for the gas mc question, was the total amount of gas 30cm3? You start with 30 of A (can't remember), 20 of B and 0 of product. The mole ratios were 1:1:1, so 20 of the product forms but as 10 of A is in excess it remains so 20+10=30??
Me too, I got the same, but I have strong feeling we have may lost a mark by not writing 0.70g as most of the values were like that.
my chemistry teacher once gave us a test in which all the values given were to 3sf, except for one which was 100cm^3. the whole class gave their answer to 3sf because they assumed that 100cm^3 was also 3sf, but he told us that it's not clear that it is so we should have answered to 1sf. (SO IF WE LOSE THE MARK FOR THIS I'M GONNA BE VERY MAD AT HIM LOL)
my chemistry teacher once gave us a test in which all the values given were to 3sf, except for one which was 100cm^3. the whole class gave their answer to 3sf because they assumed that 100cm^3 was also 3sf, but he told us that it's not clear that it is so we should have answered to 1sf. (SO IF WE LOSE THE MARK FOR THIS I'M GONNA BE VERY MAD AT HIM LOL)
I also study Physics and Biology. In both I have learnt that to round to the appropriate significant figure you find the value given in the least significant figure and use that.
I can understand how you worked it out. I used another method by working out the number of moles using moles=mass/mr of baso4 then i wrote the equation bacl2+h2so4=baso4+2hcl and everything is a 1:1 ratio so the same moles of bacl2 were used. I worked out the mass of bacl2 from this using the mass=mr x moles. Then i took this away from the mass of both bacl2 and mgcl2 to work out how much mgcl2 there was. My answer was also 35.5% 👍🏿👍🏿
I can understand how you worked it out. I used another method by working out the number of moles using moles=mass/mr of baso4 then i wrote the equation bacl2+h2so4=baso4+2hcl and everything is a 1:1 ratio so the same moles of bacl2 were used. I worked out the mass of bacl2 from this using the mass=mr x moles. Then i took this away from the mass of both bacl2 and mgcl2 to work out how much mgcl2 there was. My answer was also 35.5% 👍🏿👍🏿
I got 35.5% too. Got a bit nervous when some people said 29.5%, but i think that method was correct.
Looking at some of the answers on page one, some of the content you do at AS level is similar to highers we do in Scotland. I sat my higher chemistry exam last week.
I can understand how you worked it out. I used another method by working out the number of moles using moles=mass/mr of baso4 then i wrote the equation bacl2+h2so4=baso4+2hcl and everything is a 1:1 ratio so the same moles of bacl2 were used. I worked out the mass of bacl2 from this using the mass=mr x moles. Then i took this away from the mass of both bacl2 and mgcl2 to work out how much mgcl2 there was. My answer was also 35.5% 👍🏿👍🏿
First person I have seen to get the same answer as me! Didn't ask anyone after the exam though, to be fair.
Guys for the gas mc question, was the total amount of gas 30cm3? You start with 30 of A (can't remember), 20 of B and 0 of product. The mole ratios were 1:1:1, so 20 of the product forms but as 10 of A is in excess it remains so 20+10=30??
did the question ask for the volume of product? i thought it asked for the volume at the end of the reaction, and because it wasn't an equilibrium reaction i just added them together and said it was 50 :/
did the question ask for the volume of product? i thought it asked for the volume at the end of the reaction, and because it wasn't an equilibrium reaction i just added them together and said it was 50 :/
It asked for the volume at the end of the reaction. I thought as 20 of A reacts with 20 of B to form 20cm3 of product, at the end of the reaction there is 20cm3 of product + the excess 10cm3 of A.