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mylen [45]
3 years ago
11

Match the term with the definition. Match Term Definition Condensation A) A phase change from a gas to a liquid Melting B) A pha

se change from a liquid to a solid Sublimation C) A phase change from a solid to a liquid Freezing D) A phase change from a solid to a gas
Chemistry
1 answer:
Inga [223]3 years ago
3 0

OK, so since this question is corrupt, lets fix this.

Answer:

1. Sublimation: C

2: Melting: B

3: Freezing: D

4: Evaporation: A

Explanation:

In the lesson, we learned that Sublimation is like a solid turning into gas (vapor). For melting, we learned that it is a solid that turns into a liquid, example of melting is snow turning into water. For Freezing, we learned that it is a liquid turning into a solid, example of freezing is water turning into snow. For evaporation, it is a solid turning into a gas like vapor for example, example of evaporation: We know that when we mix Dry Ice with water, we get a chemical reaction, Dry Ice with water creates carbon dioxide smoke (fog).

Hope this answer helps!

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A small bubble rises from the bottom of a lake where the temperature and pressure are 4.0 C and 3.0 atm, to the water’s surface,
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Answer:

The final volume of the bubble is 7.13 mL.

Explanation:

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 3 atm

P_2 = final pressure of gas = 0.95 atm

V_1 = initial volume of gas = 2.1 mL=0.0021 L

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 4^oC=273.15+4K=277.15 K

T_2 = final temperature of gas = 25^oC=273.15+25 k=298 .15 K

Now put all the given values in the above equation, we get:

\frac{3 atm\times 0.0021 L}{277.15 K}=\frac{0.95 atm\times V_2}{298.15 K}

V_2=\frac{3 atm\times 0.0021 L\times 298.15 K}{277.15 K\times 0.95 atm}=0.00713 L = 7.13 mL

The final volume of the bubble is 7.13 mL.

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Which of the following conditions explains Floridas risk of muck fires​
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Explanation:

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The compound XCl is classified as ionic if X represents the element:
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If the element is a metal with one electron in its outermost energy level.
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Draw the structure of the compound C9H10O2 that might exhibit the 13C-NMR spectrum below. Impurity peaks are omitted from the pe
zhenek [66]

Complete question

Draw the structure of the compound C_{9}H_{10}O_{2} that exhibits the ^{13}C-NMR spectrum shown on the first uploaded image(on the second and third uploaded image is closer look at the ^{13}C-NMR spectrum ) . Impurity peaks are omitted from the peak list. The triplet at 77 ppm is CDC_{l3}.

Answer:

The structure that might exhibit the ^{13}C-NMR  spectrum is shown on the fifth uploaded image

Explanation:

    In order to get a good understanding of the answer above we need to know that

• Proton NMR spectrum: proton NMR spectroscopy is one of the techniques, which is useful to predict the structure of the compound.

• In ^{\rm{1}}{\rm{H NMR}}  spectroscopy, peaks are observed at the point where the wavelength of proton nuclei matched to substance nuclei wavelength.

• In same manner there are other spectroscopies are present like ^{{\rm{13}}}{\rm{C NMR}}

, IR and mass spectroscopy.

• Infrared spectroscopy is used to determine the functional groups present in a compound.

• Infrared bands observed when there is change in dipole moment occurs between the atoms. Infrared bands describe about the bond stretches, which causes due to the dipole moment present in the molecule.

Fundamentals

Double bond equivalence: number of double bonds or number of rings in the structure can be calculated by using double bond equivalence formula.

DBE = N_{c} + 1 - (\frac{N_{H}+N_{Cl}-N_{N}}{2}})

Where,

N_{c} = number of carbon atoms

N_{H}= number of hydrogen atoms

N_{Cl} = number of chlorine atoms

N_{N}=number of nitrogen atoms

The table for the ^{{\rm{13}}}{\rm{C NMR}} is shown on the fourth uploaded image

Molecular formula of the compound is {{\rm{C}}_9}{{\rm{H}}_{{\rm{10}}}}{{\rm{O}}_{\rm{2}}}

Double bond equivalence of the compound is calculated below.

  DBE = N_{c} + 1 - (\frac{N_{H}+N_{Cl}-N_{N}}{2}})

Where,

N_{c} = 9

N_{H}= 10

N_{Cl} = 0

N_{N}= 0

                    DBE = N_{c} + 1 - (\frac{(10+0) -0}{2}})

                    DBE =5

Therefore, the compound has five double bonds, which indicating that there is chance of getting aromatic rings too.

Note:

Double bond equivalence is calculated as 5 which indicates that there are 5 double bond (may rings) in the structure of the compound.

Double bond equivalence is calculated by using this formula.

           DBE = N_{c} + 1 - (\frac{N_{H}+N_{Cl}-N_{N}}{2}})

13C NMR data of the compound is explained below.

1.A peak at 166.5 ppm, which indicates the presence of ester group

2.Peaks at 132.7, 130.5, 129.5, 128.2 ppm (aromatic carbons) are indicating a mono substituted aromatic ring

3.A peak at 60.9 ppm means methylene group attached to oxygen atom

4.A peak at 14.3 ppm, which indicates the presence of methyl group

According to this data and the using the double bond equivalence, structure of the compound shown on the fifth uploaded image .

Note:

According to given spectral data, structure of the compound has been predicted. It is clear that; -ester functional group is present in the structure because there is a peak at 166.5ppm. According to given proton ^{13}C NMR data, above structure has been drawn. Therefore, the compound is ethyl benzoate.

7 0
4 years ago
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