A-level Chemistry Revision Squad!
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Original post by djmans
can some one help?
In a (i) Look for the peak with the highest value for m/z, and that value is the relative formula mass of the compound.
a(ii) we add O and H molar mass (16+1) and then deduct it from 88 =71 so the CH3 group at one end has Mr 15. Deduct it from 71=56,now divide 56 by 14 (as -CH2 has Mr 14) and you get 4. So the x=5 (4+1 in methyl group at the terminal end) and y=11 as (4*2+3 in the methyl group)
b. In three as it doesn't oxidise so its a tertiary alcohol with formula CH3-CH2-CH2-CH2-CH2-OH.
Hope you understood. Better see the markscheme!
I need help with a double indicator titration.
I've got methyl orange and phenolpthalein, but I can't figure out how you work out the number of moles of each substance.
I understand that the first end point = half of na2co3 and naoh
Second end-point = 2nd half of na2co3, but where does CO2 come into it?
Eqn: 2NaOH + CO2 --> Na2CO3 + H2O
I've got methyl orange and phenolpthalein, but I can't figure out how you work out the number of moles of each substance.
I understand that the first end point = half of na2co3 and naoh
Second end-point = 2nd half of na2co3, but where does CO2 come into it?
Eqn: 2NaOH + CO2 --> Na2CO3 + H2O
Original post by Confused_Soul_
I need help with a double indicator titration.
I've got methyl orange and phenolpthalein, but I can't figure out how you work out the number of moles of each substance.
I understand that the first end point = half of na2co3 and naoh
Second end-point = 2nd half of na2co3, but where does CO2 come into it?
Eqn: 2NaOH + CO2 --> Na2CO3 + H2O
I've got methyl orange and phenolpthalein, but I can't figure out how you work out the number of moles of each substance.
I understand that the first end point = half of na2co3 and naoh
Second end-point = 2nd half of na2co3, but where does CO2 come into it?
Eqn: 2NaOH + CO2 --> Na2CO3 + H2O
HCl + Na2CO3 --> NaHCO3 + H2O
HCl + NaHCO3 --> NaCl + CO2 + H2O
Original post by Confused_Soul_
I need help with a double indicator titration.
I've got methyl orange and phenolpthalein, but I can't figure out how you work out the number of moles of each substance.
I understand that the first end point = half of na2co3 and naoh
Second end-point = 2nd half of na2co3, but where does CO2 come into it?
Eqn: 2NaOH + CO2 --> Na2CO3 + H2O
I've got methyl orange and phenolpthalein, but I can't figure out how you work out the number of moles of each substance.
I understand that the first end point = half of na2co3 and naoh
Second end-point = 2nd half of na2co3, but where does CO2 come into it?
Eqn: 2NaOH + CO2 --> Na2CO3 + H2O
To work out the concentration of the analyte you need to work out the concentration of what you're titrating it against, so use the concentration of your standard solutions to do it.
The moles would be with this equation: No. moles = (concentration x volume (cm3))/1000 or N = C x V (dm3).
Original post by MiracleLeaf
To work out the concentration of the analyte you need to work out the concentration of what you're titrating it against, so use the concentration of your standard solutions to do it.
The moles would be with this equation: No. moles = (concentration x volume (cm3))/1000 or N = C x V (dm3).
The moles would be with this equation: No. moles = (concentration x volume (cm3))/1000 or N = C x V (dm3).
It's being titrated with HCL and i've got the volumes for both end-points.
i know that the second end-point minus the first-end point gives you the volume needed to react with half of the na2co3. Times it by two to give the volume needed for all the Na2CO3. Right?
What i dont understand is the HCl titre --> how do i calculate the number of moles of HCL and would i use the following equation to calculate the moles of CO2?
Na2CO3 + 2HCl --> 2NaCl + H2O + CO2?
thank you
Original post by djmans
can some one help?
Hi there.
Question is one of simple maths really. The formula mass of the alcohol is 88. So subtract the mass of the OH functional group from the formula mass of the alcohol. This leaves you with 71. This must be the formula mass of the alkyl group represented by carbon and hydrogen in their relative quantities. Now it is simply a question of finding how many carbon atoms will make up as much of the 71 as possible. As carbon has a relative atomic mass of 12, this means there are 5(12×5=60). This means the remaining 11 must be all hydrogens. Hence the alkyl group must have the formula C5H11
Hope this helps and good luck
With E cell if it's negative does that mean the half cell reaction is oxidising?
Original post by Ladymusiclover
With E cell if it's negative does that mean the half cell reaction is oxidising?
The one that's more negative is the one being oxidised, yes.
Can someone please help
Posted from TSR Mobile
Posted from TSR Mobile
Original post by PlayerBB
Hey!
Ive given it a shot:
1. First I calculated both the moles of HCl and NaOH..
moles of NaOH: 2.54 X 10-3
moles of HCl: 5 X 10-3
because we titrated the acid with the alkali we can take the number of mole of HCl from NaOH. so (5X10-3)- (2.54X10-3)= to get us 2.46 X 10-3 moles used.
Because it is a 2:1 ratio we divide the number of moles by 2 to get us 1.23 X10-3 in 25cm3.
We can then multiply the number of moles by 10 (to get it in 250cm3): 0.0123
Now we can calculate the mass: Mass= moles x Mr:
0.0123 x 132.1= 1.62483g
% by mass: 1.62483/3 multipied by 100: 54.12%
Original post by haj101
Hey!
Ive given it a shot:
1. First I calculated both the moles of HCl and NaOH..
moles of NaOH: 2.54 X 10-3
moles of HCl: 5 X 10-3
because we titrated the acid with the alkali we can take the number of mole of HCl from NaOH. so (5X10-3)- (2.54X10-3)= to get us 2.46 X 10-3 moles used.
Because it is a 2:1 ratio we divide the number of moles by 2 to get us 1.23 X10-3 in 25cm3.
We can then multiply the number of moles by 10 (to get it in 250cm3): 0.0123
Now we can calculate the mass: Mass= moles x Mr:
0.0123 x 132.1= 1.62483g
% by mass: 1.62483/3 multipied by 100: 54.12%
Ive given it a shot:
1. First I calculated both the moles of HCl and NaOH..
moles of NaOH: 2.54 X 10-3
moles of HCl: 5 X 10-3
because we titrated the acid with the alkali we can take the number of mole of HCl from NaOH. so (5X10-3)- (2.54X10-3)= to get us 2.46 X 10-3 moles used.
Because it is a 2:1 ratio we divide the number of moles by 2 to get us 1.23 X10-3 in 25cm3.
We can then multiply the number of moles by 10 (to get it in 250cm3): 0.0123
Now we can calculate the mass: Mass= moles x Mr:
0.0123 x 132.1= 1.62483g
% by mass: 1.62483/3 multipied by 100: 54.12%
Heey
I have solved it exactly as you said and I keep getting 54.16% which is really annoying and then on my calculator, I have put your numbers and it keeps giving me 54.16 so how you're getting 54.12% ?
Posted from TSR Mobile
Original post by PlayerBB
Heey
I have solved it exactly as you said and I keep getting 54.16% which is really annoying and then on my calculator, I have put your numbers and it keeps giving me 54.16 so how you're getting 54.12% ?
Posted from TSR Mobile
I have solved it exactly as you said and I keep getting 54.16% which is really annoying and then on my calculator, I have put your numbers and it keeps giving me 54.16 so how you're getting 54.12% ?
Posted from TSR Mobile
Loool fail😂🙈🙈 yes you're right it is 54.16% I'll try to figure it out and get back to you, sorry! X
Original post by haj101
Loool fail😂🙈🙈 yes you're right it is 54.16% I'll try to figure it out and get back to you, sorry! X
Lool thank you so much!!
I've been stuck on it for ages!Original post by PlayerBB
Lool thank you so much!! I've been stuck on it for ages!
I've been stuck on it for ages!Hey so sorry for late reply it was bugging me for ages!😂
Hope this is right:
I calculated original moles of ammonium sulphate which is 0.0227 and the moles we have already calculated which was 1.23x10-2 so then I divided 1.23x10-2 by original moles 0.0227 to get me 54.12%
Original post by haj101
Hey so sorry for late reply it was bugging me for ages!😂
Hope this is right:
I calculated original moles of ammonium sulphate which is 0.0227 and the moles we have already calculated which was 1.23x10-2 so then I divided 1.23x10-2 by original moles 0.0227 to get me 54.12%
Hope this is right:
I calculated original moles of ammonium sulphate which is 0.0227 and the moles we have already calculated which was 1.23x10-2 so then I divided 1.23x10-2 by original moles 0.0227 to get me 54.12%
Thank you so much for your effort
and I guess that's right!!Original post by PlayerBB
Thank you so much for your effort and I guess that's right!!
and I guess that's right!!You're welcome
xxAlways check the integral trace ratio and the molecular formula to get the numbers of hydrogens in each environment.
Remember the (n+1) rule for splitting only applies to non-equivalent protons, equivalent protons do not split.
'Creep' gives a clue as to which signals are coupled together (it may or may not be apparent).
Remember the (n+1) rule for splitting only applies to non-equivalent protons, equivalent protons do not split.
'Creep' gives a clue as to which signals are coupled together (it may or may not be apparent).
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