Let\'s say you had 25 ml of a 3.25M copper (||) chlorate solution. a) What is th

ID: 1004680 • Letter: L

Question

Let's say you had 25 ml of a 3.25M copper (||) chlorate solution.

a) What is the chemical formula for copper (||) chlorate?

b) How many moles of copper (||) chlorate would be in the solution?

c) How many moles of the copper (||) ion would be in the solution?

d) How many moles of the chlorate ion would be in the solution?

e)How many molecules of copper (||) chlorate would be in the solution?

f) How many atoms of oxygen would be in the solution?

g) How many grams of copper (||) chlorate would have been added to make the solution?

a) copper(II)chlorate = Cu(ClO3)2

b) moles = molarity x volume in liters = 3.25 x 0.025 = 0.08

c) Cu(ClO4)2 ----------> Cu+2 + 2 ClO3-

so 0.08 moles Cu(ClO3)2 contain 0.08 moles Cu+2 ions

d) 0.08 moles Cu(ClO3)2 contain 2 x 0.08 = 0.16 moles ClO3- ions

e) 1 mole contain 6.023 x 1023 molecules of Cu(ClO3)2

0.08 moles contain 0.08 x 6.023 x 1023 = 4.82 x 1022 molecules Cu(ClO3)2

f) 1 mole Cu(ClO3)2 contain 6 x 6.023 x 1023 O atoms

0.08 moles Cu(ClO3)2 contain 0.08 x 6 x 6.023 x 1023 = 2.89 x 1023 O atoms

g) 1 mole =230.45 g

0.08 moles = 0.08 x 230.45 = 18.44 g Cu(ClO3)2

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