If you have 16 g of manganese (ii) nitrate tetrahydrate, how much water is required to prepare 0.16 m solution from this amount of salt?

The amount of water required to prepare given amount of salt is 398.4 mL

Explanation:

To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

We are given:

Molarity of solution = 0.16 M

Given mass of manganese (II) nitrate tetrahydrate = 16 g

Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol

Putting values in above equation, we get:

Volume of water = Volume of solution = 398.4 mL

Hence, the amount of water required to prepare given amount of salt is 398.4 mL

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.

it is b that is the answer.