How many atoms of phosphorus are in 9.20 mol of copper(II) phosphate?

1 answer:

4 0

The number of atoms of phosphorus in copper (II) phosphate [Cu3 (PO4)2] is determined by multiplying first 9.20 by 2 since there are two phosphorus atoms in the compound and multiplying again with Avogadro's number. The answer is 1.108 x 10^25 atoms.

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2C3H7OH + 9O2 --> 6CO2 + 8H2O

Ne4ueva [31]

<h3>Answer:</h3>

733 g CO₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2C₃H₇OH + 9O₂ → 6CO₂ + 8H₂O

[Given] 5.55 mol C₃H₇OH

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol C₃H₇OH → 6 CO₂

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

<u>Step 3: Stoichiometry</u>

Set up conversion: $\displaystyle 5.55 \ mol \ C_3H_7OH(\frac{6 \ mol \ CO_2}{2 \ mol \ C_3H_7OH})(\frac{44.01 \ g \ CO_2}{1 \ mol \ CO_2})$
Multiply/Divide: $\displaystyle 732.767 \ g \ CO_2$