Answer:
Empirical formula is PNCl₂
Explanation:
Percent composition means that 100 g of compound has x g of each element.
In 100 g of compound x, we have 26.73 g of P, 12.09 g of N, 61.18 g of Cl.
So, let's make some rules of three:
In 100 g of compound we have 26.73 g of P, 12.09 g of N, 61.18 g of Cl
In 579.43 g of compound we have:
(579.43 . 26.73) / 100 = 155 g of P
(579.43 . 12.09) / 100 = 70 g of N
(579.43 . 61.18) / 100 = 354 g of Cl
Let's convert the mass of the elements in moles.
155 g of P / 30.97 g/mol = 5 P
70 g / 14 g/mol = 5 N
354 g / 35.45 g/mol = 10 Cl
<h3>
Answer:</h3>
4 g AgCl
<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] 2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂
[Given] 5.0 g AgNO₃
<u>Step 2: Identify Conversions</u>
[Reaction - Stoich] 2AgNO₃ → 2AgCl
Molar Mass of Ag - 107.87 g/mol
Molar Mass of N - 14.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of Cl - 35.45 g/mol
Molar Mass of AgNO₃ - 107.87 + 14.01 + 3(16.00) = 169.88 g/mol
Molar Mass of AgCl - 107.87 + 35.45 = 143.32 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>
4.21533 g AgCl ≈ 4 g AgCl
Answer:
Lead atoms have 82 electrons and the shell structure is 2.8. 18.32. 18.4.
Explanation:
The ground state electron configuration of ground state gaseous neutral lead is [Xe].