In a 3.7- gram sample, there are atoms of gold

1.1 × 10²² atoms Au

Chemistry

Atomic Structure

Math

Pre-Algebra

Order of Operations: BPEMDAS

Step 1: Define

3.7 g Au

Step 2: Identify Conversions

Avogadro's Number

Molar Mass of Au - 196.97 g/mol

Step 3: Convert

= 1.13121 × 10²² atoms Au

Step 4: Check

We are given 2 sig figs. Follow sig fig rules and round.

1.13121 × 10²² atoms Au ≈ 1.1 × 10²² atoms Au

the answer is b. dilution

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.