Question #6723

A researcher collects 0.00751 mol of an unknown gas by water displacement at a temperature of 18.6°C and 0.841 atm pressure (after the partial pressure of water vapor has been subtracted).

What volume of gas, in mL, does the researcher have?

If the volume of 2.46 x 10^21 molecules of HBr(g) at 51°C and 1.31 atm is 80.2 mL, then what is the volume of 2.46 x 10^21 molecules of CO2(g) at 51°C and 1.31 atm?

I have these 2 problems that I can not figure out...

What volume of gas, in mL, does the researcher have?

If the volume of 2.46 x 10^21 molecules of HBr(g) at 51°C and 1.31 atm is 80.2 mL, then what is the volume of 2.46 x 10^21 molecules of CO2(g) at 51°C and 1.31 atm?

I have these 2 problems that I can not figure out...

Expert's answer

PV=nRT

V=nRT/P=0.00751 * 8.31 * (273 + 18.6) / (0.841 *101.3)=0.21361

ml

volume will be the same 80.2 ml

V=nRT/P=0.00751 * 8.31 * (273 + 18.6) / (0.841 *101.3)=0.21361

ml

volume will be the same 80.2 ml

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