How many grams of cu(oh)2 will precipitate when excess naoh solution is added to 46.0 ml of 0.584 m cuso4
solution?
cuso4(aq) + 2naoh(aq) —> cu(oh)2(s) + na2so4(aq)

2.624 g

The equation for the reaction is given as;

We are required to calculate the mass of Cu(OH)₂ precipitated

Molarity = Number of moles ÷ Volume

To get the number of moles;

Moles = Molarity × volume

          = 0.584 M × 0.046 L

          = 0.0269 moles

Thus, Moles of CuSO₄ = Moles of Cu(OH)₂

Hence, moles of Cu(OH)₂ = 0.0269 moles

To get mass we multiply the number of moles with the molar mass.

Mass = Moles × Molar mass

Molar mass of Cu(OH)₂ is 97.561 g/mol

Therefore;

Mass of Cu(OH)₂ = 0.0269 moles × 97.561 g/mol

                           = 2.624 g

Thus, the mass of Cu(OH)₂ that will precipitate is 2.624 g

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ca has a crystalline lattice with positive ions (since it is a metal, metals are positively charged) and electrons flow freely in its structure since the ca atoms have a large nuclear radius so the electrons are not tightly held, therefore the electrons can flow freely since they are not being highly attracted by the nucleus. moreover, one of the properties of metals are that they are conductors of electricity, thus proving ca is the element.