Answer:
D (hope I’m right, sorry if it’s not)
Answer:

Explanation:
In multiplication and division problems, your answer can have no more significant figures than the number with the fewest significant figures.
(by my calculator)
There are three significant figures in 7.06 and two in 2.3.
You must round to
significant figures and report the answer as 1.3 × 1.0⁷.
<h3>
Answer:</h3>
4 g AgCl
<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>Stoichiometry</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN] 2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂
[Given] 5.0 g AgNO₃
<u>Step 2: Identify Conversions</u>
[Reaction - Stoich] 2AgNO₃ → 2AgCl
Molar Mass of Ag - 107.87 g/mol
Molar Mass of N - 14.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of Cl - 35.45 g/mol
Molar Mass of AgNO₃ - 107.87 + 14.01 + 3(16.00) = 169.88 g/mol
Molar Mass of AgCl - 107.87 + 35.45 = 143.32 g/mol
<u>Step 3: Stoichiometry</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 1 sig fig.</em>
4.21533 g AgCl ≈ 4 g AgCl
Since Chlorine is in excess, this is a limiting reagent problem.
1) convert 11.50 g Na to g of NaCl using the balanced equation.
2) percent yield = (actual yield)/(potential yield).
.85= (actual yield)/(g from step 1)
Solve for actual yield