Answer:
2.7 x 10^-1
Explanation:
The places you move to the left are added to the 10 in negative exponential
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.
Answer:
n = 2 moles (1 sig-fig)
Explanation:
Using the Ideal Gas Law equation (PV = nRT), solve for n (= moles) and substitute data for ...
pressure = P(atm) = 100atm
volume =V(liters) = 50L
gas constant = R = 0.08206L·atm/mol·K
temperature = T(Kelvin) = °C + 273 = (35 + 273)K = 308K
PV = nRT => n = PV/RT = (100atm)(50L)/(0.08206L·atm/mol·K)(308K)
∴ n(moles) = 1.978moles ≅ 2 moles gas (1 sig-fig) per volume data (= 50L) that has only 1 sig-fig. (Rule => for multiplication & division computations round final answer to the measured data having the least number of sig-figs).
Answer:
disposal of radioactive waste
Explanation:
if too much is released, it can wipe out large parts of the country
Answer:
0.64 mol
Explanation:
Using ideal gas equation as follows;
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
According to the information provided in this question;
P = 0.72atm
V = 22 L
T = 30°C = 30 + 273 = 303K
n = ?
Using PV = nRT
n = PV/RT
n = (0.72 × 22) ÷ (0.0821 × 303)
n = 15.84 ÷ 24.88
n = 0.64 mol
