A container filled with a mixture fluorine gas and neon gas exerts a total pressure of $$ 7.00 \mathrm{~atm} $$. There are $$ 50.0 \mathrm{~g} $$ of fluorine gas $$ \left(F_{2}\right) $$ in the container which exerts a pressure of $$ 5.00 \mathrm{~atm} $$. To use Dalton's law of partial pressures, you have to compare molar quantities. What quantity in moles of fluorine gas are present in the container? $$ \mathrm{mol} $$

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To find the quantity of fluorine gas in moles, we need to use the ideal gas law equation:

$$PV = nRT$$

Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant ($$0.0821 \frac{\mathrm{atm \cdot L}}{\mathrm{mol \cdot K}}$$)
T = temperature (in K)

We are given:
P(total) = 7.00 atm
P(fluorine) = 5.00 atm
V = unknown
n(fluorine) = unknown
R = 0.0821 $$\frac{\mathrm{atm \cdot L}}{\mathrm{mol \cdot K}}$$
T = unknown

We can rearrange the equation to solve for n:

$$n = \frac{PV}{RT}$$

Since we are comparing molar quantities, we can use the ratio of the pressures to find the ratio of the moles:

$$\frac{n(fluorine)}{n(total)} = \frac{P(fluorine)}{P(total)}$$

Substituting the given values:

$$\frac{n(fluorine)}{n(total)} = \frac{5.00 \mathrm{~atm}}{7.00 \mathrm{~atm}}$$

Simplifying:

$$\frac{n(fluorine)}{n(total)} = \frac{5}{7}$$

To find the quantity of fluorine gas in moles, we can multiply the ratio by the total number of moles:

$$n(fluorine) = \frac{5}{7} \cdot n(total)$$

We are given the mass of fluorine gas, so we can use the molar mass of fluorine ($$F_2$$) to find the total number of moles:

$$n(total) = \frac{50.0 \mathrm{~g}}{molar~mass~of~fluorine}$$

The molar mass of fluorine ($$F_2$$) is $$2 \times 19.00 \mathrm{~g/mol} = 38.00 \mathrm{~g/mol}$$.

Substituting the values:

$$n(total) = \frac{50.0 \mathrm{~g}}{38.00 \mathrm{~g/mol}}$$

Simplifying:

$$n(total) = 1.32 \mathrm{~mol}$$

Finally, substituting the value of $$n(total)$$ into the equation for $$n(fluorine)$$:

$$n(fluorine) = \frac{5}{7} \cdot 1.32 \mathrm{~mol}$$

Simplifying:

$$n(fluorine) = 0.94 \mathrm{~mol}$$

Therefore, there are 0.94 moles of fluorine gas present in the container.0.94 mol

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