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Mrrafil [7]
3 years ago
5

When a small amount of 12 M HNO3(aq) is added to a buffer solution made by mixing CH3NH2(aq) and CH3NH3Cl(aq) , the pH of the bu

ffer solution changes from 10.64 to 10.62. Which of the following equations represents the reaction that accounts for the fact that the pH does not change significantly when the HNO3(aq) is added?
a. CH3NH2(aq) + H+(aq) â CH3NH3+(aq)

b. CH3NH3+(aq) + H+(aq) â CH3NH42+(aq)

c. NO3- (aq) + H+(aq) â HNO3(aq)

d. OH- (aq) + H+(aq) â H2O(l)
Chemistry
1 answer:
fredd [130]3 years ago
5 0

Answer:

a. CH3NH2(aq) + H⁺ → CH3NH3⁺

Explanation:

The mixture of a weak base as CH3NH2 with its conjugate acid CH3NH3Cl produce a buffer. As the weak acid is in equilibrium with water, the mixture of the weak base and its conjugate base produce that the acid or base released react avoiding the change in pH.

For example, when a strong acid as HNO3 reacts, the weak base will react producing the conjugate base, that is:

CH3NH2(aq) + H⁺ → CH3NH3⁺

Right answer is:

<h3>a. CH3NH2(aq) + H⁺ → CH3NH3⁺</h3>

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Answer:

the correct option would be:

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7 0
3 years ago
What is the volume of 40.0 grams of argon gas at STP ?
MrRa [10]

Answer:

24.9 L Ar

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles
  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

<u>Aqueous Solutions</u>

  • States of Matter

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 40.0 g Ar

[Solve] L Ar

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

[PT] Molar Mass of Ar - 39.95 g/mol

[STP] 22.4 L = 1 mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 40.0 \ g \ Ar(\frac{1 \ mol \ Ar}{39.95 \ g \ Ar})(\frac{22.4 \ L \ Ar}{1 \ mol \ Ar})
  2. [DA] Divide/Multiply [Cancel out units]:                                                         \displaystyle 24.9235 \ L \ Ar

<u>Step 4: Check</u>

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

24.9235 L Ar ≈ 24.9 L Ar

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