Answer:
The molecular formula is C12H18O3
Explanation:
Step 1: Data given
The empirical formula is C4H6O
Molecular weight is 212 g/mol
atomic mass of C = 12 g/mol
atomic mass of H = 1 g/mol
atomic mass of O = 16 g/mol
Step 2: Calculate the molar mass of the empirical formula
Molar mass = 4* 12 + 6*1 +16
Molar mass = 70 g/mol
Step 3: Calculate the molecular formula
We have to multiply the empirical formula by n
n = the molecular weight of the empirical formula / the molecular weight of the molecular formula
n = 70 /212 ≈ 3
We have to multiply the empirical formula by 3
3*(C4H6O- = C12H18O3
The molecular formula is C12H18O3
<h3>
Answer:</h3>
0.0253 mol H₂O
<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>
[Given] 0.456 g H₂O (water)
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of H - 1.01 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Multiply/Divide [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.025305 mol H₂O ≈ 0.0253 mol H₂O
A logarithmic scale is a nonlinear scale used when there is a large range of quantities
Answer:
V₂ = 6.0 mL
Explanation:
Given data:
Initial volume = 9.0 mL
Initial pressure = 500 mmHg
Final volume = ?
Final pressure = 750 mmHg
Solution:
According to Boyle's Law
P₁V₁ = P₂V₂
V₂ = P₁V₁ / P₂
V₂ = 500 mmHg × 9.0 mL / 750 mmHg
V₂ = 4500 mmHg .mL / 750 mmHg
V₂ = 6.0 mL