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.
Aldehydes and Ketones. Aldehydes and ketones are classes of organic compounds that contain a carbonyl (C=O) group.
that's the correct answer
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Answer:
option (B) is correct
Explanation:
In case of nuclear reactors first the nuclear energy is emitted due to the nuclear fission of heavy elements.
This nuclear energy is emitted in the form of heat energy.
This heat energy is used to rotate the turbines, that means it is converted in the form of mechanical energy and then finally this mechanical energy is converted into electrical energy.
<span>6.38x10^-2 moles
First, let's determine how many moles of gas particles are in the two-liter container. The molar volume for 1 mole at 25C and 1 atmosphere is 24.465 liters/mole. So
2 L / 24.465 L/mol = 0.081749438 mol
Now air doesn't just consist of nitrogen. It also has oxygen, carbon dioxide, argon, water vapor, etc. and the total number of moles includes all of those other gasses. So let's multiply by the percentage of nitrogen in the atmosphere which is 78%
0.081749438 mol * 0.78 = 0.063764562 mol.
Rounding to 3 significant figures gives 6.38x10^-2 moles</span>
