Answer:
90.3 kJ/mol
Explanation:
Let's consider the following thermochemical equation.
2 NO(g) + O₂(g) → 2 NO₂(g) ∆H°rxn = –114.2 kJ
We can find the standard enthalpy of formation for NO using the following expression.
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × ΔH°f(O₂(g))
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × 0 kJ/mol
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g))
ΔH°f(NO(g)) = (2 mol × ΔH°f(NO₂(g)) - ∆H°rxn) / 2 mol
ΔH°f(NO(g)) = (2 mol × 33.2 kJ/mol + 114.2 kJ) / 2 mol
ΔH°f(NO(g)) = 90.3 kJ/mol
<span>An ionized atom has a number of protons that is unequal to the number of electrons. - I took the Penn exam, 100% sure!</span>
So I haven’t got time to answer all of it for you but the id you look at the picture of the periodic table I’ve added the top number in the red boxes are the groups and the period is how many elements down from the top it is (remember that the hydrogen and helium make up period ONE) so remember to include them when counting the elements as you go down the table
I think its A because some scientist from the Department of Physics of Northeastern University found out that is not a part of Dalton's atomic theory.
Moles of CO₂ = mass / molecular weight
Moles of CO₂ = 4.4 / (12 + 16 x 2)
Moles of CO₂ = 0.1 mol
Each mole of gas occupies 22.4 L at STP. Therefore,
Moles of NH₃ = 5.6 / 22.4
Moles of NH₃ = 0.25 mol
