Since the total amount of valence electrons is 3, it is in group 13 in the periodic table..therefore, it is specified as Boron.✅
for it to be balanced in this case would be " <em>4</em> C6H6 + <em>6</em> CI2 = <em>3</em> C6H5CI + <em>9</em> HCI" therefore it's be a <u>Double Replacement</u>
<span>The composition of a fertilizer is usually express in NPK number. NPK number is in terms of Percent by mass of the said element which are Nitrogen, Phosphorus and Potassium. A 15-35-15 fertilizer has 15%
Nitrogen, 35% Phosphorous, and 15% Potassium by mass. If you have 10 g of this
fertilizer, to get the number of moles of phosphorus, you multiply the mass by
35%, which is equal to 10*0.35 or 3.5 g phosphorus. Then you divide the
calculated mass of phosphorous by its molar mass which is 30.97 g/mol.
Therefore, you have 3.5/30.97 which is equal to 0.1130 mol Phosphorus. This is the amount of Phosphorus in moles in the fertilizer.</span>
<h3>
Answer:</h3>
3.0 × 10²³ molecules AgNO₃
<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>
- Reading a Periodic Table
- Writing Compounds
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
85 g AgNO₃ (silver nitrate)
<u>Step 2: Identify Conversions</u>
Avogadro's Number
[PT] Molar Mass of Ag - 107.87 g/mol
[PT] Molar Mass of N - 14.01 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of AgNO₃ - 107.87 + 14.01 + 3(16.00) = 169.88 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 2 sig figs.</em>
3.01313 × 10²³ molecules AgNO₃ ≈ 3.0 × 10²³ molecules AgNO₃
<span>The simplest level of organization for living things is a single organelle, which is composed of aggregates of macromolecules. so it is false</span>