Answer:
both are done due to the enviroment??
0.040 mol / dm³. (2 sig. fig.)
<h3>Explanation</h3>
in this question acts as a weak base. As seen in the equation in the question,
produces
rather than
when it dissolves in water. The concentration of
will likely be more useful than that of
for the calculations here.
Finding the value of
from pH:
Assume that
,
.
.
Solve for
:
![\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D_%5Ctext%7Bequilibrium%7D%5Ccdot%5B%28%5Ctext%7BCH%7D_3%29_3%5Ctext%7BNH%7D%5E%7B%2B%7D%5D_%5Ctext%7Bequilibrium%7D%7D%7B%5B%28%5Ctext%7BCH%7D_3%29_3%5Ctext%7BN%7D%5D_%5Ctext%7Bequilibrium%7D%7D%20%3D%20%5Ctext%7BK%7D_b%20%3D%201.58%5Ctimes%2010%5E%7B-3%7D)
Note that water isn't part of this expression.
The value of Kb is quite small. The change in
is nearly negligible once it dissolves. In other words,
.
Also, for each mole of
produced, one mole of
was also produced. The solution started with a small amount of either species. As a result,
.
,
,
.
The atomic number of an element never changes no matter how many neutrons they have so Sulfur would just have an atomic number of 16 like normal.
Answer: 120g/mol
Explanation:
The first step we are to take is to calculate the freezing point depression of the solution.
ΔT(f) = freezing point of pure solvent - freezing point of solution
ΔT(f) = 5.48 - 3.77
ΔT(f) = 1.71°C
Next we are to calculate the molal concentration of the solution using freezing point depression
ΔT(f) = K(f) * m
m = ΔT(f)/K(f)
m = 1.71/5.12
m = 0.333 molal
Now, we calculate the molecular weight of the unknown...
m = 0.333 mol = 0.333 mol X per kg of benzene
moles of X = 0.333 mol of X per kg of benzene * 0.5kg of benzene
moles of X = 0.1665
molecular weight of X = 20g of X/0.1665
molecular weight of X = 120/mol
In descending order from top:
E
F
D
A
C
B
All you really need to do is remember the symbols of each, and you’ve got it.