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

b.final

the temperature of the gas, in k, not celsius.