Answer:
The common oxide of nitrogen that has a positive ΔS°f is nitric oxide (NO)
Explanation:
Without reference to thermodynamic data, we have;
1) N₂ (g) + O₂ (g) ⇄ 2 NO (g)
1 unit of N₂ + 1 unit of O₂ (total of 2 units) gives 2 units of NO, (Increase of +0 disorder)
∴ΔS°f = +ve
2) 2NO + O₂ → 2NO₂
2 unit of NO + 1 unit of O₂ (total of 3 units) gives 2 units of NO₂, (Decrease of disorder)
∴ΔS°f = -ve
3) N₂ + 1/2 O₂ → N₂O
1 unit of N₂ + 1/2 unit of O₂ (total of 1+1/2 units) gives 2 units of NO₂, (Decrease of disorder)
∴ΔS°f = -ve
4) 4 NO₂ + O₂ → 2N₂O₅
4 unit of NO₂ + 1 unit of O₂ (total of 5 units) gives 2 units of N₂O₅, (Decrease of disorder)
∴ΔS°f = -ve
5) NO + NO₂ ⇄ N₂O₃
1 unit of NO + 1 unit of NO₂ (total of 2 units) gives 1 unit of N₂O₃, (Decrease of disorder)
∴ΔS°f = -ve
Therefore, the common oxide of nitrogen that has a positive ΔS°f without reference to thermodynamic data is nitric oxide NO.
what?? please reword this
Answer: 2.3 moles
Explanation:
Recall that based on Avogadro's law, 1 mole of any substance has 6.02 x 10^23 atoms
So if 1 mole of Aluminum = 6.02 x 10^23 atoms
Then, Z moles = 1.4 x 10^24 atoms
To get the value of Z, we cross multiply:
1 mole x 1.4 x 10^24 atoms = Z x (6.02 x 10^23 atoms)
1.4 x 10^24 atoms = Z x (6.02 x 10^23)
Hence, Z = (1.4 x 10^24 atoms) ➗ (6.02 x 10^23 atoms)
Z =2.3 moles
Thus, there are 2.3 moles in 1.4 x 10^24 atoms of aluminum.
Answer:
1) mass ZnO = 55.155 g
2) V SO2(g) = 18.289 L
Explanation:
1) Zn + H2O → ZnO + H2
∴ mass Zn = 41.6 g
∴ mm Zn = 65.38 g/mol
⇒ mol Zn = (41.6 g)(mol/61.38 g) = 0.678 mol Zn
⇒ mol ZnO = (0.678 mol Zn)(mol ZnO/mol Zn) = 0.678 mol ZnO
∴ mm ZnO = 81.38 g/mol
⇒ mass ZnO = (0.678 mol ZnO)(81.38 g/mol) = 55.155 g ZnO
2) S(s) + O2(g) → SO2(g)
∴ mass S(s) = 24 g
∴ T = 25°C ≅ 298 K
∴ P = 1 atm
∴ mm S(s) = 32.065 g/mol
⇒ mol S(s) = (24 g)(mol/32.065 g) = 0.7485 mol S(s)
⇒ mol SO2(g) = (0.7485 mol S(s))(mol SO2(g)/mol S(s)) = 0.7485 mol SO2(g)
ideal gas:
⇒ V SO2(g) = ((0.082 atm.L/K.mol)(298 K)(0.7485 mol))/(1 atm)
⇒ V SO2(g) = 18.289 L SO2(g)
