Answer:
1. 0.178 moles ; 2. 8x10²³ atoms ; 3. 7.22x10²³ molecules ; 4. 89.6 g ; 5. 1.34x10²² atoms ; 6. 1.67x10²⁵ molecules
Explanation:
1. Mass / Molar mass = Mol
5g / 28 g/m = 0.178 moles
2. 1 molecule of N₂ has 2 atoms, it is a dyatomic molecule.
4x10²³ x2 = 8x10²³ atoms
3. 1 mol of anything, has 6.02x10²³ particles
6.02x10²³ molecules . 1.2 mol = 7.22x10²³
4. 1 atom of C weighs 12 amu.
4.5x10²⁴ weigh ( 4.5x10²⁴ . 12) = 5.24x10²⁵ amu
1 amu = 1.66054x10⁻²⁴g
5.24x10²⁵ amu = (5.24x10²⁵ . 1.66054x10⁻²⁴) = 89.6 g
5. Molar mass NaCl = 58.45 g/m
1.3 g / 58.45 g/m = 0.0222 moles
1 mol has 6.02x10²³ atoms
0.0222 moles → ( 0.0222 . 6.02x10²³) = 1.34x10²²
6. Density of water is 1 g/mL, so 500 mL are contained in 500 g of water
Molar mass H₂O = 18 g/m
500 g / 18 g/m = 27.8 moles
6.02x10²³ molecules . 27.8 moles = 1.67x10²⁵
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Answer:
3. V = 0.2673 L
4. V = 2.4314 L
5. V = 0.262 L
6. V = 2.224 L
Explanation:
3. assuming ideal gas:
∴ R = 0.082 atm.L/K.mol
∴ V1 = 225 L
∴ T1 = 175 K
∴ P1 = 150 KPa = 1.48038 atm
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(175 K))/((1.48038 atm)(225 L))
⇒ n = 0.043 mol
∴ T2 = 112 K
∴ P2 = P1 = 150 KPa = 1.48038 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(112 K)(0.043 mol))/(1.48038 atm)
⇒ V2 = 0.2673 L
4. gas is heated at a constant pressure
∴ T1 = 180 K
∴ P = 1 atm
∴ V1 = 44.8 L
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(180 K))/((1 atm)(44.8 L))
⇒ n = 0.3295 mol
∴ T2 = 90 K
⇒ V2 = RT2n/P
⇒ V2 = ((0.082 atm.L/K.mol)(90 K)(0.3295 mol))/(1 atm)
⇒ V2 = 2.4314 L
5. V1 = 200 L
∴ P1 = 50 KPa = 0.4935 atm
∴ T1 = 271 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(271 K))/((0.4935 atm)(200 L))
⇒ n = 0.2251 mol
∴ P2 = 100 Kpa = 0.9869 atm
∴ T2 = 14 K
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(14 K)(0.2251 mol))/(0.9869 atm)
⇒ V2 = 0.262 L
6.a) ∴ V1 = 24.6 L
∴ P1 = 10 atm
∴ T1 = 25°C = 298 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(298 K))/((10 atm)(24.6 L))
⇒ n = 0.0993 mol
∴ T2 = 273 K
∴ P2 = 101.3 KPa = 0.9997 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(273 K)(0.0993 mol))/(0.9997 atm)
⇒ V2 = 2.224 L
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