<h3>
Answer:</h3>
2 M
<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>Unit 0</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<u>Aqueous Solutions</u>
- Molarity = moles of solute / liters of solution
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
36.7 g CaF₂
300 mL H₂O
<u>Step 2: Identify Conversions</u>
Molar Mass of Ca - 40.08 g/mol
Molar Mass of F - 19.00 g/mol
Molar Mass of CaF₂ - 40.08 + 2(19.00) = 78.08 g/mol
1000 mL = 1 L
<u>Step 3: Convert</u>
<em>Solute</em>
- Set up:
- Multiply:
<em>Solution</em>
- Set up:
- Multiply:
<u>Step 4: Find Molarity</u>
- Substitute [M]:
- Divide:
<u>Step 5: Check</u>
<em>Follow sig fig rules and round.</em> <em>We are given 1 sig fig as our lowest.</em>
1.56677 M ≈ 2 M
Answer: is the limiting reagent, 180 g of water is produced
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of require 2 moles of
Thus 5 moles of will require= of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
As 1 mole of give = 2 moles of
Thus 5 moles of give = of
Mass of
Thus 180 g of will be produced from the given masses of both reactants.
Answer:
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Molar mass SiO2 = 28 + 32 = 60
<span>so moles sand = 3.4 x 10-7 / 60</span>