The answer is 3.134 times 10 to the 9th
<span>THIS IS A GAS PHASE REACTION AND WE ARE GIVE PARTIAL PRESSURES . I WRITE IN TERMS OF P RATHER THAN CONCENTRATION :
lnPso2cl12=-kt+lnPso2cl1
initial partial pressure Pso2cl12 the rate constant k and the time t
lnPso2cl12=(4.5*10-2*s-1)*65*s+ln (375)
so lnPso2cl12=3.002
we take the base e antilog:
lnPso2cl12=e3.002
Pso2cl12=20 torr
we use the integrated first order rate
lnPso2cl12=3.002=k*t+ lnPso2cl12=3.002
we use the same rate constant and initial pressure
k=4.5*10-2*s-1
Pso2cl12=375
Pso2cl12=1* so2cl12
Pso2cl12=37.5 torr
subtract in Pso2cl12 grom both side
lnPso2cl12- lnPso2cl12=-kt
ln(x)-ln(y)=ln (x/y)
ln (Pso2cl12/Pso2cl20)=-kt
we get t
-1/k*ln(Pso2cl12/Pso2cl20)=t
t=51 s</span>
Answer:
V= 14.2m/s
Explanation:
We know that acceleration= dv/dt
So 16m/s²=dv/ dt = v dv/ds
So this wil be
Integral of 16m/s² ds [at 2,2]= integral of v dv at[ 0, v]
So 16[s (3/2)/3/2] at ( s,3) = v²/2
At s= 6m
So v² = 64/3( 6^1.5-3^1.5)
= 14.2m/s