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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Answer:
the time required for one half of a sample of a radioisotope to decay
(4) PCl5--> PCl3+ Cl2 is the correct answer because we both have 1 P atom and 5 Cl atoms on both sides, and thus the masses of the elements don't change.
Hope this would help~
