C2H4(g) + 2O2(g) → 2CO(g) + 2H2O(g) ∆H1 = −758 kJ mol−1 ....1)
2C(s) + 2H2(g) → C2H4(g) ∆H2 = +52 kJ mol−1 ....2)
H2(g) + O2(g) → H2O(g) ∆H3 = −242 kJ mol−1 ....3)
Now, enthalpy of formation of carbon monoxide is given by :
∆H = ∆H1 + ∆H2 - ∆H3
∆H = ( -758 + 52 - ( -242 ) ) kJ mol−1
∆H = −464 kJ mol−1
Therefore, the enthalpy of formation of carbon monoxide is -464 kJ mol−1.
Hence, this is the required solution.
Answer is: the maximum wavelength of light is 7.34 × 10⁻⁷ m.
E= 163 kJ/mol; bond energy of one mole of nitrogen.
E = 163000 J/mol ÷ 6.022·10²³ 1/mol.
E = 2.707·10⁻¹⁹ J; bond energy per molecule.
E = h·c/λ.
c = 3.00 × 10⁸ m/s; light speed.
h = 6.62607004·10⁻³⁴ J·s; Planck constant.
λ(photon) = h·c/E.
λ(photon) = 6.62607004·10⁻³⁴ J·s · 3.00 × 10⁸ m/s ÷ 2.707·10⁻¹⁹ J.
λ(photon) = 7.34 × 10⁻⁷ m.
λ(photon) = 734 nm.
Answer: it goes: collecting data, preforming an investigation, communicating results, asking a question, and then providing an explanation
Explanation:
edge said i got it right
0.22 moles of Al₂(CO₃)₃
Explanation:
The decomposition reaction of aluminium carbonate (Al₂(CO₃)₃):
Al₂(CO₃)₃ → Al₂O₃ + 3 CO₂
Taking account the chemical reaction we devise the following reasoning:
if 1 mole of Al₂(CO₃)₃ produces 3 moles of 3 CO₂
then X moles of Al₂(CO₃)₃ produces 0.65 moles of 3 CO₂
X = (1 × 0.65) / 3 = 0.22 moles of Al₂(CO₃)₃
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decomposition reaction
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In order to calculate the final volume of the gas, we may apply Charles Law, which states that for a fixed amount of gas subject to a fixed amount of pressure, the volume and temperature of a gas are directly proportional. This is:
V/T = constant
Using this equation,
5/298 = V/278
V = 4.66 L
The final volume of the gas is 4.66 liters