Answer the 4 questions pls
2 answers:
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Answer:
the mass of CaO present at equilibrium is, 0.01652g
Explanation:
= 3.8×10⁻²
Now we have to calculate the moles of CO₂
Using ideal gas equation,
PV =nRT
P = pressure of gas = 3.8×10⁻²
T = temperature of gas = 1000 K
V = volume of gas = 0.638 L
n = number of moles of gas = ?
R = gas constant = 0.0821 L.atm/mole.k
Now we have to calculate the mass of CaO
mass = 2.95 * 10 ⁻⁴ × 56
= 0.01652g
Therefore,
the mass of CaO present at equilibrium is, 0.01652g
<h3>Answer:</h3>
0.0463 mol KCl
<h3>General Formulas and Concepts:</h3><u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<u>Step 1: Define</u>
3.45 g KCl
<u>Step 2: Identify Conversions</u>
Molar Mass of K - 39.10 g/mol
Molar Mass of Cl - 35.45 g/mol
Molar Mass of KCl - 39.10 + 35.45 = 74.55 g/mol
<u>Step 3: Convert</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.046278 mol KCl ≈ 0.0463 mol KCl