thankyou so much! Yeah i like enthalpy too, it's just i hate the reactions in which you have to test for gases etc... and then sometimes you get it wrong and then the whole thing goes wrong! Hopefully no redox reactions this year *fingers crossed*

 

[nehaoscar]

Q = mCT
m = grams
C = J/kg/K
T = celcius
Then how is Q is Joules?
Shouldn't the temperature be in Kelvin so that the K's cancel from T and C? :S

 

[Tasneem_m98]

Someone please help (M/J 2011 paper 22 question 5)
In part b, how come the answer is dilute H2SO4 not concentrated H2SO4??

Also part e(i) , the aldehyde products of the partial oxidation of CH3(CH2)7CH--CH(CH2)7X (where X represents the rest of the molecule)
the mark scheme wrote two products, one of them being OHC(CH2)7CX .. I dont understand why is there an additional C beside the X???

Thank you :3

 

[lazyhits_5122]

Salam I need examiner reports of all past papers ... plz send me the link where i can get

 

[Zash Riyash]

Someone please help (M/J 2011 paper 22 question 5)
In part b, how come the answer is dilute H2SO4 not concentrated H2SO4??

Also part e(i) , the aldehyde products of the partial oxidation of CH3(CH2)7CH--CH(CH2)7X (where X represents the rest of the molecule)
the mark scheme wrote two products, one of them being OHC(CH2)7CX .. I dont understand why is there an additional C beside the X???

Thank you :3

Since there has to be 8 carbon in total

 

[The Sarcastic Retard]

Q = mCT
m = grams
C = J/kg/K
T = celcius
Then how is Q is Joules?
Shouldn't the temperature be in Kelvin so that the K's cancel from T and C? :S

The formula is Q = mC(delta)T. Where deltaT is read as change in temperature.
So let us take 2 of the random temperatures, let it be 23(degree)C and 45(degree)C
Now change in to kelvin by adding 273, hence 23 becomes 296K and 45 becomes 318.
Now change in temperature in both kelvin and celsius is 22. So you see no change.
To save the time people let it be in celcius.

 

[The Sarcastic Retard]

Someone please help (M/J 2011 paper 22 question 5)
In part b, how come the answer is dilute H2SO4 not concentrated H2SO4??

Also part e(i) , the aldehyde products of the partial oxidation of CH3(CH2)7CH--CH(CH2)7X (where X represents the rest of the molecule)
the mark scheme wrote two products, one of them being OHC(CH2)7CX .. I dont understand why is there an additional C beside the X???

Thank you :3

b)Dilute H2sO4, Concentrated would cause other reaction.

e)i)CH3(CH2)7CHO
OHC(CH2)7CX
Split the double bond and partially oxidise it; that is keep the end point as the Aldehyde.

 

[The Sarcastic Retard]

Salam I need examiner reports of all past papers ... plz send me the link where i can get

At last you will get specimenpapers and examiner report

 

[The Sarcastic Retard]

I want specimen papers for 2015

 

[Metanoia]

View attachment 52918

question 7, answer is C, but how do you know that?

25 Use of the Data Booklet is relevant to this question. 2.30 g of ethanol were mixed with aqueous acidified potassium dichromate(VI) and the desired organic product was collected by immediate distillation under gentle warming. The yield of product was 70.0%. What mass of product was collected?
A 1.54g
B 1.61g
C 2.10g
D 2.20g
answer is A, and I'm not getting that answer, I'm getting C...!

product is ethanal C2H4O. (not ethanoic acid), as product is obtained by "immediate distillation under gentle warming"

mass of ethanol converted = 0.7 (2.3) = 1.61 g
moles of ethanol converted = 1.61/46 = 0.035 mol
moles of ethanal produced = 0.035 mol
mass of ethanal produced = 0.035 x 44 = 1.54 g

 

[The Sarcastic Retard]

product is ethanal C2H4O. (not ethanoic acid), as product is obtained by "immediate distillation under gentle warming"

mass of ethanol converted = 0.7 (2.3) = 1.61 g
moles of ethanol converted = 1.61/46 = 0.035 mol
moles of ethanal produced = 0.035 mol
mass of ethanal produced = 0.035 x 44 = 1.54 g

That's ethanal ryt?