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lidiya [134]
4 years ago
10

After a rain a puddle of water remains on a sidewalk after a day o sunshine the puddle is gone which process is most responsible

for the disappearance of the puddle
Chemistry
1 answer:
zaharov [31]4 years ago
8 0
Evaporation or boiling
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When 2-heptyne was treated with aqueous sulfuric acid containing mercury(II) sulfate, two products, each having the molecular fo
xeze [42]

Answer:

See explanation below

Explanation:

This is a reaction of alkyne hydration. We have aqueous sulfuric acid in HgSO4. This reagent is used to convert a alkyne into a ketone. The mechanism of reaction is taking place in 4 steps.

The first step is an electrophilic adition where the atom of Hydrogen that the acid provides, is added to the alkyne in the triple bond.

In this step, as Carbon 2 and Carbon 3, do not have any atom of hydrogen, the adition of this hydrogen could take place in either carbon 2 or carbon 3, that's why at the end you can have two products in equal amounts.

The second step is the adition of one molecule of water in the carbocation that appears in the previous step.

The third step, is an acid base equilibrium, where the hydrogen atom from the water that added to the heptene, is substracted (In this case by the conjugate base of the acid).

In this way, we will have an alcohol with a double bond. This is the enolic form of the molecule.

The final step is a tautomeric equilibrium, where the double bond of the C = C will pass to a double bond C = O forming a carbonile and therefore, a ketone.

If this was done in carbon 2 from step 1, you'll have the 2-heptanone. If it were on carbon 3, it will be the 3 - Heptanone.

Picture attached has the mechanism and products.

3 0
3 years ago
An unknown compound contains only C, H, and O. Combustion of 8.50 g of this compound produced 20.0 g CO2 and 5.46 g H2O. What is
Galina-37 [17]

Answer:

The answer to your question is: C₃H₄O

Explanation:

Data

CxHyOz = 8.5 g

CO2 = 20 g

H2O = 5.46 g

Reaction

             CxHyOz + O2     ⇒    CO2   +    H2O

CO2

    MW = 44g

                       44g CO2 ----------------- 12g of C

                        20g CO2 ----------------   x

                       x = (20 x 12) / 44

                       x = 5.45 g of C

                     # of moles = n = 5.45 / 12 = 0.454 mol of C

H2O

     MW = 18 g

                       18 g H2O ------------------- 2g of H

                       5.46 g    --------------------   x

                       x = (5.46 x 2) / 18 = 0.61 g of H

                       n = 0.61 / 1 = 0.61 moles of H2

Mass of O2

              mass CxHyOz = mass CO2 + mass H2 + mass O2

              mass O2 = 8.5 - 5.45 - 0.61

              mass O2 = 2.44g

              n = 2.44 / 16 = 0.153 mol of O2

Now, divide by the lowest number of moles

0.454 mol of C/ 0.153 = 2.97 ≈ 3

0.61 moles of H2/ 0.153 = 3.99 ≈ 4

0.153 mol of O2/ 0.153 =  1

Then, the empirical formula is: C₃H₄O

7 0
3 years ago
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