The reaction for the formation of MgO(s):
2 Mg (s) + O2(g) -à
2MgO(s) ΔH = -601.24
kJ/mol
<span>The enthalpy
information is taken from: http://webbook.nist.gov/cgi/inchi?ID=C1309484&Mask=2</span>
From the equation and with an enthalpy change of -231 kJ:
-231 kJ * 2 mol Mg * (1/-601.24 kJ/mol) = 0.76841 mol Mg
Then, with the molar mass of MgO = 40.3,
0.76841 mol Mg *(2 mol MgO/2 mol Mg)* 40.3 g/mol MgO = <span>30.967 g MgO</span>
Answer:
Oxidation - reduction reaction which is called redox reaction.
Explanation:
The type of reaction that is a part of another type but treated separately is called Oxidation - reduction reaction.
Thus is because they usually involve 2 separate half reactions which are oxidation reactions that involve loss of electrons and then reduction equations which involve gain of electrons.
They are treated separately for example when sodium reacts with chlorine to form sodium chloride.
2Na + Cl2 = 2NaCl
Now, sodium undergoes oxidation by loosing elctrons and it's half reaction is;
2Na → 2Na^(+) + 2e^(-)
Meanwhile chlorine undergoes reduction by gaining electrons and its half reaction is;
Cl2 + 2e^(-) → 2Cl^(-)
The hydrogen ion concentration is calculated as follows
Ph + POh =14
ph= 14-9.85= 4.15
anti -H+
=anti (-4.15) = 7.07 x10^-5 moles per liter
2 stands for valency of ca ion
its +2
that it takes two electrons to complete its shell