<h3>
Answer:</h3>
1000 g CCl₄
<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>
- Reading a Periodic Table
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
3.93 × 10²⁴ molecules CCl₄
<u>Step 2: Identify Conversions</u>
Avogadro's Number
Molar Mass of C - 12.01 g/mol
Molar Mass of Cl - 35.45 g/mol
Molar Mass of CCl₄ - 12.01 + 4(35.45) = 153.81 g/mol
<u>Step 3: Convert</u>
- Set up:
- Multiply:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
1003.77 g CCl₄ ≈ 1000 g CCl₄
Answer: It has to be 56.10564
Explanation:1 grams KOH to mol = 0.01782 mol
10 grams KOH to mol = 0.17824 mol
20 grams KOH to mol = 0.35647 mol
30 grams KOH to mol = 0.53471 mol
40 grams KOH to mol = 0.71294 mol
50 grams KOH to mol = 0.89118 mol
100 grams KOH to mol = 1.78235 mol
200 grams KOH to mol = 3.5647
Molarity is defined as the number of moles of solute in 1 L of solution
molarity of stock solution to be prepared - 100 x 10⁻³ mol/L
volume of stock solution to be prepared - 1.2 mL
Therefore number of moles in 1.2 mL - 100 x 10⁻³ mol/L x 1.2 x 10⁻³ L
number of moles of drug - 1.2 x 10⁻⁴ mol
mass of drug required - 1.2 x 10⁻⁴ mol x 181.6 g/mol = 21. 8 mg
21.8 g of drug is required to make the stock solution
Answer:
Density independent factor
Explanation:
in dense areas where people livein very close and tight spaces
Moles of solute does not change.