What is happening in the following equation? N204 (g) → 2 NO2(g)
OA. Nothing; the reactant and the product are the same.
ОВ.
Six units of dinitrogen tetraoxide are decomposing into two units of nitrogen dioxide.
OC.
Two units of nitrogen dioxide are synthesizing into one unit of dinitrogen tetraoxide.
OD
One unit of dinitrogen tetraoxide gas is decomposing into two units of nitrogen dioxide gas.

This question is incomplete, the complete question is;

What is the heat of a reaction, in joules, with a total reaction mixture volume of 73.0 mL if the reaction causes a temperature change of 4.4 °C in a calorimeter?

Assume that the reaction mixture has a density of 1.00 g/mL and a specific heat of 4.184 J/g-oC. The calorimeter has a heat capacity of 10.0 J/oC.

the heat of the reaction is 1387.9 J

Explanation:

Given that;

Volume = 73.0 mL

density = 1.00 g/mL

specific heat of solution =  4.184 J/g-oC

calorimeter has a heat capacity of 10.0 J/oC.

Temperature change ΔT = 4.4 °C

the heat of a reaction = ?

Total heat of the reaction will be equal to the amount of heat absorbed/lost by the solution and the calorimeter.

so heat of reaction q = mC_solΔT + C_alΔT

where m is mass, C_sol is specific heat capacity of the solution, ΔT temperature change and C_cal is heat capacity of calorimeter.

so first we calculate our mass;

mass = volume × density

=  73.0 mL × 1.00 g/mL

= 73.0 g

so we substitute into our equation;

q = mC_solΔT + C_alΔT  

q = (73 × 4.184 × 4.4) + ( 10 × 4.4 )

q = 1343.9008 + 44

q = 1387.9 J

Therefore the heat of a reaction is 1387.9 J