Molarity can be defined as the number of moles of solute in 1 L solution.
Molarity of Na₂SO₄ solution - 0.200 M
this means there are 0.200 moles in 1 L solution
Molar mass of Na₂SO₄ - 142 g/mol
therefore mass of Na₂SO₄ in 1.00 L - 0.200 mol x 142 g/mol = 28.4 g
a mass of 28.4 g of Na₂SO₄ is present in 1.00 L
Answer:
Explanation:
We know we will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.
You don't tell us what the reaction is, but we can solve the problem so long as we balance the OH.
M_r: 58.32
Mg(OH)₂ + … ⟶ … + 2HOH
m/g: 58.3
(a) Moles of Mg(OH)₂
(b) Moles of H₂O
The molar ratio is 2 mol H₂O = 1 mol Mg(OH)₂.
The reaction will form of water.
Gold is one of the softest metals. So YES
<h3>
Answer:</h3>
0.000538 mol 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>
- 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>
3.24 × 10²⁰ particles Pb (lead)
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
Our final answer is in 3 sig figs, no need to round.