c is the answer i odnt know the rest im only a sophmore
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>
Answer:
Al ascender las burbujas van aumentando de tamaño.
Explanation:
Las burbujas que produce el buzo debajo del agua son pequeñas moléculas de dióxido de carbono gaseoso producto de la respiración del mismo.
Ahora, a medida que las burbujas suben a la superficie, la presión que sufren estas (Presión debido al agua), es menor conforme van ascendiendo debido a la ley de Boyle: A medida que la presión aumenta, el volumen va disminuyendo.
Esto significa que al ascender las burbujas van aumentando de tamaño debido a que la presión que sufren estas es menor que cuando están a mayores profundidades.
Remember that a conjugate acid-base pair will differ only by one proton.
None of the options you listed are conjugate acid-base pairs as none of them differ only by one proton (or H⁺)
An example of a conjugate acid-base pair would be NH₃ and NH₄⁺NH₃ + H₂O --> NH₄⁺ + OH⁻NH3 is the base, and NH₄⁺ is the conjugate acid
Answer:
295.7 mL of 24% trichloroacetic acid (tca) is needed .
Explanation:
Let the volume of 24% trichloroacetic acid solution be x
Volume of required 10% trichloroacetic acid solution =8 bottles of 3 ounces
= 24 ounces = 709.68 mL
(1 ounces = 29.57 mL)
Amount of trichloroacetic acid in 24% solution of x volume of solution will be equal to amount of trichloroacetic acid in 10% solution of volume 709.68 mL.

x = 295.7 mL
295.7 mL of 24% trichloroacetic acid (tca) is needed .