Answer:
5.702 mol K₂SO₄
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Compounds
- Moles
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[Given] 993.6 g K₂SO₄
[Solve] moles K₂SO₄
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of K: 39.10 g/mol
[PT] Molar Mass of S: 32.07 g/mol
[PT] Molar mass of O: 16.00 g/mol
Molar Mass of K₂SO₄: 2(39.10) + 32.07 + 4(16.00) = 174.27 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Divide [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 4 sig figs.</em>
5.7015 mol K₂SO₄ ≈ 5.702 mol K₂SO₄
Physical Properties<span>: </span>Physical properties<span> can be observed or measured without changing the composition of matter. </span>Physical properties<span> are used to observe and describe matter. so physical changes are the change in temperature of the land and the evaporation of water and change humidity of the air. chemical change is the ripening of the orange</span>
Answer:
0.1g (Gallon) of chlorine
Explanation:
<u>Formula</u>
1 gallon = 3.7L; the density of water is 1.0g/ml
<u>Given</u>
2g (gallon) of chlorine to sanitize = 1,000,000g (gallon) of water
<u>Solve</u>
If 2g (gallon) chlorine = 1,000,000g (gallon)
∴, ? chlorine = 40,000
The First step; set up an equation
1000000/2 = 40000/?
The Next step; divide 1 million to 2
1000000 ÷ 2 = 500000
Then, divide the result by 40000
40000 ÷ 500000 = 0.08
In the nearest unit that is 0.1
Therefore, it will take 0.1g (gallon) of chlorine to sanitize a 40,000-gallon pool.
Answer:
17.83M
Explanation:
Given parameters:
Mass of KCO₂ = 1.5kg
Volume of water = 850mL
Unknown:
Molarity of solution = ?
Solution:
Molarity is defined as the number of moles per unit volume. It is the number of moles of solute in a given volume of solution.
Molarity = 
Number of moles of solute = 
Molar mass of KCO₂ = 39 + 12 + 3(16) = 99g/mol
Number of moles =
= 15.15moles
850mL : gives 0.85L
Molarity =
= 17.83M
Answer:
Water vapor has mostly disappeared at colder elevations and its effect on weather turbulence.
Explanation: