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Alika [10]
3 years ago
7

Consider the titration of 10.00 mL of a monoprotic weak acid with 0.1234 M NaOH. If the equivalence point volume of NaOH was det

ermined to be 15.4 mL, what is the initial concentration of the unknown acid
Chemistry
1 answer:
gogolik [260]3 years ago
4 0

The initial concentration of the unknown acid is 0.1900 M.

Explanation:

Titration is a chemical method of analysis to know the concentration and volume of the unknown chemical or analyte.

The formula for the titration is:

Macid x Vacid = Mbase x V base

The volume must be in litres. The volume is given in ml it should be divided with 1000 to obtain values in litre.

Data given are:

volume of acid= 10 ml 0.01 L

Molarity of the acid = ?

volume of the NaOH or base = 15.4 ml or 0.0154 L (equivalence point of the base)

molarity of the base = 0.1234 M

Applying the formula and putting the values, we get

Macid x 0.01 = 0.1234 x 0.0154

Macid =  0.1900 M

The weak acid is having molarity of 0.1900 M against the strong base with molarity of 0.1234M.

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Carbon is made up of one type of atom.

Explanation:

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What is the fraction of the hydrogen atom's mass (11h) that is in the nucleus? the mass of proton is 1.007276 u, and the mass of
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Mass Fraction= \frac{mass 1 Proton}{mass H atom} = \frac{1,007276 u}{1,007825}=0,9995

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3 0
3 years ago
How many moles of ethanol are produced starting with 500.g glucose?
Monica [59]
<h3>Answer:</h3>

5.55 mol C₂H₅OH

<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 Tables
  • Moles

<u>Stoichiometry</u>

  • Using Dimensional Analysis
  • Analyzing Reactions RxN
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

[Given] 500. g C₆H₁₂O₆ (Glucose)

[Solve] moles C₂H₅OH (Ethanol)

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

[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH

[PT] Molar mass of C - 12.01 g/mol

[PT] Molar Mass of H - 1.01 g/mol

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

Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol

<u>Step 3: Stoichiometry</u>

  1. [DA] Set up conversion:                                                                                 \displaystyle 500 \ g \ C_6H_{12}O_6(\frac{1 \ mol \ C_6H_{12}O_6}{180.18 \ g \ C_6H_{12}O_6})(\frac{2 \ mol \ C_2H_5OH}{1 \ mol \ C_6H_{12}O_6})
  2. [DA} Multiply/Divide [Cancel out units]:                                                         \displaystyle 5.55001 \ mol \ C_2H_5OH

<u>Step 4: Check</u>

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

5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH

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