Answer : The mass of helium gas added must be 12.48 grams.
Explanation : Given,
Mass of helium (He) gas = 6.24 g
Molar mass of helium = 4 g/mole
First we have to calculate the moles of helium gas.
Now we have to calculate the moles of helium gas at doubled volume.
According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,
or,
where,
= initial volume of gas = V
= final volume of gas = 2V
= initial moles of gas = 1.56 mole
= final moles of gas = ?
Now we put all the given values in this formula, we get
Now we have to calculate the mass of helium gas at doubled volume.
Therefore, the mass of helium gas added must be 12.48 grams.
The way its structured is they're layered on top of each other and because of that it has a much faster charge rate
Answer:
6.5 liters.
Explanation:
- We can use the general law of ideal gas: PV = nRT.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If P and T are constant, and have different values of n and V:
<em>(V₁n₂) = (V₂n₁)</em>
<em></em>
V₁ = 4.5 L, n₁ = 0.80 mol,
V₂ = ??? L, n₂ = 0.35 + 0.80 = 1.15 mol.
- Applying in the above equation
<em>(V₁n₂) = (V₂n₁)</em>
<em></em>
<em>∴ V₂ = (V₁n₂)/n₁ </em>= (4.5 L)(1.15 mol)/(0.8 mol) = <em>6.469 L ≅ 6.5 L.</em>
<em></em>
<em>So, the right choice is: 6.5 liters.</em>
C. 5670 mL
1 liter is 1000 ml so 5.67 * 1000 is 5670
