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<h3>Answer:</h3>
733 g CO₂
<h3>General Formulas and Concepts:</h3><u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right<u> </u>
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
<u>Stoichiometry</u>
- Using Dimensional Analysis
<u>Step 1: Define</u>
[RxN - Balanced] 2C₃H₇OH + 9O₂ → 6CO₂ + 8H₂O
[Given] 5.55 mol C₃H₇OH
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol C₃H₇OH → 6 CO₂
Molar Mass of C - 12.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol
<u>Step 3: Stoichiometry</u>
- Set up conversion:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
732.767 g CO₂ ≈ 733 g CO₂
The molecular weight of silver bromide (AgBr) is 187.77 g/mole. The presence of the ions in solution can be shown as- AgBr (insoluble) ⇄ +
.
45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains ×1000 = 14.411 gm-mole. Thus the solubility product
of AgBr = [
]×
.
Or, 5.0× =
(the given value of solubility product of AgBr is 5.0×
and the charge of the both ions are same).
Thus S = (5.00×)
= 7.071×
g/mL.
Thus the concentration of Br or HBr is 7.071×
g/mL.