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mihalych1998 [28]
3 years ago
13

A gas cylinder of volume 5.00 l contains 1.00 g of ar and 0.500 g of ne. the temperature is 275 k. find the partial pressure of

ne.
Chemistry
1 answer:
amid [387]3 years ago
8 0
<span>11.3 kPa The ideal gas law is PV = nRT where P = Pressure V = Volume n = number of moles R = Ideal gas constant (8.3144598 L*kPa/(K*mol) ) T = Absolute temperature We have everything except moles and volume. But we can calculate moles by starting with the atomic weight of argon and neon. Atomic weight argon = 39.948 Atomic weight neon = 20.1797 Moles Ar = 1.00 g / 39.948 g/mol = 0.025032542 mol Moles Ne = 0.500 g / 20.1797 g/mol = 0.024777375 mol Total moles gas particles = 0.025032542 mol + 0.024777375 mol = 0.049809918 mol Now take the ideal gas equation and solve for P, then substitute known values and solve. PV = nRT P = nRT/V P = 0.049809918 mol * 8.3144598 L*kPa/(K*mol) * 275 K/5.00 L P = 113.8892033 L*kPa / 5.00 L P = 22.77784066 kPa Now let's determine the percent of pressure provided by neon by calculating the percentage of neon atoms. Divide the number of moles of neon by the total number of moles. 0.024777375 mol / 0.049809918 mol = 0.497438592 Now multiply by the pressure 0.497438592 * 22.77784066 kPa = 11.33057699 kPa Round the result to 3 significant figures, giving 11.3 kPa</span>
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6 0
4 years ago
If 1.76 g of an ideal gas occupy 1.0 L at standard temperature and pressure (STP), what is the molar mass of the gas?
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Explanation:

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8 0
2 years ago
6.201cm+7.4cm+0.68 cm+12.0cm
zimovet [89]
If you want the answer is centimeters is it going to be:
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If it make it easier for you to solve add 6.201 and 7.4 which will equal 13.601. Then add .68 to 13.601 which equals 14.281. Last add 12 to 14.281 which equals 26.281.

Hope this helps.

8 0
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