An impure sample of benzoic acid (C 6H 5COOH, 122.12 g/mol) is titrated with 0.2099 M NaOH. A 1.021-g sample requires 35.73 mL of titrant to reach the endpoint. What is the percent by mass of benzoic acid in the sample? C6H5COOH(aq) + NaOH(aq) → NaC6H5COO(aq) + H2O(l)

The percent by mass of benzoic acid = 89.7 %

Explanation:

Step 1: Data given

Molarity of NaOH = 0.2099 M

Mass of the sample = 1.021 grams

Volume of NaOH = 35.73 mL = 0.03573 L

Step 2: The balanced equation

C6H5COOH(aq) + NaOH(aq) → NaC6H5COO(aq) + H2O(l)

Step 3: Calculate moles NaOH

Moles NaOH = molarity NaOH * volume NaOH

Moles NaOH = 0.2099 * 0.03573 L

Moles NaOH = 0.00750 moles

Step 3: Calculate moles benzoic acid

From your balanced equation - benzoic acid reacts in 1:1 molar ratio with NaOH  

The sample must have contained 0.00750 moles benzoic acid  

Mass benzoic acid in 0.00750 moles  = 0.00750 * 122.12 = 0.9159 g benzoic acid  

Mass % acid in sample =  0.9159/1.021 * 100 = 89.7% purity.

The percent by mass of benzoic acid = 89.7 %

c- cold periods in earth's history during which glaciers covered much of the surface.

explanation:

ca has a crystalline lattice with positive ions (since it is a metal, metals are positively charged) and electrons flow freely in its structure since the ca atoms have a large nuclear radius so the electrons are not tightly held, therefore the electrons can flow freely since they are not being highly attracted by the nucleus. moreover, one of the properties of metals are that they are conductors of electricity, thus proving ca is the element.