Answer:
Standard enthalpy of formation of Carbon disulfide CS2 = 87.3 KJ/mol
Explanation:
forming CS2 means that it should in the product side
C(graphite) + O2 → CO2 ΔH = -393.5
2S(rhombic) + 2O2 → 2SO2 ΔH = -296.4 x 2
CO2 + 2SO2 → CS2 + 3O2 ΔH = -1073.6 x -1
the second reaction is multiplied by 2 so that the SO2 and O2 can cancel out.
the third reaction is reversed (multiplied by -1) so that CS2 will be on the product side.
after adding the reaction and cancelling out similarities, the final reaction is: C(graphite) + 2S(rhombic) → CS2
Add ΔH to find the enthalpy of formation of CS2
ΔHf = (-393.5) + (-296.4 x 2) + (-1073.6 x -1) = 87.3 KJ/mol
be aware of signs
Answer: Calcium carbonate is another example of a compound with both ionic and covalent bonds. Here calcium acts as the cation, with the carbonate species as the anion. These species share an ionic bond, while the carbon and oxygen atoms in carbonate are covalently bonded
Explanation:
NaOH (aq) + HCl(aq) => NaCl(aq)+ H2O(l)
Na+(aq)+ OH - (aq) + H +(aq) +Cl - (aq) + Cl- (aq)
=> Na+(aq) + Cl - (aq) + H2O(l)
H+(aq) + OH-(aq) => H2O(l)
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)
