Answer:
The
of a substrate will be "10 μM".
Explanation:
The given values are:

![[Substract] = 40 \ \mu M](https://tex.z-dn.net/?f=%5BSubstract%5D%20%3D%2040%20%5C%20%5Cmu%20M)

Reaction velocity, 
As we know,
⇒ ![Vo=\frac{K_{cat}[E_{t}][S]}{K_{m}+[S]}](https://tex.z-dn.net/?f=Vo%3D%5Cfrac%7BK_%7Bcat%7D%5BE_%7Bt%7D%5D%5BS%5D%7D%7BK_%7Bm%7D%2B%5BS%5D%7D)
On putting the estimated values, we get
⇒ 
⇒ 
⇒ 
On subtracting "40" from both sides, we get
⇒ 
⇒ 
The correct answer would be the first option. Material A having a smaller latent heat of fusion would mean that it will take only less energy to phase change into the liquid phase. Latent of heat of fusion is the amount of energy needed of a substance to phase change from solid to liquid or liquid to solid.
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,
.
,
,
.