Question

Volumes of 0.18M copper (II) sulfate, and water.

6.5ml 0.18M CuSO4

4.0ml H20

This is my chemistry finals, I need help immediately!

Answer 1

See explanation

Step-by-step explanation:

Number of moles of Copper sulphate = 0.18 M × 6.5/1000= 1.17 × 10^-3 moles of CuSO4

From;

number of moles = mass/molar mass

Molar mass of copper II sulphate = 160g/mol

1.17 × 10^-3 moles = mass/160g/mol

Mass = 1.17 × 10^-3 moles ×160

Mass = 0.187 g

If 1 mole contains 6.02 × 10^23 molecules

1.17 × 10^-3 moles contains 1.17 × 10^-3 moles × 6.02 × 10^23/1 = 7 × 10^20 molecules

From;

C1 V1=C2 V2

0.18 × 6.5 =C2 × 10.5

C2 = 0.18 × 6.5/10.5

C2=0.11 M

CuSO4(aq) + 2NaOH(aq) ----> Cu(OH)2(s) + Na2SO4(aq)

1 mole of CuSO4 yields 1 mole of Cu(OH)2

1.17 × 10^-3 moles of CuSO4 yields 1.17 × 10^-3 moles of Cu(OH)2

Mass of Cu(OH)2 = 1.17 × 10^-3 moles × 98g/mol= 0.11g of Cu(OH)2