Ascientist needs to collect a 0.050 mole sample of helium, but needs to know how large his helium container should be. what volume is needed to store the sample of helium gas at 202.6kpa and 400k? a. 657 milliliters b. 760 milliliters c. 820 milliliters d. 1,025 milliliters

We can use the ideal gas equation:

PV = nRT

P = 202.6kPa = 202600 Pa (You have to multiply by 1000)

n = 0.050 mole

R = 0.082 atm*l/(K*mol)

T = 400K

We will have to convert from Pa to atm or viceversa.

101325 Pa1 atm

202600 Pax = 2.00 atm

2atm*V = 0.050 mole*0.082 atm*l/(K*mol)* 400K

V = 0.050 mole*0.082 atm*l/(K*mol)* 400K/2atm = 0.82 liters = 820 mililiters

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

lol for the pointsss