id
int64
1
14M
domain
stringclasses
6 values
topic
stringclasses
23 values
subtopic
stringclasses
37 values
difficulty
int64
1
8
unit_type
stringclasses
3 values
title
stringlengths
14
86
content
stringlengths
203
553
key_equations
stringclasses
23 values
prerequisites
stringclasses
29 values
learning_objective
stringclasses
37 values
2,701
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of CaCO3 from mass 69.55 g
The molar mass of CaCO3 is 100.1 g/mol. A sample of mass 69.55 g therefore contains n = m / M = 69.55 / 100.1 = 0.6949 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,702
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of H2O from mass 64.17 g
The molar mass of H2O is 18.02 g/mol. A sample of mass 64.17 g therefore contains n = m / M = 64.17 / 18.02 = 3.562 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,703
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of NaCl from mass 63.05 g
The molar mass of NaCl is 58.44 g/mol. A sample of mass 63.05 g therefore contains n = m / M = 63.05 / 58.44 = 1.079 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,704
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of NaCl from mass 0.4073 g
The molar mass of NaCl is 58.44 g/mol. A sample of mass 0.4073 g therefore contains n = m / M = 0.4073 / 58.44 = 0.006969 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,705
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of C6H12O6 from mass 62.87 g
The molar mass of C6H12O6 is 180.2 g/mol. A sample of mass 62.87 g therefore contains n = m / M = 62.87 / 180.2 = 0.349 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,706
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of NH3 from mass 57.62 g
The molar mass of NH3 is 17.03 g/mol. A sample of mass 57.62 g therefore contains n = m / M = 57.62 / 17.03 = 3.383 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,707
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of CO2 from mass 15.87 g
The molar mass of CO2 is 44.01 g/mol. A sample of mass 15.87 g therefore contains n = m / M = 15.87 / 44.01 = 0.3605 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,708
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of Fe2O3 from mass 71.39 g
The molar mass of Fe2O3 is 159.7 g/mol. A sample of mass 71.39 g therefore contains n = m / M = 71.39 / 159.7 = 0.447 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,709
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of NaCl from mass 84.71 g
The molar mass of NaCl is 58.44 g/mol. A sample of mass 84.71 g therefore contains n = m / M = 84.71 / 58.44 = 1.45 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,710
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of CH4 from mass 79.28 g
The molar mass of CH4 is 16.04 g/mol. A sample of mass 79.28 g therefore contains n = m / M = 79.28 / 16.04 = 4.942 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,711
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of NH3 from mass 45.2 g
The molar mass of NH3 is 17.03 g/mol. A sample of mass 45.2 g therefore contains n = m / M = 45.2 / 17.03 = 2.654 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,712
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of NaCl from mass 26.14 g
The molar mass of NaCl is 58.44 g/mol. A sample of mass 26.14 g therefore contains n = m / M = 26.14 / 58.44 = 0.4473 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,713
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of CuSO4 from mass 82.69 g
The molar mass of CuSO4 is 159.6 g/mol. A sample of mass 82.69 g therefore contains n = m / M = 82.69 / 159.6 = 0.5181 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,714
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of H2SO4 from mass 19.95 g
The molar mass of H2SO4 is 98.07 g/mol. A sample of mass 19.95 g therefore contains n = m / M = 19.95 / 98.07 = 0.2034 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,715
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of CH4 from mass 36.82 g
The molar mass of CH4 is 16.04 g/mol. A sample of mass 36.82 g therefore contains n = m / M = 36.82 / 16.04 = 2.295 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,716
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of CH4 from mass 87.65 g
The molar mass of CH4 is 16.04 g/mol. A sample of mass 87.65 g therefore contains n = m / M = 87.65 / 16.04 = 5.464 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,717
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of CuSO4 from mass 22.53 g
The molar mass of CuSO4 is 159.6 g/mol. A sample of mass 22.53 g therefore contains n = m / M = 22.53 / 159.6 = 0.1411 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,718
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of H2O from mass 92.32 g
The molar mass of H2O is 18.02 g/mol. A sample of mass 92.32 g therefore contains n = m / M = 92.32 / 18.02 = 5.125 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,719
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of CuSO4 from mass 62.7 g
The molar mass of CuSO4 is 159.6 g/mol. A sample of mass 62.7 g therefore contains n = m / M = 62.7 / 159.6 = 0.3929 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,720
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of CO2 from mass 97.42 g
The molar mass of CO2 is 44.01 g/mol. A sample of mass 97.42 g therefore contains n = m / M = 97.42 / 44.01 = 2.214 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,721
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of C6H12O6 from mass 26.25 g
The molar mass of C6H12O6 is 180.2 g/mol. A sample of mass 26.25 g therefore contains n = m / M = 26.25 / 180.2 = 0.1457 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,722
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of NaCl from mass 5.073 g
The molar mass of NaCl is 58.44 g/mol. A sample of mass 5.073 g therefore contains n = m / M = 5.073 / 58.44 = 0.08681 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,723
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of CuSO4 from mass 42.95 g
The molar mass of CuSO4 is 159.6 g/mol. A sample of mass 42.95 g therefore contains n = m / M = 42.95 / 159.6 = 0.2691 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,724
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of H2SO4 from mass 15.01 g
The molar mass of H2SO4 is 98.07 g/mol. A sample of mass 15.01 g therefore contains n = m / M = 15.01 / 98.07 = 0.1531 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,725
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of NH3 from mass 17.1 g
The molar mass of NH3 is 17.03 g/mol. A sample of mass 17.1 g therefore contains n = m / M = 17.1 / 17.03 = 1.004 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,726
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of CuSO4 from mass 11.82 g
The molar mass of CuSO4 is 159.6 g/mol. A sample of mass 11.82 g therefore contains n = m / M = 11.82 / 159.6 = 0.07408 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,727
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of Fe2O3 from mass 43.57 g
The molar mass of Fe2O3 is 159.7 g/mol. A sample of mass 43.57 g therefore contains n = m / M = 43.57 / 159.7 = 0.2729 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,728
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of CaCO3 from mass 82.74 g
The molar mass of CaCO3 is 100.1 g/mol. A sample of mass 82.74 g therefore contains n = m / M = 82.74 / 100.1 = 0.8267 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,729
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of CaCO3 from mass 11.94 g
The molar mass of CaCO3 is 100.1 g/mol. A sample of mass 11.94 g therefore contains n = m / M = 11.94 / 100.1 = 0.1193 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,730
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of C6H12O6 from mass 19.14 g
The molar mass of C6H12O6 is 180.2 g/mol. A sample of mass 19.14 g therefore contains n = m / M = 19.14 / 180.2 = 0.1062 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,731
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of H2SO4 from mass 28.87 g
The molar mass of H2SO4 is 98.07 g/mol. A sample of mass 28.87 g therefore contains n = m / M = 28.87 / 98.07 = 0.2943 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,732
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of CaCO3 from mass 38.27 g
The molar mass of CaCO3 is 100.1 g/mol. A sample of mass 38.27 g therefore contains n = m / M = 38.27 / 100.1 = 0.3824 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,733
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of CH4 from mass 85.65 g
The molar mass of CH4 is 16.04 g/mol. A sample of mass 85.65 g therefore contains n = m / M = 85.65 / 16.04 = 5.339 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,734
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of CuSO4 from mass 21.64 g
The molar mass of CuSO4 is 159.6 g/mol. A sample of mass 21.64 g therefore contains n = m / M = 21.64 / 159.6 = 0.1356 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,735
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of H2SO4 from mass 72.7 g
The molar mass of H2SO4 is 98.07 g/mol. A sample of mass 72.7 g therefore contains n = m / M = 72.7 / 98.07 = 0.7413 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,736
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of C6H12O6 from mass 41.89 g
The molar mass of C6H12O6 is 180.2 g/mol. A sample of mass 41.89 g therefore contains n = m / M = 41.89 / 180.2 = 0.2325 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,737
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of CH4 from mass 79.27 g
The molar mass of CH4 is 16.04 g/mol. A sample of mass 79.27 g therefore contains n = m / M = 79.27 / 16.04 = 4.941 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,738
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of H2O from mass 29.09 g
The molar mass of H2O is 18.02 g/mol. A sample of mass 29.09 g therefore contains n = m / M = 29.09 / 18.02 = 1.615 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,739
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of H2SO4 from mass 23.63 g
The molar mass of H2SO4 is 98.07 g/mol. A sample of mass 23.63 g therefore contains n = m / M = 23.63 / 98.07 = 0.2409 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,740
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of CaCO3 from mass 90.74 g
The molar mass of CaCO3 is 100.1 g/mol. A sample of mass 90.74 g therefore contains n = m / M = 90.74 / 100.1 = 0.9066 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,741
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of NH3 from mass 58.33 g
The molar mass of NH3 is 17.03 g/mol. A sample of mass 58.33 g therefore contains n = m / M = 58.33 / 17.03 = 3.425 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,742
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of CaCO3 from mass 8.665 g
The molar mass of CaCO3 is 100.1 g/mol. A sample of mass 8.665 g therefore contains n = m / M = 8.665 / 100.1 = 0.08658 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,743
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of CaCO3 from mass 13.41 g
The molar mass of CaCO3 is 100.1 g/mol. A sample of mass 13.41 g therefore contains n = m / M = 13.41 / 100.1 = 0.134 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,744
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of NH3 from mass 97.44 g
The molar mass of NH3 is 17.03 g/mol. A sample of mass 97.44 g therefore contains n = m / M = 97.44 / 17.03 = 5.721 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,745
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of CH4 from mass 68.72 g
The molar mass of CH4 is 16.04 g/mol. A sample of mass 68.72 g therefore contains n = m / M = 68.72 / 16.04 = 4.283 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,746
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of CH4 from mass 70.87 g
The molar mass of CH4 is 16.04 g/mol. A sample of mass 70.87 g therefore contains n = m / M = 70.87 / 16.04 = 4.418 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,747
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of CuSO4 from mass 25.7 g
The molar mass of CuSO4 is 159.6 g/mol. A sample of mass 25.7 g therefore contains n = m / M = 25.7 / 159.6 = 0.161 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,748
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of CO2 from mass 55.74 g
The molar mass of CO2 is 44.01 g/mol. A sample of mass 55.74 g therefore contains n = m / M = 55.74 / 44.01 = 1.267 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,749
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of H2O from mass 33.35 g
The molar mass of H2O is 18.02 g/mol. A sample of mass 33.35 g therefore contains n = m / M = 33.35 / 18.02 = 1.851 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,750
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of C6H12O6 from mass 47.15 g
The molar mass of C6H12O6 is 180.2 g/mol. A sample of mass 47.15 g therefore contains n = m / M = 47.15 / 180.2 = 0.2617 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,751
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of NH3 from mass 99.67 g
The molar mass of NH3 is 17.03 g/mol. A sample of mass 99.67 g therefore contains n = m / M = 99.67 / 17.03 = 5.853 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,752
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of CuSO4 from mass 70.99 g
The molar mass of CuSO4 is 159.6 g/mol. A sample of mass 70.99 g therefore contains n = m / M = 70.99 / 159.6 = 0.4448 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,753
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of H2O from mass 0.6356 g
The molar mass of H2O is 18.02 g/mol. A sample of mass 0.6356 g therefore contains n = m / M = 0.6356 / 18.02 = 0.03528 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,754
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of Fe2O3 from mass 62 g
The molar mass of Fe2O3 is 159.7 g/mol. A sample of mass 62 g therefore contains n = m / M = 62 / 159.7 = 0.3883 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,755
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of H2SO4 from mass 97.75 g
The molar mass of H2SO4 is 98.07 g/mol. A sample of mass 97.75 g therefore contains n = m / M = 97.75 / 98.07 = 0.9968 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,756
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of NH3 from mass 13.31 g
The molar mass of NH3 is 17.03 g/mol. A sample of mass 13.31 g therefore contains n = m / M = 13.31 / 17.03 = 0.7816 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,757
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of Fe2O3 from mass 4.208 g
The molar mass of Fe2O3 is 159.7 g/mol. A sample of mass 4.208 g therefore contains n = m / M = 4.208 / 159.7 = 0.02635 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,758
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of NaCl from mass 26.1 g
The molar mass of NaCl is 58.44 g/mol. A sample of mass 26.1 g therefore contains n = m / M = 26.1 / 58.44 = 0.4467 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,759
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of Fe2O3 from mass 49.33 g
The molar mass of Fe2O3 is 159.7 g/mol. A sample of mass 49.33 g therefore contains n = m / M = 49.33 / 159.7 = 0.3089 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,760
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of NH3 from mass 87.8 g
The molar mass of NH3 is 17.03 g/mol. A sample of mass 87.8 g therefore contains n = m / M = 87.8 / 17.03 = 5.155 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,761
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of NaCl from mass 9.056 g
The molar mass of NaCl is 58.44 g/mol. A sample of mass 9.056 g therefore contains n = m / M = 9.056 / 58.44 = 0.155 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,762
chemistry
stoichiometry
mole_concept
3
worked_example
Moles of Fe2O3 from mass 43.35 g
The molar mass of Fe2O3 is 159.7 g/mol. A sample of mass 43.35 g therefore contains n = m / M = 43.35 / 159.7 = 0.2715 mol. The mole is the SI unit for amount of substance and links macroscopic mass to number of entities via Avogadro's constant.
n = m / M
atomic masses; chemical formulas
Convert between mass and moles for a pure compound.
2,763
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=1.309 mol, V=48.94 L, T=325.3 K
For an ideal gas, P V = n R T. With n = 1.309 mol, V = 48.94 L, T = 325.3 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 0.7143 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,764
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=2.501 mol, V=14.41 L, T=589.3 K
For an ideal gas, P V = n R T. With n = 2.501 mol, V = 14.41 L, T = 589.3 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 8.394 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,765
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=1.809 mol, V=45.19 L, T=372 K
For an ideal gas, P V = n R T. With n = 1.809 mol, V = 45.19 L, T = 372 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 1.222 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,766
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=1.323 mol, V=46.37 L, T=531.8 K
For an ideal gas, P V = n R T. With n = 1.323 mol, V = 46.37 L, T = 531.8 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 1.245 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,767
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=2.59 mol, V=8.667 L, T=374.3 K
For an ideal gas, P V = n R T. With n = 2.59 mol, V = 8.667 L, T = 374.3 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 9.178 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,768
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=3.687 mol, V=47.53 L, T=396.5 K
For an ideal gas, P V = n R T. With n = 3.687 mol, V = 47.53 L, T = 396.5 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 2.524 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,769
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=2.783 mol, V=32.01 L, T=569.1 K
For an ideal gas, P V = n R T. With n = 2.783 mol, V = 32.01 L, T = 569.1 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 4.06 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,770
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=2.206 mol, V=3.201 L, T=275.6 K
For an ideal gas, P V = n R T. With n = 2.206 mol, V = 3.201 L, T = 275.6 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 15.58 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,771
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=0.9239 mol, V=8.596 L, T=388.1 K
For an ideal gas, P V = n R T. With n = 0.9239 mol, V = 8.596 L, T = 388.1 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 3.423 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,772
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=2.341 mol, V=44.81 L, T=246.4 K
For an ideal gas, P V = n R T. With n = 2.341 mol, V = 44.81 L, T = 246.4 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 1.056 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,773
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=2.48 mol, V=49.45 L, T=376.6 K
For an ideal gas, P V = n R T. With n = 2.48 mol, V = 49.45 L, T = 376.6 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 1.55 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,774
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=2.809 mol, V=6.068 L, T=383.1 K
For an ideal gas, P V = n R T. With n = 2.809 mol, V = 6.068 L, T = 383.1 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 14.55 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,775
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=3.602 mol, V=17.81 L, T=259.2 K
For an ideal gas, P V = n R T. With n = 3.602 mol, V = 17.81 L, T = 259.2 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 4.303 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,776
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=1.027 mol, V=41.83 L, T=313.6 K
For an ideal gas, P V = n R T. With n = 1.027 mol, V = 41.83 L, T = 313.6 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 0.6317 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,777
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=3.752 mol, V=28.38 L, T=529 K
For an ideal gas, P V = n R T. With n = 3.752 mol, V = 28.38 L, T = 529 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 5.739 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,778
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=3.319 mol, V=49.9 L, T=285 K
For an ideal gas, P V = n R T. With n = 3.319 mol, V = 49.9 L, T = 285 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 1.555 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,779
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=0.03059 mol, V=9.438 L, T=506.7 K
For an ideal gas, P V = n R T. With n = 0.03059 mol, V = 9.438 L, T = 506.7 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 0.1348 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,780
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=4.603 mol, V=25.16 L, T=298.5 K
For an ideal gas, P V = n R T. With n = 4.603 mol, V = 25.16 L, T = 298.5 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 4.481 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,781
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=0.4806 mol, V=14.57 L, T=331.5 K
For an ideal gas, P V = n R T. With n = 0.4806 mol, V = 14.57 L, T = 331.5 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 0.897 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,782
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=4.907 mol, V=18.99 L, T=248.8 K
For an ideal gas, P V = n R T. With n = 4.907 mol, V = 18.99 L, T = 248.8 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 5.275 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,783
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=4.646 mol, V=16.52 L, T=242.4 K
For an ideal gas, P V = n R T. With n = 4.646 mol, V = 16.52 L, T = 242.4 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 5.597 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,784
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=0.1541 mol, V=35.53 L, T=384.1 K
For an ideal gas, P V = n R T. With n = 0.1541 mol, V = 35.53 L, T = 384.1 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 0.1367 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,785
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=2.404 mol, V=13.83 L, T=451.3 K
For an ideal gas, P V = n R T. With n = 2.404 mol, V = 13.83 L, T = 451.3 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 6.438 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,786
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=3.293 mol, V=2.67 L, T=331.8 K
For an ideal gas, P V = n R T. With n = 3.293 mol, V = 2.67 L, T = 331.8 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 33.58 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,787
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=3.116 mol, V=47.4 L, T=381.2 K
For an ideal gas, P V = n R T. With n = 3.116 mol, V = 47.4 L, T = 381.2 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 2.056 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,788
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=4.999 mol, V=39.95 L, T=562.8 K
For an ideal gas, P V = n R T. With n = 4.999 mol, V = 39.95 L, T = 562.8 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 5.778 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,789
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=0.6553 mol, V=40.18 L, T=545.3 K
For an ideal gas, P V = n R T. With n = 0.6553 mol, V = 40.18 L, T = 545.3 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 0.7298 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,790
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=1.422 mol, V=37.32 L, T=519.3 K
For an ideal gas, P V = n R T. With n = 1.422 mol, V = 37.32 L, T = 519.3 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 1.624 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,791
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=2.74 mol, V=5.081 L, T=440.9 K
For an ideal gas, P V = n R T. With n = 2.74 mol, V = 5.081 L, T = 440.9 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 19.51 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,792
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=1.283 mol, V=38.66 L, T=296 K
For an ideal gas, P V = n R T. With n = 1.283 mol, V = 38.66 L, T = 296 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 0.8064 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,793
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=2.001 mol, V=10.81 L, T=208.2 K
For an ideal gas, P V = n R T. With n = 2.001 mol, V = 10.81 L, T = 208.2 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 3.163 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,794
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=1.949 mol, V=32.53 L, T=361 K
For an ideal gas, P V = n R T. With n = 1.949 mol, V = 32.53 L, T = 361 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 1.774 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,795
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=4.773 mol, V=15.28 L, T=261.7 K
For an ideal gas, P V = n R T. With n = 4.773 mol, V = 15.28 L, T = 261.7 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 6.707 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,796
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=4.007 mol, V=46.12 L, T=494.4 K
For an ideal gas, P V = n R T. With n = 4.007 mol, V = 46.12 L, T = 494.4 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 3.525 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,797
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=4.52 mol, V=18.27 L, T=441.4 K
For an ideal gas, P V = n R T. With n = 4.52 mol, V = 18.27 L, T = 441.4 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 8.959 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,798
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=0.4896 mol, V=10.39 L, T=347 K
For an ideal gas, P V = n R T. With n = 0.4896 mol, V = 10.39 L, T = 347 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 1.341 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,799
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=2.233 mol, V=13.79 L, T=355.2 K
For an ideal gas, P V = n R T. With n = 2.233 mol, V = 13.79 L, T = 355.2 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 4.72 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.
2,800
chemistry
gases
ideal_gas_law
4
worked_example
Ideal-gas pressure for n=3.006 mol, V=21.42 L, T=236.1 K
For an ideal gas, P V = n R T. With n = 3.006 mol, V = 21.42 L, T = 236.1 K and R = 0.082057 L·atm·mol⁻¹·K⁻¹, the pressure is P = n R T / V = 2.719 atm. Real gases approach ideal behavior at low pressure and high temperature relative to their critical points.
P V = n R T
mole_concept
Apply the ideal-gas law to compute pressure, volume, or temperature.