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# OCR chemistry past questions confusion AS Watch

1. A student mixes 100 cm3 of 0.200 mol dm–3 NaCl(aq) with 100 cm3 of 0.200 mol dm–3 Na2CO3(aq).
What is the total concentration of Na+ ions in the mixture formed?
A 0.100 mol dm–3
B 0.200 mol dm–3
C 0.300 mol dm–3
D 0.400 mol dm–3
How would I go about working this out ? The answer is C
Which mass of substance contains the greatest number of atoms?
A 3.00 g of ammonia, NH3
B 3.00 g of chloromethane, CHCl3
C 4.00 g of hydrogen sulfide, H2S
D 4.00 g of hydrogen chloride, HCl
Which volume of oxygen gas at room temperature and pressure is required for complete combustion of 1.25x10-3 mol of propan-1-ol ?
A.105cm3
B.120cm3
C.135cm3
2. (Original post by Hmb28)
A student mixes 100 cm3 of 0.200 mol dm–3 NaCl(aq) with 100 cm3 of 0.200 mol dm–3 Na2CO3(aq).
What is the total concentration of Na+ ions in the mixture formed?
A 0.100 mol dm–3
B 0.200 mol dm–3
C 0.300 mol dm–3
D 0.400 mol dm–3
How would I go about working this out ? The answer is C
Which mass of substance contains the greatest number of atoms?
A 3.00 g of ammonia, NH3
B 3.00 g of chloromethane, CHCl3
C 4.00 g of hydrogen sulfide, H2S
D 4.00 g of hydrogen chloride, HCl
Which volume of oxygen gas at room temperature and pressure is required for complete combustion of 1.25x10-3 mol of propan-1-ol ?
A.105cm3
B.120cm3
C.135cm3
For the first question work out the number of moles of Na+ ions in each sample.
NaCl n= 0.1 x 0.2 = 0.02 moles
Na2CO3 n= 0.1 x 0.2 = 0.02 moles BUT you have 2 Na + ions so there are 0.04 moles of Na+ in Na2CO3.
So overall you have 0.06 moles of Na+ in a TOTAL volume of 200cm^3. Therefore the concentration is 0.06/0.2=0.3 moldm^-3 (C).

For the next question work out the number of moles of each. You can then multiply by Avogadro's number, but because you don't need to specify the number of atoms, you can just tell that the one with the highest number of moles will also have the highest number of atoms (they are, after all, being multiplied by the same number). The answer is NH3 with 0.176moles (so 1.06 x 10e23 atoms)

For the last question start by writing out a BALANCED equation for the complete combustion of propan-1-ol. CH3CH2CH2OH + 4.5O2 ---> 3CO2 + 4H2O. You know the number of moles of propan-1-ol so simply use the stoichiometric ratio to work out the number of moles of O2 then multiply by 24000cm^3 (because RTP) to find the volume. 1.25x10e-3 x 4.5 = 5.625 x 10e-3 moles of O2 x 24000 = 135 cm^3.
3. (Original post by victoria98)
For the first question work out the number of moles of Na+ ions in each sample.
NaCl n= 0.1 x 0.2 = 0.02 moles
Na2CO3 n= 0.1 x 0.2 = 0.02 moles BUT you have 2 Na + ions so there are 0.04 moles of Na+ in Na2CO3.
So overall you have 0.06 moles of Na+ in a TOTAL volume of 200cm^3. Therefore the concentration is 0.06/0.2=0.3 moldm^-3 (C).

For the next question work out the number of moles of each. You can then multiply by Avogadro's number, but because you don't need to specify the number of atoms, you can just tell that the one with the highest number of moles will also have the highest number of atoms (they are, after all, being multiplied by the same number). The answer is NH3 with 0.176moles (so 1.06 x 10e23 atoms)

For the last question start by writing out a BALANCED equation for the complete combustion of propan-1-ol. CH3CH2CH2OH + 4.5O2 ---> 3CO2 + 4H2O. You know the number of moles of propan-1-ol so simply use the stoichiometric ratio to work out the number of moles of O2 then multiply by 24000cm^3 (because RTP) to find the volume. 1.25x10e-3 x 4.5 = 5.625 x 10e-3 moles of O2 x 24000 = 135 cm^3.
Thank you , makes it obvious now

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