A0.1375-g sample of solid magnesium is burned in
aconstant-volume bomb calorimeter that has a heat capacity of
3,024j/oc. the temperature incresases by
1.126oc. calculate the heat given off by the burning mg,
in kj/g and inkj/mol.

a) Q = 0.025 KJ/g

b) Q = 0.602 KJ/mol

Explanation:

ΔU = Q + Wfirst law

∴ a constant volume: ΔV = 0

⇒ W = 0

⇒ ΔU = Q = CvΔT

∴ Cv = 3.024 J/°C

∴ ΔT = 1.126 °C

⇒ Q = (3.024 J/°C)×(1.126 °C ) = 3.405 J

⇒ Q = 3.405 E-3 KJ

a) Q [=] KJ/g

⇒ Q = (3.405 E-3 KJ) / (0.1375 g) = 0.025 KJ/g

b) Q [=] KJ/mol

∴ Mw Mg = 24.305 g/mol

⇒ Q = (0.025 KJ/g)×(24.305 g/mol) = 0.602 KJ/mol

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.