A 2.00L mixture of helium, nitrogen, and neon has a total pressure of 815 mmHg at a temperature of 255K. If the partial pressure of helium is 201 mmHg and the partial pressure of nitrogen is 351 mmHg, what is the partial pressure of neon in the mixture?

Answersublimation;

Use the universal gas law formula: pv = nrt now, the key word in this problem is stp, so at stp, temperature will always be 0 degrees celsius and pressure will always be 1.0 atm or 760 torr. given something is at stp, we already know the two variables pressure and temperature. the other thing we can find with the given information is moles. to find moles, just divide grams by molar mass of the element. 16.0g / 2(16.00)g/mol = 0.5 moles plug the values in to the equation pv = nrt (1.0 atm)( x ) = (0.5 )(0+273=273k) the r value is just a constant for when we use atm, liters, moles and kelvin. i'm assuming you want your answer in liters, but if you dont just move the decimal 3 to the right in your final answer. so when plugging these numbers in to your calculator and trying to find x, you get 11.20665 liters using significant figures the answer is 11.2 liters