Question 5) Give the name and symbol of three elements that exist as diatomic molecules.

Question 6 

What is the molar mass of aluminum oxide, Al₂O₃?

Question 7 

What is the mass of 5.80 mol of carbon tetrachloride, CCl₄? The molar mass of carbon tetrachloride is 153.81 g/mol.

Question 8 

A chemist wants to make 100 mL of a standard solution. The chemist first dissolves the weighed quantity of solute in about 50 mL of water in a beaker, stirring with a stirring rod. The chemist then transfers this solution to a 100 mL volumetric flask. What should the chemist do before filling the volumetric flask to the volume mark and why?

Question 9 

What mass of potassium iodide, KI, is present in 2.00 L of 1.250 mol/L KI(aq)?show calculations

Question 10

What mass of silver nitrate, AgNO₃, is needed to make 500.0 mL of 0.600 mol/L AgNO₃(aq)? show calculations

Question 5 

H₂ - Hydrogen

N₂ - Nitrogen

O₂ - Oxygen

Question 6 

101.96 g/mol

Question 7 

892.098 g

Question 8 

When working with a solid solute, one weighs the material to the desired accuracy and transfers it carefully and completely to the volumetric flask. If solute is lost in transfer, the actual concentration of the resulting solution will be lower than the calculated value. Therefore, before filling the flask to mark, rinse first the beaker used to dissolve the solid in order to wash out and include in the flask all remaining solids in the beaker. This way, there would be no loss of solids during the transfer. Once the solute is dissolved, more solvent is added to bring the volume to the mark on the flask.

Question 9 

415.007 g

Question 10

50.961 g

Step-by-step explanation

Question 6

MW of Al2O3 = 2 x MW of Al + 3 x MW of O

MW of Al2O3 = 2 x 26.981 g/mol + 3 x 15.999 g/mol

MW of Al2O3 = 101.96 g/mol

Question 7

mass of CCl4 = Molar Mass x No. of Moles

mass of CCl4 = 153.81 g/mol x 5.80 moles

mass of CCl4 = 892.098 g

Question 9

What mass of potassium iodide, KI, is present in 2.00 L of 1.250 mol/L KI(aq)?show calculations

First determine the moles of KI in the solution using the formula:

moles KI = concentration x volume of solution

moles KI = 1.250 mol/L x 2 L

moles KI = 2.5 moles KI

Now that we know the moles of KI present, we use the molar mass of KI (166.0028 g/mol) to solve for the mass:

mass of KI = molar mass of KI x no. of moles

mass of KI = 166.0028 g/mol x 2.5 mol

mass of KI = 415.007 g

Question 10

What mass of silver nitrate, AgNO3, is needed to make 500.0 mL (equal to 0.5 L) of 0.600 mol/L AgNO3(aq)? show calculations

To solve this, we first need to determine how many moles of AgNO3 is needed to make the needed solution:

mol AgNO3 = concentration x volume

mol AgNO3 = 0.600 mol/L x 0.5 L

mol AgNO3 = 0.3 moles AgNO3

Now that we know the moles needed, we can now calculate the mass using the molar mass of AgNO3 (169.87 g/mol):

mass of AgNO3 = molar mass x moles

mass of AgNO3 = 169.87 g/mol x 0.3 mol

mass of AgNO3 = 50.961 g