<span>B. S⁰(s) + 2H⁺ + 2e⁻ –--> H2S⁰(g)
by mass: 1 S and 2 H ----> 2 H and 1S True.
by charge : 0 +(2*(+1)) + 2*(-1) = 0, 0+2-2=0, 0 = 0 True.</span>
According to this formula :
㏑[A] /[Ao] = - Kt
when we have Ao = 0.3 m
and K =0.46 s^-1
t = 20min = 0.2 x 60 =12 s
So by substitution :
㏑[A] / 0.3 = - 0.46 * 12
㏑[A] / 0.3 = - 5.52
by taking e^x for both side of the equation we can get [A]
∴[A] = 0.0012 mol dm^-3
Explanation:
The structures of both acetone and propanal are shown below:
In the formula of propanal there is -CHO functional group at the end.
In acetone -CO- group is present in the middle that is on the second carbon.
The molecular formula is C3H6O.
Both have same molecular formula but different structural formulas.