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Rasek [7]
3 years ago
9

How well can you apply Charles’s law to this sample of gas that experiences changes in pressure and volume? Assume that pressure

and number of moles of gas are constant in this problem.
Chemistry
1 answer:
Phoenix [80]3 years ago
6 0

Answer:

As temperature increases the volume of given amount of gas increases while pressure and number of moles remain constant.

Explanation:

According to the charle's law,

The volume of given amount of gas is directly proportional to the temperature at constant pressure and number of moles of gas.

Mathematical expression:

V ∝ T

V = KT

V/T = K

When temperature changes from T₁ to T₂ and volume changes from V₁ to V₂.

V₁/T₁ = K        V₂/T₂ = K

or

V₁/T₁  = V₂/T₂

Thus, the ratio of volume and temperature remain constant for constant amount of gas at constant pressure.

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predict the product that forms in the synthesis reaction between potassium metal and iodine gas. provide the balanced chemical e
Doss [256]
K + I - > KI 
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5 0
4 years ago
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How many molecules of carbon dioxide are in 5.61 moles of carbon dioxide (CO2)?
gulaghasi [49]
<h3>Answer:</h3>

3.38 × 10²⁴ molecules CO₂

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

<u>Math</u>

Order of Operations: BPEMDAS

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

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 5.61 moles CO₂

[Solve] molecules CO₂

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                      \displaystyle 5.61 \ mooles \ CO_2(\frac{6.022 \cdot 10^{23} \ molecules \ CO_2}{1 \ mol \ CO_2})
  2. [DA] Multiply/Divide [Cancel out units]:                                                          \displaystyle 3.37834 \cdot 10^{24} \ molecules \ CO_2

<u>Step 4: Check</u>

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

3.37834 × 10²⁴ molecules CO₂ ≈ 3.38 × 10²⁴ molecules CO₂

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How many moles in 5 grams of gold
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