Rewrite the formula C=5/9(F-32) substituting 23 for C: 23=5/9(F-32), then multiply both sides by the reciprocal of 5/9.
(9/5)*(23)=(9/5)*5/9(F-32)
41.4=F-32; add 32 to both sides.
41.4+32=F-32+32
73.4=F
H2SO4 + 2RbOH -> Rb2SO4 + 2H2O
If you want an explanation, keep reading.
In the first portion, there are two hydrogen ions and four sulfate ions.
The second portion has one rubidium ions and one hydroxide ion.
On the other side of the equation, in order to keep those two rubidiums balanced, you'll need to add a two at the beginning of the second portion, but in that process you are giving a second hydroxide value.
Back to the right side, there is there is water (H2O).
On the first portion, there were two hydrogen ions. The second portion also has two hydroxides because of the value change (adding the two to the front).
So on the fourth portion, you'd have to add another two so you could balance the four hydrogen ions (H2 and 2OH) and the two oxygen ions (2OH).
I hope this was easy to understand.
I think it’s 7.41 because you count up all the atoms and find out how many are x (the large grey ones) and you do 2/27 x 100 which gives you 7.41 :) (sorry if i counted wrong it’s kinda hard)
