<h3>
Answer:</h3>
16.7 g H₂O
<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>Stoichiometry</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 2NaOH (s) + CO₂ (g) → Na₂CO₃ (s) + H₂O (l)
[Given] 1.85 mol NaOH
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol NaOH → 1 mol H₂O
Molar Mass of H - 1.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 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 3 sig figs.</em>
16.6685 g H₂O ≈ 16.7 g H₂O
Answer: he expected to find no darkening of the film all because it hadn't been exposed to sunlight yet.
Answer:
A control group is someone who takes control over something so a tyrant or a dictator.
Explanation:
If this isnt what your looking for let me know, there wasnt alot of context.
Explanation:
The given precipitation reaction will be as follows.
Here, AgCl is the precipitate which is formed.
It is known that molarity is the number of moles present in a liter of solution.
Mathematically, Molarity = 
It is given that volume is 1.14 L and molarity is 0.269 M. Therefore, calculate number of moles as follows.
Molarity =
0.269 M = 
no. of moles = 0.306 mol
As molar mass of AgCl is 143.32 g/mol. Also, relation between number of moles and mass is as follows.
No. of moles = 
0.307 mol = 
mass = 43.99 g
Thus, we can conclude that mass of precipitate produced is 43.99 g.
It would be carbon dioxide codd2 for the molecules
