9. Let's say, you need to make 2.00 L of 0.05 M copper (II) nitrate solution. How many grams of copper (II) nitrate will you require?

18.76 g of copper II nitrate

Explanation:

Now recall that we must use the formula;

n= CV

Where;

n= number of moles of copper II nitrate solid

C= concentration of copper II nitrate solution

V= volume of copper II nitrate solution

Note that;

n= m/M

Where;

m= mass of solid copper II nitrate

M= molar mass of copper II nitrate

Thus;

m/M= CV

C= 0.05 M

V= 2.00 L

M= 187.56 g/mol

m= the unknown

Substituting values;

m/ 187.56 g/mol = 0.05 M × 2.00 L

m= 0.05 M × 2.00 L × 187.56 g/mol

m= 18.76 g of copper II nitrate

Therefore, 18.76 g of copper II nitrate is required to make 0.05 M solution of copper II nitrate in 2.00 L volume.

a corroding iron, if that's one of your choices

there are two significant figures.