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ValentinkaMS [17]
2 years ago
5

Explain the volume of ideal gasses as related to the molar concept.

Chemistry
1 answer:
love history [14]2 years ago
5 0

Answer:

n = Initial volume/22.4L

Explanation:

The molar concept is simply one that is used to find the Number of moles and explain the relationship it has with avogadro's number, molecular mass, molar mass e.t.c.

Now, in terms of molar mass, number of moles is given by the formula;

n = mass of the sample/molar mass

In terms of avogadro's number, number of moles is;

1 mole = avogadro's number = 6.02 × 10^(23)

Now, when dealing with ideal gases, the molar volume of an ideal gas is 22.4 L.

Now the relationship between this volume and the mole concept is that the number of moles is gotten by dividing the initial volume by this molar volume.

Thus;

n = Initial volume/22.4L

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1 Na2CO3(aq) + 1 CaCl2(aq) → 1 CaCO3(s) + 2 NaCl(aq) 4. Use the balanced chemical equation from the last question to solve this
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<h3>Answer:</h3>

0.6 g NaCl

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

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<u>Atomic Structure</u>

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<u>Stoichiometry</u>

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] Na₂CO₃ (aq) + CaCl₂ (aq) → CaCO₃ (s) + 2NaCl (aq)

[Given] 0.5 g Na₂CO₃ reacted with excess CaCl₂

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

[RxN] Na₂CO₃ → 2NaCl

Molar Mass of Na - 22.99 g/mol

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of Na₂CO₃ - 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol

Molar Mass of NaCl - 22.99 + 35.45 = 58.44 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up:                    \displaystyle 0.5 \ g \ Na_2CO_3(\frac{1 \ mol \ Na_2CO_3}{105.99 \ g \ Na_2CO_3})(\frac{2 \ mol \ NaCl}{1 \ mol \ Na_2CO_3})(\frac{58.44 \ g \ NaCl}{1 \ mol \ NaCl})
  2. Multiply/Divide:                                                                                               \displaystyle 0.551373 \ g \ NaCl

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

0.551373 g NaCl ≈ 0.6 g NaCl

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