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2 years ago

Label the states of each compound in the reaction below.

Chemistry

1 answer:

devlian [24]2 years ago

8 0

Answer:

CuSO₄(aq) + 2KOH (aq) —> Cu(OH)₂(s) + K₂SO₄(aq)

Explanation:

CuSO₄ + 2KOH —> Cu(OH)₂ + K₂SO₄

To know the state of each compounds, we shall determine the complete ionic equation. This can be obtained as follow:

In solution, CuSO₄ and KOH will dissociate as follow:

CuSO₄(aq) —> Cu²⁺(aq) + SO₄²¯(aq)

KOH (aq) —> K⁺ (aq) + OH¯(aq)

CuSO₄(aq) + KOH (aq) —>

Cu²⁺(aq) + SO₄²¯(aq) + 2K⁺(aq) + 2OH¯(aq) —> Cu(OH)₂(s) + 2K⁺(aq) + SO₄²¯(aq)

Note: Cu(OH)₂ is insoluble in water.

Therefore, the elemental equation is:

CuSO₄(aq) + 2KOH (aq) —> Cu(OH)₂(s) + K₂SO₄(aq)

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2 years ago

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2 years ago

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$P^0_A$ = Vapour pressure of component A in pure form

$P^0_B$ = Vapour pressure of component B in pure form

$x_A$ =Mole fraction of component A

$x_B=$ =Mole fraction of component B

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