Lead has a specific heat of 0.13 J/g°C. If 137 T of heat energy is needed to raise

4.365g

Explanation:

We'll begin by calculating the number of mole of Co(NO3)2•6H2O present in 100mL of 0.150 M Co(NO3)2•6H2O solution. This is illustrated below:

Molarity = 0.150M

Volume of solution = 100mL

Recall: 1000mL = 1L

Therefore, 100mL = 0.1L

Mole of solute (Co(NO3)2•6H2O) =?

Molarity is simply mole per unit litre of solution. It is represented mathematically as:

Molarity = mole /Volume

Mole = Molarity x Volume

Mole of Co(NO3)2•6H2O = 0.15x0.1

Mole of Co(NO3)2•6H2O = 0.015mol

Now, we need to convert 0.015mol to grams in order to obtain the desired result. This is illustrated below:

Mole of Co(NO3)2•6H2O = 0.015mol

Molar Mass of Co(NO3)2•6H2O = 59 + 2[14 + (16x3)] + 6[(2x1) + 16] = 59 + 2[14 + 48] + 6[2 + 16] = 59 + 2[62] + 6 [18] = 59 + 124 + 108 = 291g/mol

Mass of Co(NO3)2•6H2O =?

Mass = number of mole x molar Mass

Mass of Co(NO3)2•6H2O = 0.015 x 291

Mass of Co(NO3)2•6H2O = 4.365g

Therefore, 4.365g of Co(NO3)2•6H2O is needed to prepare 100mL of 0.150 M Co(NO3)2•6H2O solution.