type the correct answer in the box. the solubility of calcium carbonate is 14 $\frac{milligrams}{liter}$…

type the correct answer in the box. the solubility of calcium carbonate is 14 $\frac{milligrams}{liter}$. this rate means that 14 milligrams of calcium carbonate can dissolve in 1 liter of water. how much water would be required to fully dissolve 11 grams of calcium carbonate? express your answer to the correct number of significant figures. one milligram is equal to 0.001 grams. it would take liters of water to fully dissolve 11 grams of calcium carbonate.

type the correct answer in the box. the solubility of calcium carbonate is 14 $\frac{milligrams}{liter}$. this rate means that 14 milligrams of calcium carbonate can dissolve in 1 liter of water. how much water would be required to fully dissolve 11 grams of calcium carbonate? express your answer to the correct number of significant figures. one milligram is equal to 0.001 grams. it would take liters of water to fully dissolve 11 grams of calcium carbonate.

Answer

Explanation:

Step1: Convert grams to milligrams

Since 1 mg = 0.001 g, then 11 g of calcium - carbonate is $11\div0.001=11000$ mg.

Step2: Set up a proportion

Let $V$ be the volume of water in liters. We know that the solubility is 14 mg per 1 liter of water. So the proportion is $\frac{14\ \text{mg}}{1\ \text{L}}=\frac{11000\ \text{mg}}{V}$.

Step3: Solve for $V$

Cross - multiply gives $14V = 11000$. Then $V=\frac{11000}{14}\approx785.714$ L. Rounding to the correct number of significant figures (2 significant figures since 11 has 2 significant figures), $V\approx790$ L.

Answer:

790