Question

help due soon

Answer 1

@razor

Answer 2

fun fun fun
for the first one the first hint is \[42H _{2}\] which is 84 H's
now \[C _{6}H _{12}O _{6}\] is the only other thing that contains H on the right side, you multiply this by a constant of 14 because 7 x 12 = 84
this is also multiplied by the whole of \(C _{6}H _{12}O _{6}\) so that means \[7C_{6}\] and 7 times 6 is 42, to match 42 C on the left side the first must have a constant of 42.
Lastly O, on the left side you have 42*2 + 42*1 = 126 oxygens
on the right side you already had \(7O_{6}\) which is 42 O's and you need 84 more O's since the constant for the last is infront of \[O _{2}\] that constant will be 42


So you put in 42, 7, 42 respectively

Answer 3

Please double check, I may have made some mistakes

Answer 4

you see how I worked that problem out? basically same steps for every problem, you are given some info about one of the elements and use that to determine your constants

Answer 5

ok