I think this is the answer you was looking for hope this helps !!!
Answer:
a) Warmer
b) Exothermic
c) -10.71 kJ
Explanation:
The reaction:
KOH(s) → KOH(aq) + 43 kJ/mol
It is an exothermic reaction since the reaction liberates 43 kJ per mol of KOH dissolved.
Hence, the dissolution of potassium hydroxide pellets to water provokes that the beaker gets warmer for being an exothermic reaction.
The enthalpy change for the dissolution of 14 g of KOH is:
<u>Where:</u>
m: is the mass of KOH = 14 g
M: is the molar mass = 56.1056 g/mol
The enthalpy change is:
The minus sign of 43 is because the reaction is exothermic.
I hope it helps you!
Answer:
0.533 mol O2
Explanation:
4 Fe3O4 + O2 -> 6 Fe2O3
1 mol O2 -> 6 mol Fe2O3
x -> 3.2 mol Fe2O3
x = (3.2 mol Fe2O3 * 1 mol O2)/ 6 mol Fe2O3
x= 0.533 mol O2
The answer is ....... 47/10^8
Answer:
42.3%
Explanation:
By the reaction given the stoichiometry between MnO₄⁻ and Fe⁺² is 1 mol of MnO₄⁻ to 5 moles of Fe⁺². When KMnO₄ dissolves, it forms the same amount of K⁺ and MnO₄⁻ (1:1:1), so the number of moles of MnO₄⁻ used is:
n = Volume*concentration
n = 0.0216 L * 0.102 mol/L
n = 2.20x10⁻³ mol
1 mol of MnO₄⁻ -------------------- 5 moles of Fe⁺
2.20x10⁻³ mol ---------------------- x
By a simple direct three rule
x = 0.011 mole
The molar mass of iron is 55.8 g/mol. The mass is the molar mass multiplied by the number of moles, thus:
m = 55.8*0.011
m = 0.614 g
Then, the percent of iron in the ore is:
(mass of iron/ mass of ore) *100%
(0.614/1.45)*100%
42.3%
