Answer:
0.535 g
Explanation:
The reaction that takes place is:
- NaCl + AgNO₃ → AgCl + NaNO₃
First we <u>calculate how many AgNO₃ moles are there in 25.0 mL of a 0.366 M solution</u>, using the <em>definition of molarity</em>:
- Molarity = moles / liters
- moles = Molarity * liters
<em>Converting 25.0 mL to L </em>⇒ 25.0 / 1000 = 0.025 L
- moles = 0.366 M * 0.025 L = 0.00915 mol AgNO₃
Then we <u>convert AgNO₃ moles into NaCl moles</u>:
- 0.00915 mol AgNO₃ *
= 0.00915 mol NaCl
Finally we<u> convert NaCl moles into grams</u>, using its <em>molar mass</em>:
- 0.00915 mol NaCl * 58.44 g/mol = 0.535 g
<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
4.0
i think it has something to do with molar ratios and finding the limiting reactant
4.0 mol NO * 2 mol NO2/2 mol NO = 4.0 moles of NO2
4.0 mol O2 * 2 mol NO2/1 mol O2 = 8.0 moles of NO2
so the limiting reactant (the reactant that runs out the quickest leaving an excess) is NO
once the limiting reactant is found, we can use that data for that substance to calculate the amount of product
4.0 mol NO * 2 mol NO2/2 mole NO = 4.0 moles of NO2