I'm not so sure but I would say Answer Choice B
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 ![[\text{OH}^{-}]](https://tex.z-dn.net/?f=%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D)
from pH:
Assume that 
,
.
![[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D10%5E%7B-%5Ctext%7BpOH%7D%7D%20%3D10%5E%7B-2.80%7D%20%3D%201.59%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
.
Solve for ![[(\text{CH}_3)_3\text{N}]_\text{initial}](https://tex.z-dn.net/?f=%5B%28%5Ctext%7BCH%7D_3%29_3%5Ctext%7BN%7D%5D_%5Ctext%7Binitial%7D)
:
![\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,
![[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}](https://tex.z-dn.net/?f=%5B%28%5Ctext%7BCH%7D_3%29_3%5Ctext%7BN%7D%5D_%5Ctext%7Binitial%7D%20%3D%20%5B%28%5Ctext%7BCH%7D_3%29_3%5Ctext%7BN%7D%5D_%5Ctext%7Bfinal%7D)
.
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,
![[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5B%28%5Ctext%7BCH%7D_3%29_3%5Ctext%7BNH%7D%5E%7B%2B%7D%5D%20%3D%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%2010%5E%7B-2.80%7D%20%3D%201.58%5Ctimes%2010%5E%7B-3%7D%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
.
![\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \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_%5Ctextbf%7Binitial%7D%7D%20%3D%20%5Ctext%7BK%7D_b%20%3D%201.58%5Ctimes%2010%5E%7B-3%7D)
,
![[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}](https://tex.z-dn.net/?f=%5B%28%5Ctext%7BCH%7D_3%29_3%5Ctext%7BN%7D%5D_%5Ctextbf%7Binitial%7D%20%3D%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%5Ctext%7BK%7D_b%7D)
,
![[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5B%28%5Ctext%7BCH%7D_3%29_3%5Ctext%7BN%7D%5D_%5Ctext%7Binitial%7D%20%3D%5Cdfrac%7B%281.58%5Ctimes10%5E%7B-3%7D%29%5E%7B2%7D%7D%7B6.3%5Ctimes10%5E%7B-5%7D%7D%20%3D%200.040%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
.
<span>Evaporation from the oceans is the primary mechanism supporting the surface-to-atmosphere portion of the water cycle</span>
Not sure but it might be A or B
Answer:
A
Explanation:
Acetals are geminal-diether derivatives of aldehydes or ketones, formed by reaction with two equivalents (or an excess amount) of an alcohol and elimination of water. Ketone derivatives of this kind were once called ketals, but modern usage has dropped that term.
To achieve effective hemiacetal or acetal formation, two additional features must be implemented. First, an acid catalyst must be used because alcohol is a weak nucleophile and second the water produced with the acetal must be removed from the reaction.
