The correct answer is:
neutrality
on the pH scale 7 is neutral above 7 is basic and below 7 is acidic.
Explanation:
The pH is an example of how acidic/basic water is. The scale goes from 0 - 14, with 7 being neutral, pH of less than 7 designates acidity, whereas a pH of greater than 7 intimates a base. pH is really a pattern of the relative amount of free hydrogen and hydroxyl ions in the water.pH is a measure of how acidic/basic water is.
It is neutral because, at this pH, the molar concentrations of Hydroxide and Hydronium are same, therefore the substance is not a base nor an acid.
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:
9.79740949850 moles
Explanation:
- 1 mole = Avogardo's Number <<6.022 E 23 <<particles, atoms, etc.>>
- This problem can be solved using dimensional analysis by multiplying atoms (5.9E24 atoms) by (1) mole and then dividing the number by Avogardo's number (6.022 E 23 atoms).
- Note: E = * 10
Side Note: Please let me know if you need any clarifications about this!
