1 half-life - 4.5 days
x half-lives - 27 days
x=(1*27)/4.5=6
6 <span>half-lives have elapsed after 27 days.</span>
Answer:- 2.39 mL are required.
Solution:- It's a dilution problem and to solve this type of problems we use the dilution equation:

Where,
and
are molarities of concentrated and diluted solutions and
and
are their respective volumes.
= 1.10M
= 5.00mM = 0.005M (since, mM stands for milli molar and M stands for molar. 1M = 1000mM)
= ?
= 525 mL
Let's plug in the given values in the formula:



So, 2.39 mL of 1.10M are needed to make 525 mL of 5.00mM solution.
<h3>
Answer:</h3>
495 g K₃N
<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>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
3.77 mol K₃N
<u>Step 2: Identify Conversions</u>
Molar Mass of K - 39.10 g/mol
Molar Mass of N - 14.01 g/mol
Molar Mass of K₃N - 3(39.10) + 14.01 = 131.31 g/mol
<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>
495.039 g K₃N ≈ 495 g K₃N