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cricket20 [7]
2 years ago
7

ILL GIVE A BRAINLY PLS HELP

Chemistry
1 answer:
Oksana_A [137]2 years ago
3 0

Atoms are the unit of the molecule of the compound. The 3.01 x 10²⁴  atoms of oxygen are present in 5 moles of water and 3 moles of carbon dioxide in the sample.

<h3>What are atoms?</h3>

Atoms are the smallest fundamental unit of the compounds that can be given by Avogadro's number.

For calculating the oxygen atoms in 5 mole water:

If 1 mole = 6.02 × 10²³

Then, 5 moles = 5 ×  6.02 × 10²³

Hence, 3.01 x 10²⁴ atoms of oxygen are present in 5 moles of water.

Moles of carbon dioxide in the sample is calculated as:

If 1 mole of carbon dioxide = 6.02 × 10²³ molecules

Then moles in 1.8 x 10²⁴ molecules will be,

(1.8 x 10²⁴ molecules) ÷ (6.02 × 10²³ molecules)  = 3 moles

Hence, 3 moles of carbon dioxide is present in the sample.

Learn more about Avogadro's number here:

brainly.com/question/5098076

#SPJ1

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How many moles do you have in 37.3 g of Co(CrO4)3​
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Answer:

0.0917 mol Co(CrO₄)₃

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

37.3 g Co(CrO₄)₃

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

Molar Mass of Co - 58.93 g/mol

Molar Mass of Cr - 52.00 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Co(CrO₄)₃ - 58.93 + 3(52.00) + 12(16.00) = 406.93 g/mol

<u>Step 3: Convert</u>

<u />37.3 \ g \ Co(CrO_4)_3(\frac{1 \ mol \ Co(CrO_4)_3}{406.93 \ g \ Co(CrO_4)_3} ) = 0.091662 mol Co(CrO₄)₃

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

0.091662 mol Co(CrO₄)₃ ≈ 0.0917 mol Co(CrO₄)₃

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