<h3>Answer:</h3>
64 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] 36 g H₂O
[Solve] x g O₂
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol O₂ → 2 mol H₂O
[PT] Molar Mass of O - 16.00 g/mol
[PT] Molar Mas of H - 1.01 g/mol
Molar Mass of O₂ - 2(16.00) = 32.00 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol
<u>Step 3: Stoichiometry</u>
- Set up conversion:

- Divide/Multiply [Cancel Units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
63.929 g O₂ ≈ 64 g O₂
Answer:
F = 50 N
Explanation:
Given data:
Mass of car = 250 Kg
Acceleration of car = 0.20 m/s²
Force required = ?
Solution:
Formula:
F = m×a
F = applied force
m = mass
a = acceleration
Now we will put the values in formula.
F = 250 Kg × 0.20 m/s²
F = 50 Kg.m/s²
Kg.m/s² = N
F = 50 N
Pure air is a mixture of several gases that are invisible and odorless. Consists of 78% nitrogen, 21% oxygen, and less than 1% of argon, carbon dioxide.
Answer:

Explanation:
Hello!
In this case, since the density is computed by dividing the mass of the substance by its occupied volume (d=m/V), we first need to realize that 0.8206 g/mL is the same to 0.8206 kg/L, which means we first need to compute the volume in L:

Then, solving for the mass in d=m/V, we get m=d*V and therefore the mass of gasoline in that full tank turns out:

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