I believe it’s C , People are burning too many fossil fuels .
Answer:
how can I solve this ?4Al+3O2 produce 2Al2O3 find a) oxygen atoms needed to react with 5.4 g of aluminium b) grams of oxygen needed to react with 0.6 mol of aluminium?
(A) n=m/M,
n(Al)=5.4/27=0.2 moles
n(O2)=n(Al)*3/4=0.2*3/4=0.15 moles
Number of oxygen atoms= n(O2)*Avogadro's number
=0.15*6.02*10^23=9.03*10^22 oxgyen atoms
(B)
n=m/M
n(Al)=0.6/27=0.02222 moles
n(O2)=n(Al)*3/4=0.016666 moles
m=n*M
m(O2)=0.0166666*32=0.53333 grams
The rate constant : k = 9.2 x 10⁻³ s⁻¹
The half life : t1/2 = 75.3 s
<h3>Further explanation</h3>
Given
Reaction 45% complete in 65 s
Required
The rate constant and the half life
Solution
For first order ln[A]=−kt+ln[A]o
45% complete, 55% remains
A = 0.55
Ao = 1
Input the value :
ln A = -kt + ln Ao
ln 0.55 = -k.65 + ln 1
-0.598=-k.65
k = 9.2 x 10⁻³ s⁻¹
The half life :
t1/2 = (ln 2) / k
t1/2 = 0.693 : 9.2 x 10⁻³
t1/2 = 75.3 s
