Answer:
C co2 2co enthalpy
2 Answers. Ernest Z. The standard enthalpy of formation of carbon monoxide is -99 kJ/mol.
Answer:
Option B. 4.25×10¯¹⁹ J
Explanation:
From the question given above, the following data were obtained:
Frequency (f) = 6.42×10¹⁴ Hz
Energy (E) =?
Energy and frequency are related by the following equation:
Energy (E) = Planck's constant (h) × frequency (f)
E = hf
With the above formula, we can obtain the energy of the photon as follow:
Frequency (f) = 6.42×10¹⁴ Hz
Planck's constant (h) = 6.63×10¯³⁴ Js
Energy (E) =?
E = hf
E = 6.63×10¯³⁴ × 6.42×10¹⁴
E = 4.25×10¯¹⁹ J
Thus, the energy of the photon is 4.25×10¯¹⁹ J
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Answer:
ΔHorxn = - 11.79 KJ
Explanation:
2 SO 2 ( g ) + O 2 ( g ) ⟶ 2 SO 3 ( g )
The standard enthalpies of formation for SO 2 ( g ) and SO 3 ( g ) are Δ H ∘ f [ SO 2 ( g ) ] = − 296.8 kJ / mol Δ H ∘ f [ SO 3 ( g ) ] = − 395.7 kJ / mol
From the reaction above, 2 mol of SO2 reacts to produce 2 mol of SO3. Assuming ideal gas behaviour,
1 mol = 22.4l
x mol = 2.67l
Upon cross multiplication and solving for x;
x = 2.67 / 22.4 = 0.1192 mol
0.1192 mol of SO2 would react to produce 0.1192 mol of SO3.
Amount of heat is given as;
ΔHorxn = ∑mΔHof(products) − ∑nΔHof(reactants)
Because O2(g) is a pure element in its standard state, ΔHοf [O2(g)] = 0 kJ/mol.
ΔHorxn = 0.1192 mol * (− 395.7 kJ / mol) - 0.1192 mol * ( − 296.8 kJ / mol)
ΔHorxn = - 47.17kj + 35.38kj
ΔHorxn = - 11.79 KJ
Answer:
653.45
Explanation:
pi * 4 by the power of 2 * 13 ~653.45
