What is the volume (in cm^ 3 ) of 1.2 moles of water vapor at rtp? calculate the mass of this water vapor.

Volume = 29400 cm³

Mass =  21.6 g

Explanation:

Given data:

Number of moles of water vapors = 1.2 mol

Mass of water vapor = ?

Volume of water vapors = ?

Solution:

Room temperature = 25 °C (25+273 = 298 K)

Room pressure = 1 atm

PV = nRT

V =  nRT/P

V = 1.2 mol × 0.0821 atm. K⁻¹. mol⁻¹ . L × 298 K/ 1 atm

V = 29.4 L

In cm³:

29.4 × 1000 = 29400 mL

mL = cm³

Mass of water vapors:

Mass = number of moles × molar mass

Mass = 1.2 mol × 18 g/mol

Mass =  21.6 g

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.