Answer:
The answer to your question is: % error = 0.4
Explanation:
Data
real value = 22.48%
estimated value = 22.57 %
Formula
% error = |real value - estimated value|/real value x 100
%error = |22.48 - 22.57|/22.48 x 100
% error = |-0.09|/22.48 x 100
%error = 0.09/22.48 x 100
% error = 0.004 x 100
% error = 0.4
<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
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Answer:
solubility of X in water at 17.0
is 0.11 g/mL.
Explanation:
Yes, the solubility of X in water at 17.0
can be calculated using the information given.
Let's assume solubility of X in water at 17.0
is y g/mL
The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.
So, solubility of X in 1 mL of water = y g
Hence, solubility of X in 36.0 mL of water = 36y g
So, 36y = 3.96
or, y =
= 0.11
Hence solubility of X in water at 17.0
is 0.11 g/mL.
Answer:
2
Explanation:
There are 3 carbons on the right side
there are only two on the left side....you need one more ...so add one more... change the '1' coefficient in front of C to '2'