Answer:
the volume occupied by 3.0 g of the gas is 16.8 L.
Explanation:
Given;
initial reacting mass of the helium gas, m₁ = 4.0 g
volume occupied by the helium gas, V = 22.4 L
pressure of the gas, P = 1 .0 atm
temperature of the gas, T = 0⁰C = 273 K
atomic mass of helium gas, M = 4.0 g/mol
initial number of moles of the gas is calculated as follows;

The number of moles of the gas when the reacting mass is 3.0 g;
m₂ = 3.0 g

The volume of the gas at 0.75 mol is determined using ideal gas law;
PV = nRT

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.
Answer:
it is soluble in water
Explanation:
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Answer:
187.34 atm
Explanation:
From the question,
PV = nRT.................. Equation 1
Where P = Pressure, V = Volume, n = number of mole, R = molar gas constant, T = Temperature.
make P the subject of the equation
P = nRT/V.............. Equation 2
n = mass(m)/molar mass(m')
n = m/m'............... Equation 3
Substitute equation 3 into equation 2
P = (m/m')RT/V............ Equation 4
Given: m = 46 g, T = 25°C = (25+273) = 298 K, V = 3.00 L
Constant: m' = 2 g/mol, R = 0.082 atmL/K.mol
Substitute these values into equation 4
P = (46/2)(0.082×298)/3
P = (23×0.082×298)/3
P = 187.34 atm
C. mass is protons and neutrons. Both are in the nucleus