A 500 gram price of metal has a volume of 2.75. What is the density
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Q1)
we can use the ideal gas law equation to find the total pressure of the system ;
PV = nRT
where P - pressure
V - volume - 7 x 10⁻³ m³
n - number of moles
total number of moles - 0.477 + 0.265 + 0.115 = 0.857 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in K - 273 + 25 °C = 298 K
substituting the values in the equation
P x 7 x 10⁻³ m³ = 0.857 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
P = 303.33 kPa
1 atm = 101.325 kPa
Therefore total pressure - 303.33 kPa / 101.325 kPa/atm = 2.99 atm
Q2)
partial pressure is the pressure exerted by the individual gases in the mixture.
partial pressure for each gas can be calculated by multiplying the total pressure by mole fraction of the individual gas.
total number of moles - 0.477 + 0.265 + 0.115 = 0.857 mol
mole fraction of He -
mole fraction of Ne -
mole fraction of Ar -
partial pressure - total pressure x mole fraction
partial pressure of He - 2.99 atm x 0.557 = 1.67 atm
partial pressure of Ne - 2.99 atm x 0.309 = 0.924 atm
partial pressure of Ar - 2.99 atm x 0.134 = 0.401 atm
we can use the ideal gas law equation to find the total pressure of the system ;
PV = nRT
where P - pressure
V - volume - 7 x 10⁻³ m³
n - number of moles
total number of moles - 0.477 + 0.265 + 0.115 = 0.857 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in K - 273 + 25 °C = 298 K
substituting the values in the equation
P x 7 x 10⁻³ m³ = 0.857 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
P = 303.33 kPa
1 atm = 101.325 kPa
Therefore total pressure - 303.33 kPa / 101.325 kPa/atm = 2.99 atm
Q2)
partial pressure is the pressure exerted by the individual gases in the mixture.
partial pressure for each gas can be calculated by multiplying the total pressure by mole fraction of the individual gas.
total number of moles - 0.477 + 0.265 + 0.115 = 0.857 mol
mole fraction of He -
mole fraction of Ne -
mole fraction of Ar -
partial pressure - total pressure x mole fraction
partial pressure of He - 2.99 atm x 0.557 = 1.67 atm
partial pressure of Ne - 2.99 atm x 0.309 = 0.924 atm
partial pressure of Ar - 2.99 atm x 0.134 = 0.401 atm
2074.74 kg is the mass of siderite in which 1x10^3 kg of iron is present.
Explanation:
Mass of iron is given as 1x10^3 Kg
mass of iron in siderite FeCO3 can be known
number of moles in 1x10^3 Kg iron is calculated by the formula.
number of moles =
atomic weight of iron is 55.84
number of moles =
n= 17908.309 moles of iron in 1000 kg
1 mole of FeCO3 has 1 mole of Fe
x moles of FeCO3 will have moles
=
x = 17908.309 moles of FeCO3 is there.
so mass can be calculated as number of moles x molar mass of one mole of the substance. (molar mass of siderite is 115.854 grams/mole)
17908.309 x 115.854
= 2074749.230 grams of siderite
(the molar mass is converted to kg/mole as the mass of iron is given by kg.
0.1158 kg is the molar mass of siderite)
2074749.230 grams of siderite is converted to kilogram
2074.74 kg is the mass of siderite in which 1x10^3 kg of iron is present.