Explanation:
The given data is as follows.
= 98.70 kPa = 98700 Pa,
T =
= (30 + 273) K = 303 K
height (h) = 30 mm = 0.03 m (as 1 m = 100 mm)
Density = 13.534 g/mL = 
= 13534 
The relation between pressure and atmospheric pressure is as follows.
P = 
Putting the given values into the above formula as follows.
P = 
= 
= 102683.05 Pa
= 102.68 kPa
thus, we can conclude that the pressure of the given methane gas is 102.68 kPa.
Explanation:
Vapor pressure is defined as the pressure exerted by vapors or gas on the surface of a liquid.
It is known that at standard condition, vapor pressure is 760 mm Hg.
And, it is given that methanol vapor pressure in air is 88.5 mm Hg.
Hence, calculate the volume percentage as follows.
Volume percentage = 
= 
= 11.65%
Thus, we can conclude that the maximum volume percent of Methanol vapor that can exist at standard conditions is 11.65%.
Pressure in the submarine when the temperature is changed to 293 K is 108.9 K Pa
Explanation:
Pressure in the submarine = 108.9 kPa
Volume, V = 2.4 * 10^5 L
Pressure, P = 116k Pa
Temperature, T = 312 K
Ideal gas law: PV = nRT or n = PV / RT
So, moles of gas, n =116 KPa * 2.4 * 10 ^5L / 8.314 LK Pa K^-1 *312 K
= 1.073 *10^4 mol
when temperature is changed to 293K,
PV = nRT or P = nRT / V
=1.073 *10^4 mol *8.314 LK Pa mol^-1 K^-1 *293 K / 2.4*10^5L
=108.9 K Pa
Pressure in the submarine when the temperature is changed to 293 K is 108.9 K Pa
It becomes a acceptor because in an ionic bonding the element who gives out is a donor while the atom which accept is a acceptor