Answer:
Explanation:1) ΔrH = 2mol·ΔfH(NO) - (ΔfH(O₂) + ΔfH(N₂)).
ΔrH = 2 mol · 90.3 kJ/mol - (0 kJ/mol + 0 kJ/mol).
ΔrH = 180.6 kJ.
2) ΔS = 2mol·ΔS(NO) - (ΔS(O₂) + ΔS(N₂)).
ΔS = 2mol · 210.65 J/mol·K - (1mol · 205 J/mol·K + 1 mol · 191.5 J/K·mol).
ΔS = 24.8 J/K.
3) ΔG = ΔH - TΔS.
55°C: ΔG = 180.6 kJ - 328.15 K · 24.8 J/K = 172.46 kJ.
2570°C: ΔG = 180.6 kJ - 2843.15 K · 24.8 J/K = 110.09 kJ.
3610°C: ΔG = 180.6 kJ - 3883.15 K · 24.8 J/K = 84.29 kJ.
Answer:
Se
Explanation:
Se is in geoup 6, period 4
<u>Answer:</u>
NO ---> N +2 and O -2
<u>Explanation:</u>
Oxidation numbers are assigned to the elements of a compound to keep a track of the number of electrons each atom has.
Here we have a compound NO (Nitrogen Oxide). The Nitrogen is assigned an oxidation number of +2 while Oxygen in this compound is assigned an oxidation number of -2.
So the algebraic sum of the oxidation numbers of the elements in the compound NO is equal to zero.
Answer:
Explanation:
Assuming that temperature is constant
According to Boyle's Law, at constant temperature pressure is inversly proportional to the volume and mathematically it can be expressed as:
..........1
from the first equation after putting all the value
we get,
