Using the molar volume of a gas at STP , what is the volume (in L) occupied by 36.2 g of neon at STP?

At STP, we know that the is $22.42L/(mol)$, therefore, now we can calculate the number of mole of the gas from its volume $V=0.225L$ at STP. $=>n=0.225cancel(L)xx(1mol)/(22.42cancel(L))=0.01 mol$ From the...

Elemental neon has a molar mass of $20.18$ $g$ $mol^-1$. To get the mass, simply multiply the molar mass by the molar quantity: $0.0450$ $mol$ $xx$ $20.18$ $g*mol^-1$ $=$ $??g$

The final pressure will double as well. If you use the equation, $PV = nRT$, to express the initial and the final states of your gas sample, you can...

The holds that $(P_1V_1)/T_1=(P_2V_2)/T_2$. And thus $V_2=(P_1xxV_1xxT_2)/(T_1xxP_2)$ $=$ $(2.7*atmxx2.25*Lxx273.15*K)/(285.15*Kxx1*atm)$ $~=$ $6*L$ This volume increase is reasonable as we have only marginally changed temperature, but have reduced pressure by a third....

$n=(PV)/(RT)$ $=$ $(3.5xxkPaxx42.5*L)/(8.314*L*kPa*K^-1* mol^-1xx298*K)$ $=$ $6.00xx10^-2*mol$ Thus $"mass neon"$ $=$ $6.00xx10^-2*molxx20.18*g*mol^-1$ $=$ $??g$

We know that at room temperature and pressure, $1$ $mol$ of ideal gas occupies a volume of $25.4*L$. We assume (reasonably) that $Ne$ behaves ideally. Thus the number of moles...

We can use Graham's law of effusion to solve this: $(r_A)/(r_B) = sqrt((M_B)/(M_A))$ where $(r_A)/(r_B)$ is a number representing the ratio of effusion rates of gas A to...

So neon occupies 33.6 g First we find the number of moles of Ne= $(mass)/"molar mass"$ No. of moles= $(33.6)/"20"$ No. of moles= 1.68 mol Then, we can find volume...

Considering molar mass of $ Ar->40" g/mol"$ $ Ne->20" g/mol"$ $ "Molar mass of unknown gas"->m" g/mol"$ The total number of moles Ar , Ne and unknown gas in the...

Multiply the percentage in decimal form of each gas by $"99000 kPa"$. Nitrogen $0.68xx99000color(white)(.)"kPa"$$=$$"67320 kPa"$ Oxygen $0.23xx99000color(white)(.)"kPa"$$=$$"22770 kPa"$ Neon $0.09xx99000color(white)(.)"kPa"$$=$$"8910 kPa"$ The sum of the partial pressures equals the total...