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
Answer:
B) the chemicals are gaining energy from the surroundings.
Explanation:
The positive sign of the energy difference in a chemical reaction would indicate that the chemicals are gaining energy from the surroundings. This is what happens in an endothermic reaction.
In an endothermic reaction, heat is absorbed from the surroundings hence the surrounding becomes colder at the end of the changes.
- Here the energy change is assigned a positive value.
- This is because the heat energy level of the final state is higher than that of the initial state.
- So, the difference gives a positive value.
Moles of PF₃ : 4
<h3>Further explanation</h3>
A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.
Reaction
1.25 moles of P₄(s) is reacted with 6 moles of F₂(g)
Limiting reactant : the smallest ratio (mol divide by coefficient)
P₄ : F₂ =
mol PF₃ based on mol of limiting reactant(F₂), so mol PF₃ :
Answer:
c
Explanation:
use %composition given to calculate emperical 4mular
