A. Chloroplasts
B. The cell wall and the vacuole
C. Vacuoles
D. The mitochondrion
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
Answer:
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Answers:
(a) 30.55 °C
(b) 298 K and 77°F
(c) 204.44 °C and 477.44 K
(d) -320.8 °F and -196 °C
Explanation:
Converting °C into °F;
°F = °C × 1.8 + 32
Converting °F into °C;
°C = °F - 32 ÷ 1,8
Converting °C into K;
K = °C + 273
Converting K into °C;
°C = K - 273
I think answer should be d. Please give me brainlest let me know if it’s correct or not okay thanks bye