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Ka (mol/L ) = <span>0.00002340</span>
Answer:
reactivaty
Explanation:
here you go for the answer
Answer : The pressure of gas will be, 3.918 atm and the combined gas law is used for this problem.
Solution :
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,

where,
= initial pressure of gas = 3 atm
= final pressure of gas = ?
= initial volume of gas = 1.40 L
= final volume of gas = 0.950 L
= initial temperature of gas = 
= final temperature of gas = 
Now put all the given values in the above equation, we get the final pressure of gas.


Therefore, the pressure of gas will be, 3.918 atm and the combined gas law is used for this problem.
<h3>
Answer:</h3>
0.6 g NaCl
<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] Na₂CO₃ (aq) + CaCl₂ (aq) → CaCO₃ (s) + 2NaCl (aq)
[Given] 0.5 g Na₂CO₃ reacted with excess CaCl₂
<u>Step 2: Identify Conversions</u>
[RxN] Na₂CO₃ → 2NaCl
Molar Mass of Na - 22.99 g/mol
Molar Mass of C - 12.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of Cl - 35.45 g/mol
Molar Mass of Na₂CO₃ - 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol
Molar Mass of NaCl - 22.99 + 35.45 = 58.44 g/mol
<u>Step 3: Stoichiometry</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 1 sig fig.</em>
0.551373 g NaCl ≈ 0.6 g NaCl