Calculate the enthalpy of combustion of butane, c4h10, for the formation of h2o and co2 the enthalpy of formation of butane is -126 kj/mol.

The enthalpy of the combustion of butane is -2,875.5 kJ/mol

Explanation:

To find the enthalpy of the combustion of butane, first we need to write the balanced equation. A combustion is when a substance reacts with oxygen and, as it says in the task, forms water and carbon dioxide. Since enthalpies are usually calculated at 1 bar and 298 K (standard conditions), formed water is in the liquid state.

The balanced equation is:

C₄H₁₀(g) + 6.5 O₂(g) = 4 CO₂(g) +  5 H₂O(l)

Now we can find the enthalpy of this reaction using the formula:

ΔH°r = Σ n(p).ΔH°f(p) - Σ n(r).ΔH°f(r)

where,

ΔH°r is the standard enthalpy of the reaction (in this case the reaction is the combustion, so this the data we are looking for).

n represents the number of moles of reactants and products (which can be found in the balanced equation).

ΔH°f are the standard enthalpies of formation of reactant and products (and can be found in tables).

To apply this formula we need to search the ΔH°f, which are:

Replacing this data in the formula:

ΔH°r = [4mol.(-393.5 kJ/mol) + 5mol.(-285.5kJ/mol)] - [1mol.(-126 kJ/mol)+6.5.(0 kJ/mol)]

ΔH°r = -2,875.5 kJ

Since this enthalpy corresponds to the combustion of 1 mol of butane (according to the balanced equation), we can say that the enthalpy of the combustion of butane is -2,875.5 kJ/mol.

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally

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