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.
Stir the water continuously, this is the only logical answer. Adding powdered sugar, decreasing the volume, increasing the amount of suga, cooling, all don't make the sugar dissolve quicker.
Answer:
Answer below
Explanation:
First to show combustion, you use O₂.
So C₆H₁₀O₅ + O₂ is the reaction. Assuming complete combustion, all you get as products are CO₂ and H₂O. Then you have to balance the full reaction.
C₆H₁₀O₅ + 6O₂ -----> 6CO₂ + 5H₂O
6 Carbons on each side
10 Hydrogens on each side
17 Oxygens on each side
Oxygen is used up and glucose is broken down.
