A. Two electrodes separated by an electrolyte that can generate an electrical current.
<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
In order to find molarity, you must first find the number of moles that was dissolved.
Now, Moles = Mass ÷ Molar Mass
⇒ Moles of NaCl = 2.922 g ÷ 58.44 g/mol
= 0.05 moles
∴ the Molarity of the NaCl is 0.05 M [Option 1]
Answer:
M of Al=33.09g or 0.0331kg
Explanation:
Heat Energy= specific heat*mass*change in temperature
H=M*C*T
make M subject of the formula
M=H/CT
M=685J/0.90J/g°C*(45°C-22°C)
M=685J/0.90J/g°C*23°C
M=685J/20.7J/g
M=33.09g or 0.0331kg