According to the equation of molarity:
Molarity= no.of moles / volume per liter of Solution
when we have the molarity=0.58 M and the beaker at 150mL so V (per liter) = 150mL/1000 = 0.150 L
by substitution:
∴ No.of moles = Molarity * Volume of solution (per liter)
= 0.58 * 0.150 = 0.087 Moles
<h3>Answer:</h3>
100 g O₂
<h3>General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O
[Given] 2 mol CH₄
[Solve] x g O₂
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol CH₄ → 2 mol O₂
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of O₂ - 2(16.00) = 32.00 g/mol
<u>Step 3: Stoichiometry</u>
- Set up conversion:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 1 sig fig.</em>
128 g O₂ ≈ 100 g O₂
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Answer:
Explanation:
1 Mole of Aluminum with mass 26.98g contains 6.02*10^23 atoms.
In 2.88g of Aluminum, there are 2.88/26.98*6.02*10^23 = 6.426*10^22 atoms.