Question

What mass of aluminum nitrate do you need to prepare 3.58 L of a 1.77 M solution?

Answer 1

Answer:

177 grams CaCI2

Explanation:

Use the equation for Molarity = moles / L and rearrange to remedy for moles = M x L. You're given each those values. Moles = 1.56 M x 1.032 L = 1.61 moles Now find the mass of 1 mole of CaCl2 Ca = forty grams Cl = 35 x 2 = 70 grams brought together they complete one hundred ten grams per mole. Multiply this via the number of moles you need to get the grams (mass) of CaCl2. One hundred ten grams/mole x 1.61 moles = 177 grams CaCl2

Answer 2

Answer:

Hope it helps you

Explanation:

Use the equation for Molarity = moles / L and rearrange to remedy for moles = M x L. You're given each those values. Moles = 1.56 M x 1.032 L = 1.61 moles Now find the mass of 1 mole of CaCl2 Ca = forty grams Cl = 35 x 2 = 70 grams brought together they complete one hundred ten grams per mole. Multiply this via the number of moles you need to get the grams (mass) of CaCl2. One hundred ten grams/mole x 1.61 moles = 177 grams CaCl