The units of molar mass is g/mol that is the letter B.
Answer : The normal boiling point of ethanol will be, 
or 
Explanation :
The Clausius- Clapeyron equation is :
where,
= vapor pressure of ethanol at 
= 98.5 mmHg
= vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg
= temperature of ethanol = 
= normal boiling point of ethanol = ?
= heat of vaporization = 39.3 kJ/mole = 39300 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:
Hence, the normal boiling point of ethanol will be, or
Answer:
We need 78.9 mL of the 19.0 M NaOH solution
Explanation:
Step 1: Data given
Molarity of the original NaOH solution = 19.0 M
Molarity of the NaOH solution we want to prepare = 3.0 M
Volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
Step 2: Calculate volume of the 19.0 M NaOH solution needed
C1*V1 = C2*V2
⇒with C1 = the concentration of the original NaOH solution = 19.0 M
⇒with V1 = the volume of the original NaOH solution = TO BE DETERMINED
⇒with C2 = the concentration of the NaOH solution we want to prepare = 3.0 M
⇒with V2 = the volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
19.0 M * V2 = 3.0 M * 0.500 L
V2 = (3.0 M * 0.500L) / 19.0 M
V2 = 0.0789 L
We need 0.0789 L
This is 0.0789 * 10^3 mL = 78.9 mL
We need 78.9 mL of the 19.0 M NaOH solution
You kind of need to show more, but the independent variable would be something that you are changing to then measure the reaction, like the time left in for example.
Answer:
The 
for the reaction 
will be 4.69.
Explanation:
The given equation is A(B) = 2B(g)
to evaluate equilibrium constant for
![K_c=[B]^2[A]](https://tex.z-dn.net/?f=K_c%3D%5BB%5D%5E2%5BA%5D)
= 0.045
The reverse will be 
Then, ![K_c = \frac{[A]}{[B]^2}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cfrac%7B%5BA%5D%7D%7B%5BB%5D%5E2%7D)
= 
= 
The equilibrium constant for will be
= 4.69
Therefore, for the reaction will be 4.69.
