Nul a

Asked by maham237 @ 03/01/2021 in Mathematics viewed by 153 People

Find an explicit description of Nul A by listing vectors that span the null space.
A=[ 10 3 1 5 4 0 −2 ]

Answered by maham237 @ 03/01/2021

Answer:

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:

C₄H₁₀(g) -126 kJ/mol

O₂(g) 0 kJ/mol (by convention, all elements in its most stable state have enthalpy of formation equal to zero).

CO₂(g) -393.5 kJ/mol
H₂O(l) -285.5 kJ/mol

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.

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Nul a

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