Possibly different But here I hope this helps Recall that neutrons, along with protons, make up the nucleus. Since neutrons have no electrical charge, they do not affect the chemical behavior. ... Atoms of the same element with different numbers of neutrons are called isotopes.
Answer:
F. 2NO + 02 —> 2NO
H. 4NH3 + 502 —> 4NO + 6H20
Explanation:
The law of conservation of mass states that matter can neither be created nor destroyed during a chemical reaction but can be convert from one form to another.
2NO + 02 —> 2NO
From the above, the total number of N on the left balance the total number on the right i.e 2 atoms of N on both side of the equation.
The total number of O on the left balance the total number on the right i.e 2 atoms of O on both side of the equation. This is certified by the law of conservation of mass.
4NH3 + 502 —> 4NO + 6H20
From the above, the total number of N on the left balance the total number on the right i.e 4 atoms of N on both side of the equation.
The total number of O on the left balance the total number on the right i.e 10 atoms of O on both side of the equation.
The total number of H on the left balance the total number on the right i.e 12 atoms of O on both side of the equation.
This is certified by the law of conservation of mass.
The rest equation did not conform to the law of conservation of mass as the atoms on the left side did not balance those on the right side
<span>4 Al + 3 O2 → 2 Al2O3
(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al
(19.0 g O2) / (31.99886 g O2/mol) = 0.59377 mol O2
0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess.
((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) =
10.1 g O2 left over</span><span>
</span>
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
