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
Hey there!:
H is always +1 so the H's have a +3 charge.
O is always -2 so the O's have a -8 charge .
Now, suppose oxidation state for P = X , then :
+3 + X + (-8) = 0 (because of neutral molecule)
x = 8 - 3
x = + 5
So, X = +5 oxidation state.
Answer C
Hope that helps!
Answer: C
Explanation:
The one closest to the atomic center, there is a single 1s orbital that can hold 2 electrons. At the next energy level, there are four orbitals.
Answer:
If 500.0 mL of 0.450 M sodium phosphate is reacted with an excess of iron (II) nitrate solution, how many grams of iron (II) phosphate are produced?
idk
Explanation:
