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:
Component 1, since it has the greatest mass.
Explanation:
Hello,
In this case, considering that the by-mass percent of a substance is computed:

In such a way, given the form of the equation one could infer that the higher the mass of the i component, the higher its by-mass percent, it means that the component with the greatest mass will have the highest by-mass percent for the mixture. In such a way, since the component 1 was found to have a mass of 1.12 g, it will have the highest by-mass percent.
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The substance of which physical objects are composed, not counting engergies or force fields which are genearlly not counted as matter, although can contribute to the mass of an object.
Or it may be solid, liquid, mass or the 4th state of matter,plasma
Electrons attract objects and pull toward
- 0.05 L needs to be added to the original 0.12 L solution in order to dilute it from 0.13 M to 0.23 M.
The dilution problem uses the equation :

The initial molarity (concentration)
0.13 M
The initial volume
= 0.12 L
The desired molarity (concentration)
= 0.23 M
The volume of the desired solution
= ( 0.12 + x L )
Substituting values in above equation;
(0.13 M ) (0.12 L) = (0.23 M ) (0.12 L + x L)
0.0156 M L = 0.0276 M L + 0.23 x M L
- 0.012 M L = 0.23 x M L
x = - 0.05
Therefore, - 0.05 L needs to be added to the original 0.12 L solution in order to dilute it from 0.13 M to 0.23 M.
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