Answer:
1. 7 (a neutral solution)
Answer: 10-7= 0.0000001 moles per liter
2. 5.6 (unpolluted rainwater)
Answer: 10-5.6 = 0.0000025 moles per liter
3. 3.7 (first acid rain sample in North America)
Answer: 10-3.7 = 0.00020 moles per liter
The concentration of H+ in the Hubbard Brook sample is 0.00020/0.0000025, which is 80 times higher than the H+ concentration in unpolluted rainwater.
Explanation:
Answer:
A. Water particles barely move forward; they move in a circular pattern.
Answer:
The correct answer is the imperial system.
Explanation:
Only three countries in the world, that is, the United States, Myanmar, and Liberia use the imperial system. These include measurements in the form of inches, ounces, Fahrenheit, and feet. In the imperial system, the distances, height, weight, or area measurements are used eventually that traced back to everyday items or parts of the body.
In comparison to other metric systems, the units used in the imperial system are not further differentiated easily into parts of hundreds or thousands, and are thus, regarded of less use in comparison to other metric systems by some. The real follower of the imperial system at present in the world is the United States.
<h3>
Answer:</h3>
1.83 × 10⁻⁷ mol Au
<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
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
3.60 × 10⁻⁵ g Au (Gold)
<u>Step 2: Identify Conversions</u>
Molar Mass of Au - 196.97 g/mol
<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>
1.82769 × 10⁻⁷ mol Au ≈ 1.83 × 10⁻⁷ mol Au
The amount, in mL, of the concentrated acid required, would be 1.1875 mL
<h3>Dilution</h3>
From the dilution equation:
m1v1=m2v2 where m1 and m2 = molarity before and after dilution, and v1 and v2 = volume before and after dilution.
m2 = 0.285M, m1 = 12.0M v2 = 50.0 mL
v1 = m2v2/m1 = 0.285x50/12 = 1.1875 mL
Thus, 1.1875 mL of the acid would be taken and diluted with water up to the 50 mL mark.
More on dilution can be found here: brainly.com/question/13949222
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