Question

What is the mass of 8.56 x 10^23 formula units of BaBr2? (3 sig figs in your answer) ​

Answer 1

Answer:

296 g BaBr₂

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets

Parenthesis

Exponents

Multiplication

Division

Addition

Subtraction

Left to Right

Chemistry

Atomic Structure

Reading a Periodic Table

Moles

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

[Given] 8.56 × 10²³ formula units BaBr₂

[Solve] grams BaBr₂

Step 2: Identify Conversions

Avogadro's Number

[PT] Molar Mass of Ba - 137.33 g/mol

[PT] Molar Mass of Br - 35.45 g/mol

Molar Mass of BaBr₂ - 137.33 + 2(35.45) = 208.23 g/mol

Step 3: Convert

[DA} Set up:                                                                                                      

[DA] Multiply/Divide [Cancel out units]:                                                          

Step 4: Check

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

295.99 g BaBr₂ ≈ 296 g BaBr₂

Answer 2

Answer:

296 g BaBr₂

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Chemistry

Atomic Structure

• Reading a Periodic Table
• Moles
• Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

• Using Dimensional Analysis

Explanation:

Step 1: Define

[Given] 8.56 × 10²³ formula units BaBr₂

[Solve] grams BaBr₂

Step 2: Identify Conversions

Avogadro's Number

[PT] Molar Mass of Ba - 137.33 g/mol

[PT] Molar Mass of Br - 35.45 g/mol

Molar Mass of BaBr₂ - 137.33 + 2(35.45) = 208.23 g/mol

Step 3: Convert

1. [DA} Set up:                                                                                                     $\displaystyle 8.56 \cdot 10^{23} \ formula \ units \ BaBr_2(\frac{1 \ mol \ BaBr_2}{6.022 \cdot 10^{23} \ formula \ units \ BaBr_2})(\frac{208.23 \ g \ BaBr_2}{1 \ mol \ BaBr_2})$
2. [DA] Multiply/Divide [Cancel out units]:                                                         $\displaystyle 295.99 \ g \ BaBr_2$

Step 4: Check

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

295.99 g BaBr₂ ≈ 296 g BaBr₂

Thank you agenthammerx for helping me with this question!