Answer:
146 g
Explanation:
Step 1. Calculate the <em>molar mass</em> of NaNO₃
Na = 22.99
N = 14.01
3O = 3 × 16.00 = 48.00
Total = 85.00 g/mol
Step 2. Calculate the <em>mass</em> of NaNO₃
Mass of NaNO₃ = 1.72 × 85.00/1
Mass of NaNO₃ = 146 g
<h3>
Answer:</h3>
5.2 mol 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>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 6HCl + Fe₂O₃ → 2FeCl₃ + 3H₂O
[Given] 10.4 mol HCl
<u>Step 2: Identify Conversions</u>
[RxN] 6 mol HCl = 3 mol H₂O
<u>Step 3: Stoichiometry</u>
- Set up:

- Multiply/Divide:

Answer:
6.75 × 10⁻⁸is the value of the equilibrium constant at this temperature.
Explanation:
2H₂O(g) ⇄ 2H₂(g) + O₂(g)
Partial pressure of H₂O = 0.0500 atm
Partial pressure of H₂ = 0.00150 atm
Partial pressure of O₂ = 0.00150 atm
The expression of Kp for the given chemical equation is:
![K_p = \frac{[H_2]^2[O_2]}{H_2O}](https://tex.z-dn.net/?f=K_p%20%3D%20%5Cfrac%7B%5BH_2%5D%5E2%5BO_2%5D%7D%7BH_2O%7D)

6.75 × 10⁻⁸is the value of the equilibrium constant at this temperature
<h3>
Answer:</h3>
2.49 × 10⁻¹² moles Pb
<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>
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
1.50 × 10¹² atoms Pb
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- Set up:

- Multiply:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
2.49087 × 10⁻¹² moles Pb ≈ 2.49 × 10⁻¹² moles Pb