Answer: False
Explanation:
Since the given equation is not balanced properly.
Since oxygen and hydrogen atoms are not balanced.
There should be 6 H2O (g) molecules and 14 mol H2 (g)
Answer:
The ΔG° is 29 kJ and the reaction is favored towards reactant.
Explanation:
Based on the given information, the ΔH°rxn or enthalpy change is 41.2 kJ, the ΔS°rxn or change in entropy is 42.1 J/K or 42.1 * 10⁻³ kJ/K. The temperature given is 289 K. Now the Gibbs Free energy change can be calculated by using the formula,
ΔG° = ΔH°rxn - TΔS°rxn
= 41.2 kJ - 289 K × 42.1 × 10⁻³ kJ/K
= 41.2 kJ - 12.2 kJ
= 29 kJ
As ΔG° of the reaction is positive, therefore, the reaction is favored towards reactant.
Balance Chemical Equation,
2 CO + O₂ → 2CO₂
Acc. to this reaction,
88 g (2 mole) of CO₂ was produced when = 56 g (2 mole)of CO was reacted
So,
24.7 g of CO₂ will be produced by reacting = X g of CO
Solving for X,
X = (56 g × 24.7 g) ÷ 88 g
X = 2.26 g ÷ 88 g
X = 0.0257 g of CO
Result:
0.0257 g of CO is required to be reacted with excess of O₂ to produce 24.7 g of CO₂.
Answer:
See explanation
Explanation:
When the complex ion Co(H2O)6 2+(aq) is placed in solution and chloride ions are added, the following equilobrium is set up;
Co(H2O)62+(aq) + 4 Cl-(aq) <=> CoCl42-(aq) + 6 H2O(g)
Co(H2O)6 2+(aq) solution is pink in colour while CoCl42-(aq) solution is blue in colour.
Since the solubility of CoCl42-(aq) is endothermic, heating the solution will move the equilibrium position towards the right (more CoCl42-(aq) is formed and the solution is blue in colour).
When the solution is cooled, more Co(H2O)62+(aq) is formed and the equilibrium position shifts towards the left and the solution becomes pink in colour.
