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Gre4nikov [31]
4 years ago
8

A 25.0 mL solution of a monoprotic weak acid is titrated with 0.10 M NaOH. The equivalence point is reached after 38.52 mL of Na

OH has been added. What was the initial concentration of the weak acid
Chemistry
1 answer:
Sveta_85 [38]4 years ago
8 0

Answer:

0.154 M

Explanation:

In case of titration , the following formula can be used -

M₁V₁ = M₂V₂

where ,

M₁ = concentration of acid ,

V₁ = volume of acid ,

M₂ = concentration of base,

V₂ = volume of base .

from , the question ,

M₁ = ? M

V₁ = 25.0mL

M₂ = 0.10 M

V₂ = 38.52 mL

Using the above formula , the molarity of acid , can be calculated as ,

M₁V₁ = M₂V₂  

Putting the respective values ,

M₁ * 25.0mL =  0.10 M * 38.52 mL

M₁ = 0.154 M

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Indicate whether each of the statements below is true or false. Match the words in the left column to the appropriate blanks in
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Answer:

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4) CBr₄ is more volatile than CCl₄: False

Explanation:

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3 years ago
What is the mass of 8.56 x 10^23 formula units of BaBr2? (3 sig figs in your answer) ​
cupoosta [38]
<h3>Answer:</h3>

296 g BaBr₂

<h3>General Formulas and Concepts:<u> </u></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
  • Moles
  • 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] 8.56 × 10²³ formula units BaBr₂

[Solve] grams BaBr₂

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

Avogadro's Number

[PT] Molar Mass of Ba - 137.33 g/mol

[PT] Molar Mass of Br - 35.45 g/mol

Molar Mass of BaBr₂ - 137.33 + 2(35.45) = 208.23 g/mol

<u>Step 3: Convert</u>

  1. [DA} Set up:                                                                                                     \displaystyle 8.56 \cdot 10^{23} \ formula \ units \ BaBr_2(\frac{1 \ mol \ BaBr_2}{6.022 \cdot 10^{23} \ formula \ units \ BaBr_2})(\frac{208.23 \ g \ BaBr_2}{1 \ mol \ BaBr_2})
  2. [DA] Multiply/Divide [Cancel out units]:                                                         \displaystyle 295.99 \ g \ BaBr_2

<u>Step 4: Check</u>

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

295.99 g BaBr₂ ≈ 296 g BaBr₂

Thank you agenthammerx for helping me with this question!

6 0
3 years ago
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