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Marina CMI [18]
2 years ago
15

Jon does an experiment in which he varies the amount of calcium chloride (CaCl2) powder used for the chemical reaction with sodi

um bicarbonate (NaHCO3). He plans to observe whether the amount of CaCl2 powder has an effect on reaction rate.
Make the most logical prediction for the results of his experiment using the drop-down menus.

I predict that using a larger amount of calcium chloride powder will
______
a. increase
b. decrease
c. have no effect on

the reaction rate because the calcium chloride molecules will ______
a. have more kinetic energy
b. balance the number of product molecules
c. collide more often with other reactant molecules
Chemistry
1 answer:
asambeis [7]2 years ago
8 0

Answer:

a.) increase

c.) collide more often with other reactant molecules

Explanation:

Generally, when you increase the quantity of reactants, the rate of the reaction increases (the forward reaction). When there are more reactants, there are more chances for them to interact with each other. More interactions means more reactions.

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675 g of carbon tetrabromide is equivalent to how many
VARVARA [1.3K]
<h3>Answer:</h3>

2.04 mol CBr₄

<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>Organic</u>

  • Writing Organic Compounds
  • Writing Covalent Compounds
  • Organic Prefixes

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

675 g CBr₄

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

Molar Mass of C - 12.01 g/mol

Molar Mass of Br - 79.90 g/mol

Molar Mass of CBr₄ - 12.01 + 4(79.90) = 331.61 g/mol

<u>Step 3: Convert</u>

<u />\displaystyle 675 \ g \ CBr_4(\frac{1 \ mol \ CBr_4}{331.61 \ g \ CBr_4}) = 2.03552 \ mol \ CBr_4<u />

<u />

<u>Step 4: Check</u>

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

2.03552 mol CBr₄ ≈ 2.04 mol CBr₄

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