Calcium carbonate is most likely to dissolve in water with:
e. Lots of Carbon dioxide and colder temperature.
Explanation:
- Calcium carbonate is very sparingly soluble in water. However, it has been observed to dissolve in cold water with higher concentration of Carbon dioxide.
- The reason behind the observation is the formation of Calcium bicarbonate which is soluble in water.
- Higher concentration of Carbon dioxide in water turns it acidic. When this acidic water reacts with calcium carbonate it forms Calcium bicarbonate which is soluble in water.
- So,Calcium carbonate is most likely to dissolve in water with lots of Carbon dioxide and colder temperature.
Answer:
See explanation
Explanation:
If we look at the models, we will see that the three fluorine atoms in CF3COOH are attached to the carbon that is next to the -COOH group.
As a result of the electron withdrawing effect of the three fluorine atoms, CF3COOH is much more acidic (104 times more acidic) than CH3COOH. This is reflected in the value of the Ka for each acid.
This electron withdrawing effect of the three fluorine atoms also stabilizes CF3COO- much more than CH3COO-.
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:
1/9 moles
Explanation:
No of moles = mass/molar mass
No of moles = 2g/18gmol-1
No of moles = 1/9 moles
Use the specific heat equation for each one:
Q = mCT
m is mass
T is change in temp
C is spec heat
You will not only have to calculate three different specific heats, but also do the heat of vaporization and heat of fusion.
Heat of vaporization is 3.33 * 10^5 J/Kg
Heat of Fusion is 2.26 * 10^6 J/Kg
You’ll want to make sure all your units match. So change all Kg to g (or g to Kg)
-20 to 0 = mC(Ice)T
Melt ice (fusion) = m(heat of fusion)
0 to 100 = mC(Water)T
Boil (Vaporization) = m(heat of vaporization)
100 to 250 = mC(Steam)T
Then add each one together
