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Leto [7]
2 years ago
12

Why do you add hydroxide to your hexanediamine solution? what would occur if you did not add it?.

Chemistry
1 answer:
MA_775_DIABLO [31]2 years ago
5 0

In order to react with the hydrochloric acid (HCl) created when the hexanediamine combines with the adiptic acid, NaOH was added to the hexanediamine solution. By combining with the HCl to create NaCl and water, the NaOH preserves the pH of the solution. HCl is taken out of the solution and the reaction is pushed in the direction of the products.

<h3>Hexanediamine solution</h3>

A C6 alkane-alpha, omega-diamine is hexanediamine. It serves as a xenobiotic metabolite for humans. It comes from a hydride.

  • Weight in Formula: 116.21
  • Indication: 1,6-NH2(CH2)6NH2
  • Risk Information: Corrosive
  • Density: 0.854 (g/mL).
  • Heat of Boiling (°C): 205
  • Temp. Freeze (°C): 39–42
  • Water and Alcohol Are Both Slightly Solubilized
  • Synonyms include 1,6-diaminohexane and hexamethyleneediamine.
  • Shelf Life: 36 months

What is the purpose of hexanediamine?

  • Hexanediamine is mostly used to make adhesive resins, resins for paper manufacturing, resins for water treatment, and unaltered resins.
  • Hexanediamine is only used in industrial products; it is unlikely that consumers will come into touch with it.

learn more about hexanediamine solution here

<u>brainly.com/question/17112756</u>

#SPJ4

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In a lab experiment 80.0 g of ammonia [NH3] and 120 g of oxygen are placed in a reaction vessel. At the end of the reaction 72.2
valentinak56 [21]

The percent yield of the reaction : 89.14%

<h3>Further explanation</h3>

Reaction of Ammonia and Oxygen in a lab :

<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>

mass NH₃ = 80 g

mol NH₃ (MW=17 g/mol):

\dfrac{80}{17}=4.706

mass O₂ = 120 g

mol O₂(MW=32 g/mol) :

\tt \dfrac{120}{32}=3.75

Mol ratio of reactants(to find limiting reatants) :

\tt \dfrac{4.706}{4}\div \dfrac{3.75}{5}=1.1765\div 0.75\rightarrow O_2~limiting~reactant(smaller~ratio)

mol of H₂O based on O₂ as limiting reactants :

mol H₂O :

\tt \dfrac{6}{5}\times 3.75=4.5

mass H₂O :

4.5 x 18 g/mol = 81 g

The percent yield :

\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{72.2}{81}\times 100\%=89.14\%

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I believe this question has the following five choices to choose from:

 

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The correct answer is:

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