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3 years ago

Find the empirical formula of a compound containing: 19.32% Ca, 34.30% Cl, and 46.38% O

Chemistry

1 answer:

Bess [88]3 years ago

5 0

40×19.32/100=7.7=8×2=16Ca
35.5×34.30/100=12.1=12×2=24Cl
16×46.38/100=7.4=7×2=14O

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What is 93,000,000 in scientific notation?

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Which compound is classified as a hydrocarbon? A) butanal B) butyne C) 2-butanol D) 2-butanone

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Pairs

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Atoms are made up of protons, neutrons, and electrons.

Energy can be neither created nor destroyed.

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<h3>What are the correct pairing of the given terms?</h3>

The correct pairing of the given terms are obtained from the given options as follows:

Atoms of elements contains three sub-atomic particles which are protons, neutrons, and electrons.

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2 years ago

How many moles of ethanol are produced starting with 500.g glucose?

Monica [59]

<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

Left to Right<u> </u>

<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: $\displaystyle 500 \ g \ C_6H_{12}O_6(\frac{1 \ mol \ C_6H_{12}O_6}{180.18 \ g \ C_6H_{12}O_6})(\frac{2 \ mol \ C_2H_5OH}{1 \ mol \ C_6H_{12}O_6})$
[DA} Multiply/Divide [Cancel out units]: $\displaystyle 5.55001 \ mol \ C_2H_5OH$

<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

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3 years ago