We can use combined gas laws to solve for the volume of the gas

where P - pressure, V - volume , T - temperature and k - constant

parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
T1 - temperature in Kelvin - 20 °C + 273 = 293 K
T2 - 40 °C + 273 = 313 K
substituting the values

V = 17.8 L
volume of the gas is 17.8 L
Answer:
B) 0.32 %
Explanation:
Given that:

Concentration = 1.8 M
Considering the ICE table for the dissociation of acid as:-

The expression for dissociation constant of acid is:
![K_{a}=\frac {\left [ H^{+} \right ]\left [ {CH_3COO}^- \right ]}{[CH_3COOH]}](https://tex.z-dn.net/?f=K_%7Ba%7D%3D%5Cfrac%20%7B%5Cleft%20%5B%20H%5E%7B%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20%7BCH_3COO%7D%5E-%20%5Cright%20%5D%7D%7B%5BCH_3COOH%5D%7D)


Solving for x, we get:
<u>x = 0.00568 M</u>
Percentage ionization = 
<u>Option B is correct.</u>
Mass equals density times volume
Because D=m/v
Multiply by v
Dv=m
The concentration of the solution in mass percent is 60%.
<h3>What is the mass percent?</h3>
The concentration of a substance can be expressed in mass percent. This refers to the percentage of the solute that is contained in the solution.
Thus we can write;
Mass percent = Mass of solute/ Mass of solution * 100/1
Mass percent = 1.5kg/1.5kg + 1 Kg * 100/1
Mass percent = 60%
Hence, the concentration of the solution in mass percent is 60%.
Learn more about mass percent:brainly.com/question/5394922
#SPJ1
Answer: wait im gonna search some info.
Explanation:I come back with an asnwer for you