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earnstyle [38]
3 years ago
5

1 pc

Chemistry
1 answer:
Alex3 years ago
5 0

Answer:0.085

Explanation:

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What is the unit of temperature on a solubility graph
Andrej [43]

Answer:

Khfxgkhxxkfhkugxxiufxhfixiyf

Explanation:

Xlufy8fxtukuxfxtu

6 0
3 years ago
Read 2 more answers
Determine the pH of 0.050 M HCN solution. HCN is a weak acid with a Ka equal to 4.9 x 10-10<br> DONE
nadezda [96]

Answer:

The pH of the solution is 5.31.

Explanation:

Let "\alpha is the dissociation of weak acid - HCN.

The dissociation reaction of HCN is as follows.

                  HCN+H_{2}O\rightarrow H_{3}O^{+}+CN^{-}

Initial                  C                         0            0

Equilibrium        c(1- \alpha)              c\alpha c\alpha

Dissociation constant = Ka= c\alpha \times \frac{c\alpha}{c(1-\alpha)}

=\frac{c\alpha^{2}}{(1-\alpha)}

In this case weak acids \alpha is very small so, (1-\alpha ) is taken as 1.

Ka=C\alpha^{2}

\alpha=\sqrt\frac{ka}{c}

From the given the concentration = 0.050 M

Substitute the given value.

\alpha=\sqrt\frac{4.9\times 10^{-10}}{0.05}=9.8\times 10^{-4}

[H_{3}O^{+}]=c\alpha

[H_{3}O^{+}]=0.05\times 9.8\times 10^{-4}= 4.9\times10^{-6}

pH= -log[H_{3}O^{+}]

=-log[4.9\times10^{-6}]

=6-log 4.9= 5.31

Therefore, The pH of the solution is 5.31.

7 0
3 years ago
Read 2 more answers
Match each pH value with the correct description. 7, 2, 8, 13, and 6 :)
bazaltina [42]
1.strong acid: 2
2.weak acid:6
3.strong base:13
4.weak base:8
5.neutral:7
8 0
4 years ago
Telluric acid (H2TeH4O6) is a diprotic acid with Ka1 = 2.0x10-8 and Ka2 = 1.0x10-11. A 0.25 M H2TeH4O6 contains enough HCl so th
tensa zangetsu [6.8K]

Answer:

5x10⁻⁶ = [HTeH₄O₆⁺]

Explanation:

The first dissociation equilibrium of the telluric acid in water is:

H₂TeH₄O₆ + H₂O ⇄ HTeH₄O₆⁺ + H₃O⁺

Using H-H equation for telluric acid:

<em>pH = pKa + log₁₀ [HTeH₄O₆⁺] / [H₂TeH₄O₆]</em>

pKa of telluric acid is -logKa1

pKa = -log 2.0x10⁻⁸

pKa = 7.699

As concentration of [H₂TeH₄O₆] is 0.25M, replacing in H-H equation:

3.00 = 7.699+ log₁₀ [HTeH₄O₆⁺] / [0.25M]

-4.699 = log₁₀ [HTeH₄O₆⁺] / [0.25M]

2x10⁻⁵ = [HTeH₄O₆⁺] / [0.25M]

<h3>5x10⁻⁶ = [HTeH₄O₆⁺]</h3>

7 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
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