Respuesta :
Answer:
0.1498 g of O2.
Explanation:
The Behavior of Gases => Ideal Gas Law.
The ideal gas law is a single equation that relates the pressure, volume, temperature, and the number of moles of an ideal gas, which is:
[tex]PV=nRT,[/tex]where P is pressure in atm, V is the volume in liters, n is the number of moles, R is the ideal gas constant (0.082 L*atm/mol*K), and T is the temperature in the Kelvin scale.
So we have to convert pressure from 700.0 mmHg to atm, volume from 120.0 mL to L, and 15 °C to K.
Let's convert pressure taking into account that 1 atm equals 760 mmHg, like this:
[tex]700.0\text{ mmHg}\cdot\frac{1\text{ atm}}{760\text{ mmHg}}=0.9211\text{ atm.}[/tex]Remember that 1 L equals 1000 mL, so 120.0 mL would be equal:
[tex]120.0\text{ mL}\cdot\frac{1\text{ L}}{1000\text{ mL}}=0.1200\text{ L.}[/tex]And the conversion from °C to K is just sum °C with 273, so 15 °C in K is:
[tex]K=\degree C+273=15\degree C+273=288\text{ K.}[/tex]Finally, we can use the ideal gas formula, solving for 'n' (number of moles) and replacing the data that we have, as follows:
[tex]\begin{gathered} n=\frac{PV}{RT}, \\ \\ n=\frac{0.9211\text{ atm}\cdot0.1200\text{ L}}{0.082\frac{L\cdot atm}{mol\cdot K}\cdot288\text{ K}}, \\ \\ n=4.680\cdot10^{-3}\text{ moles.} \end{gathered}[/tex]Now, the final step is to convert 4.680 x 10⁻³ moles of O2 to grams using the molar mass of O2 that can be calculated using the periodic table, which is 32 g/mol. The conversion will look like this:
[tex]4.68\cdot10^{-3\text{ }}moles\text{ O}_2\cdot\frac{32\text{ g O}_2}{1\text{ mol O}_2}=0.1498\text{ g O}_2.[/tex]The answer would be that there are 0.1498 g of O2.
