Anything that has mass and takes up space: Matter Nucleus Element Atom
1 answer:
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<h3>Answer:</h3>
495 g K₃N
<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> </u>
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
<u>Stoichiometry</u>
- Using Dimensional Analysis
<u>Step 1: Define</u>
3.77 mol K₃N
<u>Step 2: Identify Conversions</u>
Molar Mass of K - 39.10 g/mol
Molar Mass of N - 14.01 g/mol
Molar Mass of K₃N - 3(39.10) + 14.01 = 131.31 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>
495.039 g K₃N ≈ 495 g K₃N
<u>Answer:</u> The cell potential of the cell is +0.118 V
<u>Explanation:</u>
The half reactions for the cell is:
<u>Oxidation half reaction (anode):</u>
<u>Reduction half reaction (cathode):</u>
In this case, the cathode and anode both are same. So, will be equal to zero.
To calculate cell potential of the cell, we use the equation given by Nernst, which is:
where,
n = number of electrons in oxidation-reduction reaction = 2
= ?
=
= 1.0 M
Putting values in above equation, we get:
Hence, the cell potential of the cell is +0.118 V