<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:

- [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>
5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH
Water combines with carbon dioxide to produce slightly acidic groundwater
that dissolves limestone and forms caves.
This is because the reaction between water and carbondioxide to form
bicarbonate ions( HCO₃⁻). The bicarbonate ions dissociate into Hydrogen
atoms thereby increasing the acidity.
The acidic environment results in the formation of acidic groundwater that
dissolves limestone and forms caves.
Read more on brainly.com/question/25385913
13% of 0.900 is 0.117
there are 0.117 milliliters of alcohol in a 0.900-L bottle of wine.
Answer:
a. 53.5 g/mol
b. 80.06 g/mol
c. 133.33 g/mol
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Molar Mass - 1 mol per <em>x</em> grams substance
Explanation:
<u>Step 1: Define</u>
a. NH₄Cl
b. NH₄NO₃
c. AlCl₃
<u>Step 2: Find masses</u>
Molar Mass of N - 14.01 g/mol
Molar Mass of H - 1.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of Al - 26.98 g/mol
Molar Mass of Cl - 35.45 g/mol
<u>Step 3: Calculate compound masses</u>
Molar Mass of NH₄Cl - 14.01 g/mol + 4(1.01 g/mol) + 35.45 g/mol = 53.5 g/mol
Molar Mass of NH₄NO₃ - 2(14.01 g/mol) + 4(1.01 g/mol) + 3(16.00 g/mol) = 80.06 g/mol
Molar Mass of AlCl₃ - 26.98 g/mol + 3(35.45 g/mol) = 133.33 g/mol