<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>
It hink that it is the 2nd one
1/s' = 1/f - 1/s
= 1/12 - 1/8
= 2/24 - 3/24
= 2-3/24
= - 1/24
s'= -24
M= |s'/s|
M = |24/8|
M = 3
Answer: 14139.19 m
Explanation:
This situation is related to parabolic motion and can be solved using the following equations:
(1)
(2)
Where:
is the horizontal distance (where the artillery shell lands)
is the initial velocity
is the angle
is the time
is the final height
is the initial height
is the acceleration due gravity, always directed downwards
So, let's begin by isolating from (2):
(3)
(4)
Substituting (4) in (1):
(5)
Rewriting (5) and taking into account 
:
(6)
(7)
Finally:
