Need ! chemistry moles and grams questions

Part 1

a) 4.2 x 10⁻⁵ mole

b) 8.5 moles

c) 0.0083 moles

d) 0.505 moles

e) 0.0034

Part 2

a) 1103 g

b) 787 g

c) 678 g

d) 202 g

e) 147 g

Explanation:

Part 1

Calculation of number of moles for the following

a) 0.0082 g of Gold

Data given:

mass of Gold (Au) = 0.0082 g

no. of moles (Au) = ?

Formula to be Used

               No. of moles = mass in grams / molar mass

Molar mass of Au = 197 g/mol

Put values in the formula

               No. of moles = 0.0082 g / 197 g/mol

               No. of moles = 4.2 x 10⁻⁵

No. of moles of (Au) = 4.2 x 10⁻⁵moles

b) 812 g of Molybdenum

Data given:

mass of Molybdenum (Mo) = 812 g

no. of moles (Mo) = ?

Formula to be Used

               No. of moles = mass in grams / molar mass

Molar mass of Mo = 96 g/mol

Put values in the formula

               No. of moles = 812 g / 96 g/mol

               No. of moles = 8.5

No. of moles of (Mo) = 8.5 moles

c) 2.02 g of Americium

Data given:

mass of Americium (Am) = 2.02 g

no. of moles (Am) = ?

Formula to be Used

               No. of moles = mass in grams / molar mass

Molar mass of Am = 243 g/mol

Put values in the formula

               No. of moles = 2.02 g / 243 g/mol

               No. of moles = 0.0083

No. of moles of (Am) =

d) 10.09 g of Neon

Data given:

mass of Neon (Ne) = 10.09 g

no. of moles (Ne) = ?

Formula to be Used

               No. of moles = mass in grams / molar mass

Molar mass of Ne = 20 g/mol

Put values in the formula

               No. of moles = 10.09 g / 20 g/mol

               No. of moles = 0.505

No. of moles of (Ne) = 0.505 moles

e) 0.705 g of Bismuth

Data given:

mass of Bismuth (Bi) = 0.705 g

no. of moles (Bi) = ?

Formula to be Used

               No. of moles = mass in grams / molar mass

Molar mass of Bi = 209 g/mol

Put values in the formula

               No. of moles = 0.705 g / 209 g/mol

               No. of moles = 0.0034

No. of moles of (Bi) = 0.505 moles

Part 2

Calculation of number of grams for the following

a) 5.6 mole of Au

Data given:

moles of Gold (Au) = 5.6 mol

mass of Au = ?

Formula to be Used

             mass in grams =  No. of moles x molar mass

Molar mass of Au = 197 g/mol

Put values in the formula

                mass in grams = 5.6 mol x 197 g/mol

                mass in grams = 1103 g

mass in grams of (Au) = 1103 g

b) 8.2 mole of Mo

Data given:

moles of Mo = 8.2 mol

mass of Mo = ?

Formula to be Used

             mass in grams =  No. of moles x molar mass

Molar mass of Mo = 96 g/mol

Put values in the formula

                mass in grams = 8.2 mol x 96 g/mol

                mass in grams = 787 g

mass in grams of (Mo) = 787 g

c) 2.79 mole of Am

Data given:

moles of Am = 2.79 mol

mass of Am = ?

Formula to be Used

             mass in grams =  No. of moles x molar mass

Molar mass of Am = 243 g/mol

Put values in the formula

                mass in grams = 2.79 mol x 243 g/mol

                mass in grams = 678 g

mass in grams of (Am) = 678 g

d) 10.09 mole of Ne

Data given:

moles of Ne = 10.09 mol

mass of Ne = ?

Formula to be Used

             mass in grams =  No. of moles x molar mass

Molar mass of Ne = 20 g/mol

Put values in the formula

                mass in grams = 10.09 mol x 20 g/mol

                mass in grams =  202 g

mass in grams of (Ne) = 202 g

e) 0.705  mole of Bi

Data given:

moles of Bi = 0.705 mol

mass of Bi = ?

Formula to be Used

             mass in grams =  No. of moles x molar mass

Molar mass of Bi = 209 g/mol

Put values in the formula

                mass in grams = 0.705 mol x 209 g/mol

                mass in grams =  147 g

mass in grams of (Bi) = 147 g

the dependant variable is spilled water

the independent is the heat

it would be a gas formation