Answer:
Número de moles de cloro en número de moles de NaCl
Explanation:
espero que si sea la correcta
Answer:
C. 26.4 kJ/mol
Explanation:
The Chen's rule for the calculation of heat of vaporization is shown below:
![\Delta H_v=RT_b\left [ \frac{3.974\left ( \frac{T_b}{T_c} \right )-3.958+1.555lnP_c}{1.07-\left ( \frac{T_b}{T_c} \right )} \right ]](https://tex.z-dn.net/?f=%5CDelta%20H_v%3DRT_b%5Cleft%20%5B%20%5Cfrac%7B3.974%5Cleft%20%28%20%5Cfrac%7BT_b%7D%7BT_c%7D%20%5Cright%20%29-3.958%2B1.555lnP_c%7D%7B1.07-%5Cleft%20%28%20%5Cfrac%7BT_b%7D%7BT_c%7D%20%5Cright%20%29%7D%20%5Cright%20%5D)
Where,
is the Heat of vaoprization (J/mol)
is the normal boiling point of the gas (K)
is the Critical temperature of the gas (K)
is the Critical pressure of the gas (bar)
R is the gas constant (8.314 J/Kmol)
For diethyl ether:



Applying the above equation to find heat of vaporization as:
![\Delta H_v=8.314\times307.4 \left [ \frac{3.974\left ( \frac{307.4}{466.7} \right )-3.958+1.555ln36.4}{1.07-\left ( \frac{307.4}{466.7} \right )} \right ]](https://tex.z-dn.net/?f=%5CDelta%20H_v%3D8.314%5Ctimes307.4%20%5Cleft%20%5B%20%5Cfrac%7B3.974%5Cleft%20%28%20%5Cfrac%7B307.4%7D%7B466.7%7D%20%5Cright%20%29-3.958%2B1.555ln36.4%7D%7B1.07-%5Cleft%20%28%20%5Cfrac%7B307.4%7D%7B466.7%7D%20%5Cright%20%29%7D%20%5Cright%20%5D)

The conversion of J into kJ is shown below:
1 J = 10⁻³ kJ
Thus,

<u>Option C is correct</u>
The new pH is 7.69.
According to Hendersen Hasselbach equation;
The Henderson Hasselbalch equation is an approximate equation that shows the relationship between the pH or pOH of a solution and the pKa or pKb and the ratio of the concentrations of the dissociated chemical species. To calculate the pH of the buffer solution made by mixing salt and weak acid/base. It is used to calculate the pKa value. Prepare buffer solution of needed pH.
pH = pKa + log10 ([A–]/[HA])
Here, 100 mL of 0.10 m TRIS buffer pH 8.3
pka = 8.3
0.005 mol of TRIS.
∴ ![8.3 = 8.3 + log \frac{[0.005]}{[0.005]}](https://tex.z-dn.net/?f=8.3%20%3D%208.3%20%2B%20log%20%5Cfrac%7B%5B0.005%5D%7D%7B%5B0.005%5D%7D)
<em> </em>inverse log 0 = ![\frac{[B]}{[A]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BB%5D%7D%7B%5BA%5D%7D)
![\frac{[B]}{[A]} = 1](https://tex.z-dn.net/?f=%5Cfrac%7B%5BB%5D%7D%7B%5BA%5D%7D%20%3D%201)
Given; 3.0 ml of 1.0 m hcl.
pka = 8.3
0.003 mol of HCL.
![pH = 8.3 + log \frac{[0.005-0.003]}{[0.005+0.003]}\\pH = 8.3 + log \frac{[0.002]}{[0.008]}\\\\pH = 8.3 + log {0.25}\\\\pH = 8.3 + (-0.62)\\pH = 7.69](https://tex.z-dn.net/?f=pH%20%3D%208.3%20%2B%20log%20%5Cfrac%7B%5B0.005-0.003%5D%7D%7B%5B0.005%2B0.003%5D%7D%5C%5CpH%20%3D%208.3%20%2B%20log%20%5Cfrac%7B%5B0.002%5D%7D%7B%5B0.008%5D%7D%5C%5C%5C%5CpH%20%3D%208.3%20%2B%20log%20%7B0.25%7D%5C%5C%5C%5CpH%20%3D%208.3%20%2B%20%28-0.62%29%5C%5CpH%20%3D%207.69)
Therefore, the new pH is 7.69.
Learn more about pH here:
brainly.com/question/24595796
#SPJ1