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shusha [124]
3 years ago
14

A student prepares a standardized solution of sodium hydroxide by the procedure outlined. The student first prepares a saturated

solution of sodium hydroxide using freshly distilled or deionized water. She measures about 5 mL of the saturated solution with a graduated cylinder and dilutes to 1 L in a plastic bottle with freshly distilled or deionized water to make a solution that is about 0.1 M. To determine the exact concentration, she puts the NaOH solution in a buret and titrates a carefully-weighed pre-dried solid acid of known molar mass. She records the volume of NaOH needed to neutralize it and repeats the process several times to determine the concentration of the NaOH solution. She calculates the concentration and uncertainty in the concentration of the NaOH from the titration data.
Chemistry
1 answer:
skad [1K]3 years ago
8 0

NaOH has the property of deliquescent hence it cannot be weighed directly instead a saturated solution is diluted and desired concentration is obtained.

Explanation:

Sodium hydroxide undergoes deliquescent.

Deliquescence is the process in which substance absorbs moisture and carbon dioxide from the atmosphere to the extent of getting dissolved and form a solution.

Deliquescence happens when vapour pressure of solution formed is less than the partial pressure of water vapour in the air.

This is the reason that student did not measure the sodium hydroxide directly as the weight will change and proper solution cannot be made. Because of the deliquescence standard solution of NaOH is not prepared.

The student made a solution without weighing the NaOH instead she prepared a solution till the NaOH got dissolved and then diluted the solution.

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A certain reaction has an activation energy of 39.5 kJ/mol. As the temperature is increased from 25.0°C to a higher temperature,
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Answer:

Explanation:

We shall apply Arrhenius equation which is given below .

ln\frac{k_2}{k_1} = \frac{E_a}{R} [\frac{1}{T_1} -\frac{1}{T_2} ]

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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}})

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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.

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