Hard Chemistry Calculation questions anybody? I've got a test coming up tomorrow on GCSE chemistry calculation questions and please include answers if you have them???
0.957mol/dm3 of H2SO4 was neutralised by 858cm3 of 0.642mol/dm3 NaOH. Calculate the volume of H2SO4 neutralised.1) Write the balanced equation. H2SO4 2NaOH → Na2SO4 2H2O2) Calculate the number of mols of NaOH. mol = concentration x volume(divide by 1000 if shown in cm3 to get dm3) =0.642*0.858=0.540564 mols of NaOH present3) Implement ratio/proportion of element. (Numbers in front of substances e.g. 2NaOH denotes 2 mols of NaOH present) NaOH2SO4 2:1 0.540564/2=0.270282 mols of H2SO4. 4) Calculate volume by using volume = mols/concentration 0.270282/0.957 = 0.2824263323dm3 of H2SO4 used
If you really want to challenge, I'd strongly recommend looking at the AS-style questions. What i've seen from the new specification papers is that they're very in lieu of AS-style papers.
Have a look at OCR/AQA AS Papers for the new a level and also old a level specification. (Some questions from the very old a level specification were sampled in the new specimen papers for GCSE).
Obviously, if you see something clearly beyond what you know don't do it - but if you do want "hard" questions, they'll be a good source for you (and also the exam board making the papers).
Hard Chemistry Calculation questions anybody? I've got a test coming up tomorrow on GCSE chemistry calculation questions and please include answers if you have them???
0.957mol/dm3 of H2SO4 was neutralised by 858cm3 of 0.642mol/dm3 NaOH. Calculate the volume of H2SO4 neutralised.
2) Calculate the number of mols of NaOH. mol = concentration x volume(divide by 1000 if shown in cm3 to get dm3) =0.642*0.858=0.540564 mols of NaOH present
3) Implement ratio/proportion of element. (Numbers in front of substances e.g. 2NaOH denotes 2 mols of NaOH present) NaOH2SO4 2:1 0.540564/2=0.270282 mols of H2SO4.
4) Calculate volume by using volume = mols/concentration 0.270282/0.957 = 0.2824263323dm3 of H2SO4 used