Answer:

Explanation:
Hello,
In this case, given the acid, we can suppose a simple dissociation as:

Which occurs in aqueous phase, therefore, the law of mass action is written by:
![Ka=\frac{[H^+][A^-]}{[HA]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
That in terms of the change
due to the reaction's extent we can write:

But we prefer to compute the Kb due to its exceptional weakness:

Next, the acid dissociation in the presence of the base we have:
![Kb=\frac{[OH^-][HA]}{[A^-]}=1x10^{6}=\frac{x*x}{0.1-x}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BOH%5E-%5D%5BHA%5D%7D%7B%5BA%5E-%5D%7D%3D1x10%5E%7B6%7D%3D%5Cfrac%7Bx%2Ax%7D%7B0.1-x%7D)
Whose solution is
which equals the concentration of hydroxyl in the solution, thus we compute the pOH:
![pOH=-log([OH^-])=-log(0.0999)=1](https://tex.z-dn.net/?f=pOH%3D-log%28%5BOH%5E-%5D%29%3D-log%280.0999%29%3D1)
Finally, since the maximum scale is 14, we can compute the pH by knowing the pOH:

Regards.
Answer:
See explanation and image attached
Explanation:
The reaction of 1-bromo-2-tert-butylcyclohexane with potassium tert-butoxide is an elimination reaction that occurs by E2 mechanism.
The E2 reaction proceeds faster when the hydrogens are in an antiperiplanar position at an angle of 180 degrees.
This is only attainable in the trans isomer of 1-bromo-2-tert-butylcyclohexane. Hence trans 1-bromo-2-tert-butylcyclohexane reacts faster with potassium tert-butoxide
1. Salt is KNO₃<span>
<span>This is a </span>strong acid - strong base<span> <span>reaction. </span></span>HNO</span>₃ is the strong acid<span> <span>and </span></span>KOH is the strong
base<span>. </span><span>
H</span>⁺<span> in the HNO₃<span>
and </span></span>OH⁻<span> <span>of the KOH pair up and make </span></span>H₂O(l)<span>. </span><span>
NO</span>₃⁻<span> <span>and </span></span>K⁺<span> <span>pair up to make </span></span>KNO₃ salt<span>. </span><span>
<span>When writing chemical formulas </span>positive ion comes first<span> <span>and second
is negative ion. The charges should be switched. Since </span></span>positive ion has +1 and negative
ion has -1<span> <span>after
the switching off charges </span>the </span>product should be KNO</span>₃.<span>
Balance
equation is </span><span>
HNO</span>₃<span>(aq) + KOH(aq) → H</span>₂O(l) + KNO<span>₃(aq)</span><span>
<span>
2. Salt is Ca(NO</span></span>₃)₂<span>
</span>This is a strong acid - strong
base<span> reaction. </span>HNO₃ is the strong acid<span> and </span>Ca(OH)₂ is the strong base<span>. </span><span>
<span>
H</span></span>⁺<span> in the HNO₃ and </span>OH⁻<span> of the Ca(OH)₂
pair up and make </span>H₂O(l)<span>. </span><span>
Ca²⁺
and </span>NO₃⁻<span> pair up to make </span>Ca(NO₃)₂ salt<span>. </span><span>
<span>
</span><span>Positive ion is </span>Ca²⁺<span>
which has </span></span>+2 charge<span> and negative ion is</span> NO₃⁻<span> <span>which has </span></span>-1 charge<span>. From switching the charges </span>Ca²⁺ gets 1<span> <span>while </span></span>NO₃⁻ gets 2.<span> Hence, the salt should be </span>Ca(NO₃)₂.<span>
Balanced equation
is
</span>2HNO₃<span>(aq) + Ca(OH)</span>₂<span>(aq) → 2H</span>₂O(l) + Ca(NO<span>₃)₂(aq)</span><span>
<span>
3. Salt is CaCl</span></span>₂<span>
This is a strong acid - strong base<span> reaction. </span>HCl is the
strong acid<span> and </span>Ca(OH)</span>₂ is the strong base<span>. </span><span>
<span>
H</span></span>⁺<span> in the HCl and </span>OH⁻<span> of the Ca(OH)₂
pair up and make </span>H₂O(l)<span>. </span><span>
Ca²⁺
and </span>Cl⁻<span> pair up to make </span>CaCl₂ salt<span>. </span><span>
<span>
</span><span>Positive ion is </span>Ca²⁺
which has </span>+2
charge<span> and negative ion is</span> Cl⁻<span> which has </span>-1
charge<span>. By switching the charges </span>Ca²⁺ gets 1<span> while </span>NO₃⁻ gets 2.<span> Hence, the salt should be </span>CaCl₂.<span>
Balance
equation is
</span><span>2HCl(aq) + Ca(OH)</span>₂<span>(aq) → 2H</span>₂O(l) + CaCl₂<span>(aq)
4. Salt is KCl<span>
</span>This is a strong acid - strong base<span> reaction. </span>HCl is the
strong acid<span> and </span>KOH is
the strong base<span>. </span>
<span>
H</span></span>⁺<span> in the HCl and </span>OH⁻<span> of the KOH pair up and make </span>H₂O(l)<span>. </span><span>
K</span><span>⁺ and </span>Cl⁻<span> pair up to make </span>KCl salt<span>. </span><span>
<span>
</span><span>Positive ion is K</span></span><span>⁺ which has </span>+1
charge<span> and negative ion is</span> Cl⁻<span> which has </span>-1
charge<span>. By switching the charges </span>K⁺ gets 1<span> and </span>Cl⁻ also gets 1.<span> Hence, the salt should be </span>KCl.<span>
Balance
equation is
</span><span>HCl(aq) + KOH(aq) → H</span>₂<span>O(l) + KCl(aq)</span>
The balanced chemical equation is given as:
2CH3CH2OH(l) → CH3CH2OCH2CH3(l) + H2O(l)
We are given the yield of CH3CH2OCH2CH3 and the amount of ethanol to be used for the reaction. These values will be the starting point for the calculations.
Theoretical amount of product produced:
329 g CH3CH2OH ( 1 mol / 46.07 g ) ( 1 mol CH3CH2OCH2CH3 / 2 mol CH3CH2OH ) (74.12 g / mol ) = 264.66 g CH3CH2OCH2CH3
% yield = .775 = actual yield / 264.66
actual yield = 205.11 g CH3CH2OCH2CH3