Calculate the enthalpy change for the reaction mn3o4(s)+co(g)⟶3mno(s)+co2(g) from the following:

mn3o4(s)+4co(g)⟶3mn(s)+4co2(g) δh=255.6kj

mno(s)+co(g)⟶mn(s)+co2(g) δh=102.1kj

express your answer using one decimal place and include the appropriate units.

- 50.7 kJ.

Explanation:

Mn₃O₄(s) + CO(g) ⟶ 3MnO(s) + CO₂(g).

The first reaction be as it is:

Mn₃O₄(s) + 4CO(g) ⟶ 3Mn(s) + 4CO₂(g), ΔH₁ = 255.6 kJ.

The second reaction should be reversed and multiplied by 3 and also the value of its ΔH must multiplied by (- 3):

3Mn(s) + 3CO₂(g) ⟶ 3MnO(s) + 3CO(g), ΔH₂ = (- 3)(102.1 kJ) = - 306.3 kJ.

Mn₃O₄(s) + CO(g) ⟶ 3MnO(s) + CO₂(g).

∴ ΔH rxn = ΔH₁ + ΔH₂ = (255.6 kJ) + (- 306.3 kJ) = - 50.7 kJ.

in 1783 cavendish was working with his hydrogen gas and found that when he burned it and condensed the vapor it was water. this was very significant since it disproved the greek theory of elements completely since water was not a simple substance but the product of two gases. it was lavoisier who hearing of the experiment gave cavendish's gas the name hydrogen ("water-producer") and added that hydrogen burned by combining with oxygen and that water was a hydrogen-oxygen combination.

explanation:

c. is the answer

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