Hi Guys, i got stuck on the part 2(a)(ii) of the question in the attachment . should i use the formula for activity to calculate t? but i don't have enough information to use the formula
you don't need to give a monkeys about what the actual number of atoms is... but you ARE interested in the ratio (which is in the question)
so try N0 = 1 N = 0.375
though any values where N=0.375 N0 will give the exact same answer
How did you realise that? The question only mentioned that the new Ca has 0.375 as many Ca-14 atoms as an equal mass of living wood? Did you get it from this part of the question?
How did you realise that? The question only mentioned that the new Ca has 0.375 as many Ca-14 atoms as an equal mass of living wood? Did you get it from this part of the question?
well it's a disguised ratio - only thinly disguised, but the examiner is hoping you'll bottle it if you see something that's not exactly what you're used to.
well it's a disguised ratio - only thinly disguised, but the examiner is hoping you'll bottle it if you see something that's not exactly what you're used to.
just practice I suppose...
So basically if varaible A has 0.375 as many atom as varaible B of the same mass , the atom's ratio of A over B is 1/ 0.375?