Answer: 5.39
Explanation: you are adding the balloon's volume and the container's volume
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 
The answer would be D carbon dioxide
Explanation:
Expression for the
speed is as follows.

where,
= root mean square speed
k = Boltzmann constant
T = temperature
M = molecular mass
As the molecular weight of oxygen is 0.031 kg/mol and R = 8.314 J/mol K. Hence, we will calculate the value of
as follows.

= 
= 498.5 m/s
Hence,
for oxygen atom is 498.5 m/s.
For nitrogen atom, the molecular weight is 0.028 kg/mol. Hence, we will calculate its
speed as follows.

= 
= 524.5 m/s
Therefore,
speed for nitrogen is 524.5 m/s.
The answer would be .5 mols because you take the total amount of grams, which is 20, and you had up the molar mass of sodium hydroxide, which would be 40. After you have this you would set this up as a stochiometry equation. With 1 mol on top you dived 20/40 to cancel out your grams. This leaves you with .5 mols