Answer:
148.04 kJ/mol
Explanation:
Let's consider the following thermochemical equation.
NO(g) + 1/2 O₂(g) → NO₂(g) ΔH°rxn = -114.14 kJ/mol
We can find the standard enthalpy of formation (ΔH°f) of NO(g) using the following expression.
ΔH°rxn = 1 mol × ΔH°f(NO₂(g)) - 1 mol × ΔH°f(NO(g)) - 1/2 mol × ΔH°f(O₂(g))
ΔH°f(NO(g)) = 1 mol × ΔH°f(NO₂(g)) - ΔH°rxn - 1/2 mol × ΔH°f(O₂(g)) / 1 mol
ΔH°f(NO(g)) = 1 mol × 33.90 kJ/mol - (-114.14 kJ) - 1/2 mol × 0 kJ/mol / 1 mol
ΔH°f(NO(g)) = 148.04 kJ/mol
it is already balanced
REACTANTS
Barium sulfide (BaS) + platinum (Ii) fluoride
PRODUCT
Barium fluoride (BaF2) + Cooperite (PtS)
Hope this answer helps you dear! take care
Answer:
the motion of molecules increases
Answer: Empirical formula is 
Explanation: We are given the masses of elements present in a sample of compound. To evaluate empirical formula, we will be following some steps.
<u>Step 1 :</u> Converting each of the given masses into their moles by dividing them by Molar masses.

Molar mass of Carbon = 12.0 g/mol
Molar mass of Hydrogen = 1.0 g/mol
Molar mass of Oxygen = 16.0 g/mol
Moles of Carbon = 
Moles of Hydrogen = 
Moles of Oxygen = 
<u>Step 2: </u>Dividing each mole value by the smallest number of moles calculated above and rounding it off to the nearest whole number value
Smallest number of moles = 13.76 moles



<u>Step 3:</u> Now, the moles ratio of the elements are represented by the subscripts in the empirical formula
Empirical formula becomes = 