What would the result be if the solution was not basic when it was heated to for

ID: 874899 • Letter: W

Question

What would the result be if the solution was not basic when it was heated to form CuO?

Some of the [Cu(H2O)6]2+ would not be converted to Cu(OH)2 and would remain in solution, causing a lower recovery Cu(s)

The [Cu(H2O)6]2+ solution would be directly converted back to Cu(s), in a one step reaction, causing a very high recovery of Cu(s)

Too much of the [Cu(H2O)6]2+ would be converted to CuO(s) causing an increased recovery Cu(s)

The acidity level of the solution is irrelevant

Some of the [Cu(H2O)6]2+ would not be converted to Cu(OH)2 and would remain in solution, causing a lower recovery Cu(s)

The [Cu(H2O)6]2+ solution would be directly converted back to Cu(s), in a one step reaction, causing a very high recovery of Cu(s)

Too much of the [Cu(H2O)6]2+ would be converted to CuO(s) causing an increased recovery Cu(s)

The acidity level of the solution is irrelevant

The result obtained if the solution was not basic when it was heated to form CuO is -

A. Some of the [Cu(H2O)6]2+ would not be converted to Cu(OH)2 and would remain in solution, causing a lower recovery Cu(s)

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