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Quick Help please :)
- Thread starter SexyFag
- Start date
- Messages
- 89
- Reaction score
- 149
- Points
- 43
I did them last night. I suck at moles. But a friend told me how to attempt them.
concentration= no. of moles/volume
we have conc of FA2 which is 0.50 and volume which we took for example 33.5 cm^3. You have to change volume from cm3 to dm3. So
moles= concentration of FA2x volume of FA3 taken
moles= 0.50x33.5x10^-3
moles= 0.017 mol.
This is the first part. Let me look at others.
concentration= no. of moles/volume
we have conc of FA2 which is 0.50 and volume which we took for example 33.5 cm^3. You have to change volume from cm3 to dm3. So
moles= concentration of FA2x volume of FA3 taken
moles= 0.50x33.5x10^-3
moles= 0.017 mol.
This is the first part. Let me look at others.
- Messages
- 89
- Reaction score
- 149
- Points
- 43
For part 2, you have to use ratio method.
250 cm3 makes 0.017 moles
17.75(supposed) will make X moles.
So the anser would be like 1.207x10^-3 moles of HC.
For part 3 again we'll use ratio method.
1 mole of Na2Co3 is reacted with 2 moles of HCl
x moles of NaCO3 will give X moles.
So,
1:2
x:1.207x10^-3
cross multiply them and you'll get 6.035x10^-4 mol.
250 cm3 makes 0.017 moles
17.75(supposed) will make X moles.
So the anser would be like 1.207x10^-3 moles of HC.
For part 3 again we'll use ratio method.
1 mole of Na2Co3 is reacted with 2 moles of HCl
x moles of NaCO3 will give X moles.
So,
1:2
x:1.207x10^-3
cross multiply them and you'll get 6.035x10^-4 mol.