<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: CF4
Explanation:
Calculate the molar mass of each compound. Divide the molar mass of Carbon by the molar mass of each compound, then multiply the answer by 100 to get the percentage.
CF4= 12+(19X 4)
=12+76= 88 g/mol
%C= 12/88 x 100= 13.64%
CO2= 12+(16 X 2)
12+32= 44 G/MOL
%C= 12/44 x 100= 27.3%
CH4= 12+ (1 X4)
=12+4
=16 G/MOL
%C= 12/16 X 100= 75%
C204
(12X2) + (16X4)
24+64
= 88 g/mol
%C= 24/88 x 100
= 27.3%
<u>Answer:</u> The standard cell potential of the cell is -0.71 V
<u>Explanation:</u>
The half reactions follows:
<u>Oxidation half reaction:</u>
( × 3)
<u>Reduction half reaction:</u>
( × 2)
The balanced cell reaction follows:

To calculate the
of the reaction, we use the equation:

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
Putting values in above equation, we get:

Hence, the standard cell potential of the cell is -0.71 V