Answer:
-1,103.39KJ/mol
Explanation:
We use the subtract the standard enthalphies of formation of the reactants from that of the products. It must be taken into consideration that the enthalpy of formation of elements and their molecules alone are not taken into consideration. Hence, what we would be considering are the standard enthalpies of formation of H2S, H2O and SO2.
In places where we have more than one mole, we multiply by the number of moles as seen in the balanced chemical equations.
The standard enthalpies of the molecules above are as follows:
H2S = -20.63KJ/mol
H2O = -285.8KJ/mol
SO2 = -296.84KJ/mol
O2 = 0KJ/mol
ΔrH⦵ = [2ΔfH⦵(H2O) + 2 ΔfH⦵(SO2)] − [ΔfH⦵(H2S) + 3
ΔfH⦵(O2)]
ΔrH⦵ =[(2 × -285.8) + (2 × -296.84)]
-[ 3 × -20.63)]
= (-571.6 - 593.68 + 61.89) = -1,103.39KJ/mol
A) 2H₂(g) + O₂(g) → 2H₂O(l) + 285.83 kJ
Exothermic
B) 2Mg + O₂ → 2MgO + 1200kJ
Exothermic
Answer:
Did you mean solution??
In chemistry, a solution is considered as a special type of homogeneous mixture composed of two or more substances. In such a mixture, a solute is a substance dissolved in another substance, known as a solvent.
Eg: sweet tea or coffee (sugar added to solution)
bleach (sodium hypochlorite dissolved in water)
HOPE IT HELPS :)
7. An exothermic reaction
8. The bonds are forming
Answer:
The procedure you will use in this exercise exploits the difference in acidity and solubility just described.
(a) you will dissolve your unknown in ethyl acetate (an organic solvent). All of the possible compounds are soluble in ethyl acetate.
(b) you will extract with sodium bicarbonate to remove any carboxylic acid that is present.
(c) you will extract with sodium hydroxide to remove any phenol that is present.
(d) you will acidify both of the resulting aqueous solutions to cause any compounds that were extracted to precipitate.
