Two atoms that have the same atomic numbers but different atomic mass numbers are called

Nitric oxide (NO) reacts readily with chlorine gas as follows.2 NO(g) + Cl2(g) equilibrium reaction arrow 2 NOCl(g)At 700. K the

Yanka [14]

Answer:

For a: The mixture will need to produce more reactants to reach equilibrium.

For b: The mixture will need to produce more reactants to reach equilibrium.

For c: The mixture will need to produce more products to reach equilibrium.

Explanation:

For the given chemical equation:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

The expression of $K_{p}$ for above equation follows:

$K_{p}=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$ .....(1)

We are given:

Value of $K_p$ = 0.26

There are 3 conditions:

When $K_{p}>Q_p$ ; the reaction is product favored.
When $K_{p}$ ; the reaction is reactant favored.
When $K_{p}=Q_p$ ; the reaction is in equilibrium.

For the given options:

For a:

We are given:

$p_{NOCl}=0.11atm\\p_{NO}=0.16atm\\p_{Cl_2}=0.30atm$

Putting values in expression 1, we get:

$Q_p=\frac{(0.11)^2}{(0.16)^2\times 0.30}=1.57$

As, $K_{p}$ ; the reaction is reactant favored

Hence, the mixture will need to produce more reactants to reach equilibrium.

For b:

We are given:

$p_{NOCl}=0.048atm\\p_{NO}=0.12atm\\p_{Cl_2}=0.10atm$

Putting values in expression 1, we get:

$Q_p=\frac{(0.048)^2}{(0.12)^2\times 0.10}=1.6$

As, $K_{p}$ ; the reaction is reactant favored

Hence, the mixture will need to produce more reactants to reach equilibrium.

For c:

We are given:

$p_{NOCl}=5.20\times 10^{-3}atm\\p_{NO}=0.15atm\\p_{Cl_2}=0.15atm$

Putting values in expression 1, we get:

$Q_p=\frac{(5.20\times 10^{-3})^2}{(0.15)^2\times 0.15}=0.008$

As, $K_{p}>Q_p$ ; the reaction is product favored.

Hence, the mixture will need to produce more products to reach equilibrium.

8 0

3 years ago

If the enantiomeric excess of a mixture is 75%, what are the % compositions of the major and minor enantiomer?

Maru [420]

Let us say that R is the major enantiomer, while S is the minor enantiomer, therefore the formula for enantiomeric excess (ee) is:

ee = (R – S) * 100%

Let us further say that the fraction of R is x (R = x), and therefore fraction of S is 1 – x (S = 1 – x), therefore:

75 = (x – (1 – x)) * 100

75 = 100 x – 100 + 100 x

200 x = 175

x = 0.875

Summary of answers:

R = major enantiomer = 0.875 or 87.5%

S = minor enantiomer = (1 – 0.875) = 0.125 or 12.5%

7 0

4 years ago

As the ph ________, the concentration of hydrogen ions increases, and the solution becomes ____________.

Sauron [17]

As the ph Decreases, the concentration of hydrogen ions increases,and the solution becomes acidic.

5 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

3 years ago

Al is

Brut [27]

Answer:

surface water

Explanation:

I hope its correct

6 0

3 years ago