Spheres representing Oxygen and Nitrogen are different colors so that it is easier for you to see.
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:
8 OH⁻(aq) + Mn(s) ⇒ MnO₄⁻(aq) + 4 H₂O(l) + 7 e⁻
Explanation:
Let's consider the following oxidation half-reaction that takes place in basic aqueous solution.
Mn(s) ⇒ MnO₄⁻(aq)
First, we will perform the mass balance. We will add 4 H₂O to the products side and 8 OH⁻ to the reactants side.
8 OH⁻(aq) + Mn(s) ⇒ MnO₄⁻(aq) + 4 H₂O(l)
Finally, we will perform the charge balance by adding 7 electrons to the products side.
8 OH⁻(aq) + Mn(s) ⇒ MnO₄⁻(aq) + 4 H₂O(l) + 7 e⁻
Niobium has Highest ionization energy.
<u>Explanation:</u>
The next ionization energy removes an electron from the same electron shell, which increases "ionization energy" due to "increased net charge"of the ion from which the electron is being removed.
Let's compare each of the metals first ionization energies,
Rubidium has its first ionization energy as 403 kJ / mol.
Zirconium has its first ionization energy as 660 KJ / mol.
Niobium has its first ionization energy as 664 KJ / mol.
From the given data above, we can infer that Rubidium has the least Ionization energy whereas Niobium has the highest Ionization energy. We can give the increasing order of the given elements as follows-
Rubidium < Zirconium < Niobium
Answer:
its D Her speed is 4.4 m/s, and her velocity is 4.4 m/s.
Explanation:
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