Which of the following laboratory procedures best illustrates the law of conservation of mass?
calculating the number of atoms in 11 g of na
using 250 g of impure cu to obtain 200 g of pure cu
heating 32 g of s and 56 g of fe to produce 88 g of fes and unused reactants
burning 2.4 g of mg in an open crucible to produce 2 g of mgo and unused reactants

it would be heating of 32g of s and 56g of Fe to produce 88g of FeS and unused reactants

Option (c) is the correct answer.

Explanation:

According to law of conservation of mass, in an isolated system mass cannot be created or destroyed as it can be transformed from one form to another.

For example, heating 32 g of S and 56 g of Fe to produce 88 g of FeS and unused reactants.

So, total mass of reactants is 32 g + 56 g = 88 g. And, total mass of product formed is also 88 g.

Therefore, mass is neither destroyed nor created as it simply transformed from one from form to another.

Thus, we can conclude that heating 32 g of S and 56 g of Fe to produce 88 g of FeS and unused reactants, best illustrates the law of conservation of mass.

the correct answer would be d. hope this !

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3