Answer:
there is no question to answer :(
Answer:
-162,5 kJ/mol
Explanation:
Cl(g) + 2O2(g) --> ClO(g) + O3(g) ΔH = 122.8 kJ/mol (as we used the reaction in the opposite direction, it will turn the enthalpy from exothermic to endothermic)
2O3(g) --> 3O2(g) ΔH = -285.3 kJ/mol
Cl(g) + O2(g) --> ClO(g) + O3(g) ΔH = 122.8 kJ
+ 2O3 (g) --> 3O2(g) ΔH = - 285.3 kJ
O3(g) + Cl(g) --> ClO(g) + 2O2(g) ΔH = 122.8 + (-285.3) = -162,5 kJ
Answer:
b. 2 mol of KI in 500. g of water
Explanation:
We have to apply the colligative property of freezing point depression.
The formula is: ΔT = Kf . m . i
As the (Kf . m . i) is higher, then the freezing temperature will be lower.
i refers to the Van't Hoff factor (number of ions dissolved in the solution)
KI → K⁺ + I⁻ (i =2)
Kf is constant so, we have to search for the highest m (molality)
Molality means the moles of solute in 1kg of solvent.
The highest m is option b → 2 mol of KI / 0.5 kg = 4 mol/kg
a. 1 mol of KI / 0.5 kg = 2 mol/kg
c. 1 mol of KI / 1kg = 1 mol/kg
d. 2 mol of KI / 1kg = 2 mol/kg
1000 g = 1kg. In order to determine molality we need to convert the mass (g) of solvent to kg
The maximum percent of yield should be 100%, so It wasn't possible to make a higher than 100% yield like 105% in this problem. If that happens, it will contradict the law of conservation of energy as you can't create mass.
The increased mass might be produced by impurities. One of the examples would be if you rinse the container and left some water. The water is not calculated as a substrate but will increase the results weight.
There also possibilities that some gas in the air react and produce a solid/liquid material. If you don't weight the gas it would seem that the mass increased.
