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3 years ago

A gas has a volume of 4 liters at 50 oC. What will its volume be (in liters) at 100oC

Chemistry

1 answer:

san4es73 [151]3 years ago

5 0

Answer:

The volume of the gas at 100°C is 4.6189 liters.

Explanation:

For this problem we are going to use Charles' law. Charles' law states that the volume is directly proportional to temperature given that the pressure is constant. In order to use the equation, the unit of temperature should be in Kelvin.

The working equation is:

where V1 and T1 are the initial volume and temperature while V2 and T2 are the final conditions.

Let us convert first the temperatures before solving for the final volume.

To convert Celsius to Kelvin just add 273.15 to temperature in Celsius.

50°C + 273.15 = 323.15 K

100°C + 273.15 = 373.15 K

Solving for the final volume:

V₂ =

V₂ = 4.6189 L

Therefore the final volume of the gas at 100°C is 4.6189 L.

Explanation:

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<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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