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padilas [110]
2 years ago
14

3. Why do gases deviate from the ideal gas law at high pressures? A. Molecules have finite volume. B. Cohesive forces increase t

he volume from the ideal. C. Increasing pressure increases the temperature of the gas. D. Collisions between molecules occur more frequently as pressure increases.
Chemistry
1 answer:
ipn [44]2 years ago
3 0

Answer:

A. Molecules have finite volume.

Explanation:

Gases deviate from the ideal gas law at high pressures because its molecules have a finite volume.

Real gases have a finite volume which enables more interaction between the molecules while ideal gases are assumed not to have a finite volume or occupy space which is why it lacks any form of interaction between its molecules.

This difference is the deviation between the real and ideal gases.

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How many atoms doesCo(NO3)2 have
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2AlF3 + 3K2O → 6KF + Al2O3<br><br> How many grams of AlF3would it take to make 15.524 g of KF?
mr_godi [17]
<h3>Answer:</h3>

7.4797 g AlF₃

<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

<u>Chemistry</u>

<u>Stoichiometry</u>

  • Reading a Periodic Table
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN] 2AlF₃ + 3K₂O → 6KF + Al₂O₃

[Given] 15.524 g KF

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

[RxN] 2 mol AlF₃ = 6 mol KF

Molar Mass of Al - 26.98 g/mol

Molar Mass of F - 19.00 g/mol

Molar Mass of K - 39.10 g/mol

Molar Mass of AlF₃ - 26.98 + 3(19.00) = 83.98 g/mol

Molar Mass of KF - 39.10 + 19.00 = 58.10 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up:                              \displaystyle 15.524 \ g \ KF(\frac{1 \ mol \ KF}{58.10 \ g \ KF})(\frac{2 \ mol \ AlF_3}{6 \ mol \ KF})(\frac{83.98 \ g \ AlF_3}{1 \ mol \ AlF_3})
  2. Multiply/Divide:                                                                                                    \displaystyle 7.47966 \ g \ AlF_3

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 5 sig figs.</em>

7.47966 g AlF₃ ≈ 7.4797 g AlF₃

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