How many particles of CuCr2O7 are present in a 64.5 gram sample?

1.39 × 10²³ particles CuCr₂O₇

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Atomic Structure

Stoichiometry

Step 1: Define

[Given] 64.5 g CuCr₂O₇

[Solve] particles CuCr₂O₇

Step 2: Identify Conversions

Avogadro's Number

[PT] Molar Mass of Cu - 63.55 g/mol

[PT] Molar Mass of Cr - 52.00 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of CuCr₂O₇ - 63.55 + 2(52.00) + 7(16.00) = 279.55 g/mol

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

1.38944 × 10²³ particles CuCr₂O₇ ≈ 1.39 × 10²³ particles CuCr₂O₇

43.452

step-by-step explanation:

284 times .153 = 43.452