How many moles were required to prepare 500mL of a 0.45M
calcium chloride solution?

0.225 moles

Explanation:

Calcium chloride has a chemical formula of CaCl2

Given that,

Amount of moles of CaCl2 (n) = ?

Volume of aqeous solution (v) = 500mL

[Convert volume in milliliters to liters

If 1000mL = 1L

500mL = Z

cross multiply

Z x 1000mL = 1L x 500mL

Z = 500mL•L / 1000mL

Z = 0.5L]

Concentration of CaCl2 solution (c) = 0.45M

Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence

c = n / v

make n the subject formula

n = c x v

n = 0.45M x 0.5L

n = 0.225 moles

Thus, 0.225 moles were required to prepare 500mL of a 0.45M calcium chloride solution

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.