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kotegsom [21]
3 years ago
12

The ester below can be separated into phenol and an acetate salt, via saponification: heating the ester with a strong base, such

as sodium hydroxide, will produce phenol and sodium acetate. the two chemicals can then be separated by heat and filtration.
a. calculate the unsaturation number of the ester above.
b. calculate the molecular weight of the ester above

Chemistry
1 answer:
erastovalidia [21]3 years ago
4 0
I found a similar question online which will help me answer your incomplete question. To make it easier, show all the elements of the compound given. It is shown in the second picture attached.

a.) The formula for unsaturation number is shown in the 3rd picture attached. Following this, 
n = 8
m = 2(8) + 2 + 0 + 0 - 0 = 18
Thus,
x = Unsaturation number = (18 - 8)/2 = 5
<em>The unsaturation number is 5.</em>

b.) The molar mass of C is 12.01 g/mol; H is 1 g/mol; O is 16 g/mol. So, the molecular weight is:
Molecular weight = 12.01(8) + 8(1) + 2(16) = 136.08 g/mol
<em>The molecular weight of the ester is 136.08 g/mol.</em>

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20.
sergiy2304 [10]
I think the anwser is A
8 0
3 years ago
Read 2 more answers
Can someone help me?
kupik [55]

Answer:

#1)2.23hrs

Explanation:

= 134/60

=2.23

=2.23hrs

4 0
3 years ago
7. Which of the following compound is the most soluble in CC14? C. NH3 A. HF B. NaCl D. C10H22​
Mashutka [201]

C10H22 is a compound which is the most soluble in CC14 because both are non-polar in nature.

<h3>Why polar solute soluble in polar solvent?</h3>

We know that like dissolve like which means that polar solutes only  dissolve in polar solvent while on the other hand, non-polar solutes only dissolve in non-polar solvent.

So we can conclude that C10H22 is a compound which is the most soluble in CC14 because both are non-polar in nature.

Learn more about soluble here: brainly.com/question/23946616

#SPJ1

4 0
2 years ago
19 grams of sodium and 34 grams of chlorine are measured out and put into separate glass vials that weigh 10 grams each. Then so
vaieri [72.5K]

Answer:

63.02 g.

Explanation:

  • Na reacts with Cl₂ according to the balanced equation:

<em>2Na + Cl₂ → 2NaCl,</em>

It is clear that 2 mole of Na react with 1 mole of Cl₂ to  produce 2 moles of NaCl.

  • Firstly, we need to calculate the no. of moles of Na and Cl₂:

no. of moles of Na = (mass/atomic mass) = (19.0 g/22.9897 g/mol) = 0.826 g.

no. of moles of Cl₂ = (mass/atomic mass) = (34.0 g/70.906 g/mol) = 0.48 g.

  • From the stichiometry, Na reacts with Cl₂ with a molar ratio (2:1).

<em>So, 0.826 mol of Na "the limiting reactant" reacts completely with 0.413 mol of Cl₂ "left over reactant".</em>

The no. of moles of Cl₂ remained after the reaction = 0.48 mol - 0.413 mol = 0.067 mol.

∴ The mass of Cl₂ remained after the reaction = (no. of moles of Cl₂ remained after the reaction)(molar mass of Cl₂) = (0.067 mol)(70.906 g/mol) = 4.75 g.

  • To get the no. of grams of produced NaCl:

<u><em>using cross multiplication:</em></u>

2 mol of Na produce  → 2 mol of NaCl, from the stichiometry.

∴ 0.826 mol of Na produce  → 0.826 mol of NaCl.

∴ The mass of NaCl produced after the reaction = (no. of moles of NaCl)(molar mass of NaCl) = (0.826 mol)(58.44 g/mol) = 48.27 g.

∴ The total weight of the glass vial containing the final product = the weight of the glass vial + the weight of the remaining Cl₂ + the weight of the produced NaCl = 10.0 g + 4.75 g + 48.27 g = 63.02 g.

7 0
3 years ago
Read 2 more answers
Look at the following chemical reaction and determine what is true about Mg. Click 2 that apply.
Anna007 [38]

Answer:

The answer to your question is b. Mg is being oxidized and c. Mg is the

                                                    reducing agent.

Explanation:

Reduction is the gain of electrons

Oxidation is the loss of electrons

Reducing agent is the element that oxidizes itself and reduces another element.

Oxidizing agent is the element the reduces itself and oxidizes another element.

 In this reaction:

Magnesium loses electrons so it is being oxidized and it is the reducing agent.              

4 0
3 years ago
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