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nikitadnepr [17]
3 years ago
13

5. Consider the following experiment: A block of iron with an initial mass of 50 g is placed in a sealed container of pure oxyge

n. After several days, the block is red in color and has a mass of 74 g. Are these observations a violation of the law of conservation of mass? Justify your answer.
Chemistry
1 answer:
Vera_Pavlovna [14]3 years ago
8 0

The observation in the illustration remains in accordance with the law of conservation of mass assuming that moisture is present in the container.

<h3>What is the law of conservation of mass?</h3>

The law state that mass can neither be created nor destroyed. However, mass can be converted from one form to another.

In this case, the ion was placed in a container of pure oxygen. Assuming that moisture is present, iron reacts with oxygen to form rust according to the following equation:

4Fe + 3O_2 + 6H_2O --- > 4Fe(OH)_3

The rust formed will weigh more than the initial weight of the iron.

Without the presence of moisture, one can effectively conclude that the observation violates the law of conservation of mass.

More on the law of conservation of mass can be found here: brainly.com/question/13383562

#SPJ1

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How many moles are in a 13.7g block of calcium?
goldenfox [79]
<h3>Answer:</h3>

0.342 mol Ca

<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> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

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

<u>Step 1: Define</u>

13.7 g Ca

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

Molar Mass of Ca - 40.08 g/mol

<u>Step 3: Convert</u>

  1. Set up:                               \displaystyle 13.7 \ g \ Ca(\frac{1 \ mol \ Ca}{40.08 \ g \ Ca})
  2. Multiply/Divide:                 \displaystyle 0.341816 \ mol \ Ca

<u>Step 4: Check</u>

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

0.341816 mol Ca ≈ 0.342 mol Ca

6 0
3 years ago
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