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Tom [10]
3 years ago
14

When the temperature of a sample of pure water is raised above 25 °C,(1)the hydronium ion concentration will be greater than the

hydroxide ion concentration.(2)the hydronium ion concentration will be less than the hydroxide ion concentration.(3)the value of Kw will increase.(4)the hydronium ion concentration could change to 1.0 × 10−10 M.(5)the hydroxide ion concentration could change to 1.0 × 10−10 M.
Chemistry
1 answer:
Ludmilka [50]3 years ago
5 0

Answer:

The correct answer is option 3.

Explanation:

H_2O\rightleftharpoons H^++OH^-

The expression of K_w is given by :

K_w=[H^+][OH^-]

K_w\propto [H^+]

K_w\propto [OH^-]

Rise in temperature will result in more dissociation of water molecules into hydrogen ions and hydroxide ions.

With an increase in concentration of hydrogen ions and hydroxide ions value of K_w will also increase.

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How many molecules are in 85g of silver nitrate?
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<h3>Answer:</h3>

3.0 × 10²³ molecules AgNO₃

<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
  • Writing Compounds
  • 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>

85 g AgNO₃ (silver nitrate)

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

Avogadro's Number

[PT] Molar Mass of Ag - 107.87 g/mol

[PT] Molar Mass of N - 14.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of AgNO₃ - 107.87 + 14.01 + 3(16.00) = 169.88 g/mol

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 85 \ g \ AgNO_3(\frac{1 \ mol \ AgNO_3}{169.88 \ g \ AgNO_3})(\frac{6.022 \cdot 10^{23} \ molecules \ AgNO_3}{1 \ mol \ AgNO_3})
  2. Multiply/Divide:                                                                                                \displaystyle 3.01313 \cdot 10^{23} \ molecules \ AgNO_3

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

3.01313 × 10²³ molecules AgNO₃ ≈ 3.0 × 10²³ molecules AgNO₃

6 0
2 years ago
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