<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:
Using the grid, graph, and given data, estimate the distance traveled by the walker during the two hour interval from t= 0 t = 0 to t= 2. t = 2. You should use time intervals of width Δt= 0.5.
Explanation:
Using the grid, graph, and given data, estimate the distance traveled by the walker during the two hour interval from t= 0 t = 0 to t= 2. t = 2. You should use time intervals of width Δt= 0.5.
Hi there!
According to Newton's second law:
∑F = m · a, where:
∑F = net force (N = kgm/s²)
m = mass (kg)
a = acceleration (m/s²)
Rearrange to solve for acceleration:
F/m = a
20N / 4.0kg = 5 m/s²
Momentum is (mass) times (speed), so nothing that is at rest has any momentum. If the battleship is at rest, then a mosquito in flight, a leaf falling from a tree, and your speedy baseball each have more momentum than the ship has.
<h2>it is round and in a ovalish shape it has the sun and moon surrounding it. it spins. when its facing the sun its giving light to half the world while its nighttime at other countries and then it spins again. it spins once in a day and once at nighttime </h2>
I hope this helps :)