At constant number of mole $n$ and Temperature $T$, using the $PV=nRT$ we can write: $P_1V_1=P_2V_2$ We should calculate the pressure of pure $O_2$ in a $2.01L$ volume: $P_1=(P_2V_2)/V_1=(785" torr"xx1.92cancel(L))/(2.01cancel(L))=750"...
1 Answers 1 viewsThe idea here is that the vapor pressures of benzene and toluene will contribute to the total vapor pressure of the solution proportionally to their respective mole fraction - this...
1 Answers 1 views$PV = nRT$ $P$ is pressure in atm (760 torr = 1 atm); $V$ is volume in liters; $n$ is the number of moles; $R$ is the constant 0.0821; $T$...
1 Answers 1 viewsThe volume of the gas at STP is 25.4 mL When you examine the question, you see that the pressure does not change. Standard pressure is 760 Torr. The only...
1 Answers 1 viewsFirst, recall that “Standard” temperature and pressure are 0’C and 1 atm (760Torr), as opposed to “Normal” temperature and pressure of 20’C and 1 atm. In the calculations all...
1 Answers 1 viewsThe idea here is that the exerted by a gas that's part of a gaseous mixture is proportional to its mole fraction. This means that in order to find...
1 Answers 1 viewsConsider Raoult's law compared with Dalton's law of partial pressures: $P_i = overbrace(chi_(i(l))P_i^"*")^"Raoult's Law" = underbrace(chi_(i(v))P_"tot")_"Dalton's Law"$, where: $P_i$ is the partial vapor pressure of component $i$, i.e....
1 Answers 1 viewsFrom Dalton's Law of , the total pressure of a mixture of gases is the sum of the partial pressures of each gas. Therefore, the partial pressure of $"CO"_2$ is...
1 Answers 1 viewsSTP conditions are defined as a pressure of $"100 kPa" = 100 color(red)(cancel(color(black)("kPa"))) * (1color(red)(cancel(color(black)("atm"))))/(101.325color(red)(cancel(color(black)("kPa")))) * "760 torr"/(1color(red)(cancel(color(black)("atm")))) ="750.1 torr"$ and a temperature of $0^@"C" =...
1 Answers 1 viewsI got $0.294$. Raoult's law states: $P_j = chi_j^lP_j^"*"$, where: $P_j$ is the partial vapor pressure above the solution coming from the component $j$ in solution....
1 Answers 1 views