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Marrrta [24]
3 years ago
7

Was the acetylacetone miscible with water? What about if NH3 is added to the solution? Why?

Chemistry
1 answer:
luda_lava [24]3 years ago
4 0

Answer:

See explanation

Explanation:

The compound is acetylacetone. This compound is immiscible with water because it is completely nonpolar.

However the introduction of ammonia, leads to the ionization of the specie. In the presence of a base(such as NH3), the molecule acetylacetone is deprotonated and its corresponding anion is formed. This anion can now interact with water and become miscible with it.

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<h3>Answer:</h3>

0.424 J/g °C

<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  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

<u>Chemistry</u>

<u>Thermochemistry</u>

Specific Heat Formula: q = mcΔT

  • q is heat (in Joules)
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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] m = 38.8 g

[Given] q = 181 J

[Given] ΔT = 36.0 °C - 25.0 °C = 11.0 °C

[Solve] c

<u>Step 2: Solve for Specific Heat</u>

  1. Substitute in variables [Specific Heat Formula]:                                             181 J = (38.8 g)c(11.0 °C)
  2. Multiply:                                                                                                             181 J = (426.8 g °C)c
  3. [Division Property of Equality] Isolate <em>c</em>:                                                         0.424086 J/g °C = c
  4. Rewrite:                                                                                                             c = 0.424086 J/g °C

<u>Step 3: Check</u>

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

0.424086 J/g °C ≈ 0.424 J/g °C

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