Electron - negligible mass, negative charge, orbits the nucleus
Proton - 1 AMU, positive charge, in the nucleus
Neutron, 1 AMU, no charge, in the nucleus
Answer:
The pressure inside the container is 6.7 atm
Explanation:
We have the ideal gas equation: P x V = n x R x T
whereas, P (pressure, atm), V (volume, L), n (mole, mol), R (ideal gas constant, 0.082), T (temperature, Kelvin)
Since the container is evacuated and then sealed, the volume of the body of gas is the volume of the container.
So we can calculate the pressure by
P = n x R x T / V
where as,
n = 41.1 g / 44 g/mol = 0.934 mol
Hence P = 0.934 x 0.082 x 298 / 3.4 L = 6.7 atm
<h3>
Answer:</h3>
5.55 mol C₂H₅OH
<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 Tables
- Moles
<u>Stoichiometry</u>
- Using Dimensional Analysis
- Analyzing Reactions RxN
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
[Given] 500. g C₆H₁₂O₆ (Glucose)
[Solve] moles C₂H₅OH (Ethanol)
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH
[PT] Molar mass of C - 12.01 g/mol
[PT] Molar Mass of H - 1.01 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol
<u>Step 3: Stoichiometry</u>
- [DA] Set up conversion:
![\displaystyle 500 \ g \ C_6H_{12}O_6(\frac{1 \ mol \ C_6H_{12}O_6}{180.18 \ g \ C_6H_{12}O_6})(\frac{2 \ mol \ C_2H_5OH}{1 \ mol \ C_6H_{12}O_6})](https://tex.z-dn.net/?f=%5Cdisplaystyle%20500%20%5C%20g%20%5C%20C_6H_%7B12%7DO_6%28%5Cfrac%7B1%20%5C%20mol%20%5C%20C_6H_%7B12%7DO_6%7D%7B180.18%20%5C%20g%20%5C%20C_6H_%7B12%7DO_6%7D%29%28%5Cfrac%7B2%20%5C%20mol%20%5C%20C_2H_5OH%7D%7B1%20%5C%20mol%20%5C%20C_6H_%7B12%7DO_6%7D%29)
- [DA} Multiply/Divide [Cancel out units]:
![\displaystyle 5.55001 \ mol \ C_2H_5OH](https://tex.z-dn.net/?f=%5Cdisplaystyle%205.55001%20%5C%20mol%20%5C%20C_2H_5OH)
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH
The mass of 100 pills would be 78000 mg and 0.078 kilograms.
Hope this helps you :)