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2 years ago

If a gas has a volume of 15l and a temperature of 125K, and the temperature is increased to 225K, what is the new volume

Chemistry

1 answer:

aalyn [17]2 years ago

6 0

Answer:

V₂ = 27 L

Explanation:

Given data:

Initial volume = 15 L

Initial temperature = 125 K

Final temperature = 225 K

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 15 L × 225K / 125 k

V₂ = 3375 L.K / 125 K

V₂ = 27 L

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<h3>Answer:</h3>

0.0113 mol Ba(ClO₃)₂

<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>Chemistry</u>

<u>Atomic Structures</u>

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.45 g Ba(ClO₃)₂

<u>Step 2: Identify Conversions</u>

Molar Mass of Ba - 137.33 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Ba(ClO₃)₂ - 137.33 + 2(35.45) + 6(16.00) = 304.33 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 3.45 \ g \ Ba(ClO_3)_2(\frac{1 \ mol \ Ba(ClO_3)_2}{304.33 \ g \ Ba(ClO_3)_2})$
Multiply/Divide: $\displaystyle 0.011336 \ mol \ Ba(ClO_3)_2$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

0.011336 mol Ba(ClO₃)₂ ≈ 0.0113 mol Ba(ClO₃)₂

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