Answer:
b- 4.4 * 10^-12.
Explanation:
Hello.
In this case, as the reaction:
A + 2B → 3C
Has an equilibrium expression of:
![K_1=\frac{[C]^3}{[A][B]^2}=2.1x10^{-6}](https://tex.z-dn.net/?f=K_1%3D%5Cfrac%7B%5BC%5D%5E3%7D%7B%5BA%5D%5BB%5D%5E2%7D%3D2.1x10%5E%7B-6%7D)
If we analyze the reaction:
2A + 4B → 6C
Which is twice the initial one, the equilibrium expression is:
![K_2=\frac{[C]^6}{[A]^2[B]^4}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BC%5D%5E6%7D%7B%5BA%5D%5E2%5BB%5D%5E4%7D)
It means that the equilibrium constant of the second reaction is equal to the equilibrium constant of the first reaction powered to second power:
Thus, the equilibrium constant of the second reaction turns out:
Therefore, the answer is b- 4.4 * 10^-12.
Best regards.
Answer:
x(t) = −39e
−0.03t + 40.
Explanation:
Let V (t) be the volume of solution (water and
nitric acid) measured in liters after t minutes. Let x(t) be the volume of nitric acid
measured in liters after t minutes, and let c(t) be the concentration (by volume) of
nitric acid in solution after t minutes.
The volume of solution V (t) doesn’t change over time since the inflow and outflow
of solution is equal. Thus V = 200 L. The concentration of nitric acid c(t) is
c(t) = x(t)
V (t)
=
x(t)
200
.
We model this problem as
dx
dt = I(t) − O(t),
where I(t) is the input rate of nitric acid and O(t) is the output rate of nitric acid,
both measured in liters of nitric acid per minute. The input rate is
I(t) = 6 Lsol.
1 min
·
20 Lnit.
100 Lsol.
=
120 Lnit.
100 min
= 1.2 Lnit./min.
The output rate is
O(t) = (6 Lsol./min)c(t) = 6 Lsol.
1 min
·
x(t) Lnit.
200 Lsol.
=
3x(t) Lnit.
100 min
= 0.03 x(t) Lnit./min.
The equation is then
dx
dt = 1.2 − 0.03x,
or
dx
dt + 0.03x = 1.2, (1)
which is a linear equation. The initial condition condition is found in the following
way:
c(0) = 0.5% = 5 Lnit.
1000 Lsol.
=
x(0) Lnit.
200 Lsol.
.
Thus x(0) = 1.
In Eq. (1) we let P(t) = 0.03 and Q(t) = 1.2. The integrating factor for Eq. (1) is
µ(t) = exp Z
P(t) dt
= exp
0.03 Z
dt
= e
0.03t
.
The solution is
x(t) = 1
µ(t)
Z
µ(t)Q(t) dt + C
= Ce−0.03t + 1.2e
−0.03t
Z
e
0.03t
dt
= Ce−0.03t +
1.2
0.03
e
−0.03t
e
0.03t
= Ce−0.03t +
1.2
0.03
= Ce−0.03t + 40.
The constant is found using x(t) = 1:
x(0) = Ce−0.03(0) + 40 = C + 40 = 1.
Thus C = −39, and the solution is
x(t) = −39e
−0.03t + 40.
<span>Glucose and oxygen react together in cells to produce carbon dioxide and water and releases energy.</span>
Charge # = protons - electons
Mass # = protons + neutrons
so that would be
3-3= charge#
3+4= mass#
D-sublevel can occupy 10 electrons whereas s-sublevel can occupy 2 electrons...
