Question 4 (2 points)

For the following gases, correct the volumes to the conditions indicated. A.654 cm3 at 6 degrees Celsius and 65.3 kPa to 4 degre

leva [86]

Initial pressure of the gas = 65.3 kPa

Initial volume of the gas = 654 cm³

Initial temperature of the gas = 6⁰C = 273 + 6 = 279 K

Final pressure of the gas = 108.7 kPa

Final temperature of the gas = 4⁰C = 273 + 4 = 277 K

Using the combined gas law for ideal gases:

P₁V₁/T₁ = P₂V₂/T₂

where P₁, V₁ and T₁ are the pressure, volume and temperature for the initial state and P₂, V₂ and T₂ are the pressure, volume and temperature for the final state.

Plugging the given data into the combined gas law we have,

(65.3 kPa x 654 cm³) / (279 K) = (108.7 kPa x V₂)/(277 K)

V₂ = (65.3 kPa x 654 cm³ x 277 K) / (279 K x 108.7 kPa)

V₂ = 390.1 cm³

8 0

3 years ago

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

Br2 + 2LiF → 2LiBr + F2

N76 [4]

1 grams F2 to mol = 0.02632 mol

10 grams F2 to mol = 0.26318 mol

20 grams F2 to mol = 0.52636 mol

30 grams F2 to mol = 0.78954 mol

40 grams F2 to mol = 1.05272 mol

50 grams F2 to mol = 1.3159 mol

100 grams F2 to mol = 2.6318 mol

200 grams F2 to mol = 5.2636 mol

6 0

3 years ago

All members of the alkene series of hydrocarbons have the general formula:

LuckyWell [14K]

Answer:

D: CnH2n

Explanation:

Unsaturated hydrocarbons have double and/or triple bonds between carbon atoms. Those with double bond assuming they have non-cyclic structures are called alkenes with the general formula C_nH_2n. Meanwhile the hydrocarbons that possess triple bonds are called alkynes and possess the general formula CnH2n-2.

Looking at the options, for alkenes, the correct answer is option D

7 0

3 years ago

What are sister chromatids

Pavel [41]

Sister chromatids are two identical copies of a single chromosome that are connected by a centromere. They occur as a result of a chromosome that duplicated during the S phase of the cell cycle.

4 0

3 years ago