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Oksanka [162]
2 years ago
9

Why wasn’t sodium hydroxide used in the synthesis of divanillyl oxalate, instead of triethylamine?

Chemistry
2 answers:
kupik [55]2 years ago
4 0

Sodium hydroxide wasn't used in the synthesis of divanillyl oxalate because it's a strong base.

<h3>What is sodium hydroxide?</h3>

It should be noted that sodium hydroxide simply means a corrosive cystaline solid which readily absorbs moisture until it dissolves.

In this case, sodium hydroxide wasn't used in the synthesis of divanillyl oxalate because it's a strong base and will decrease the yield of the product.

Learn more about sodium hydroxide on:

brainly.com/question/27072350

ValentinkaMS [17]2 years ago
4 0

Divanillyl oxalate is a chemical that is produced by triethylamine. Sodium hydroxide is not used in its synthesis as it is a strong base.

<h3>What is sodium hydroxide?</h3>

Sodium hydroxide or caustic soda is an ionic compound with ions of sodium and hydroxide. The formula is shown as, \rm NaOH. It is crystalline and absorbs moisture from the surrounding.

It is not used in the synthesis of divanillyl oxalate as it is a strong base with pH 14 that affects the yield of the product in the reaction.

Therefore, sodium hydroxide is not used in the synthesis reaction.

Learn more about sodium hydroxide here:

brainly.com/question/15456619

#SPJ4

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Answer:

  • The molarity of the student's sodium hydroxide solution is 0.0219 M

Explanation:

<u>1) Chemical reaction.</u>

a) Kind of reaction: neutralization

b) General form: acid + base → salt + water

c) Word equation:

  • sodium hydroxide + oxalic acid → sodium oxalate + water

d) Chemical equation:

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b) Balanced chemical equation:

  • 2NaOH + H₂C₂O₄ → Na₂C₂O₄ + 2H₂O

<u>2) Mole ratio</u>

  • 2mol Na OH : 1 mol H₂C₂O₄ :1 mol Na₂C₂O₄ : 2 mol H₂O

<u>3) Starting amount of oxalic acid</u>

  • mass = 28 mg = 0.028 g
  • molar mass = 90.03 g/mol
  • Convert mass in grams to number of moles, n:

        n = mass in grams / molar mass = 0.028 g / 90.03 g/mol =  0.000311 mol

<u>4) Titration</u>

  • Volume of base: 28.4 mL = 0.0248 liter
  • Concentration of base: x (unknwon)

  • Number of moles of acid: 2.52 mol (calculated above)
  • Proportion using the theoretical mole ratio (2mol Na OH : 1 mol H₂C₂O₄)

\frac{2}{1} =\frac{x}{2.52}\\ \\ \\x=0.000311(2)=0.000622

That means that there are 0.000622 moles of NaOH (solute)

<u>5) Molarity of NaOH solution</u>

  • M = n / V (liter) = 0.000622 mol / 0.0284 liter = 0.0219 M

That is the correct number using <em>three signficant figures</em>, such as the starting data are reported.

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