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Dominik [7]
3 years ago
13

Compound B, C6H12O2, was found to be optically active, and it was oxidized to an optically active carboxylic acid A, by Ag (aka,

Tollens Reagent). Oxidation of B by PCC gave an optically inactive compound X that reacted with Zn amalgam/HCl to give 3-methylpentane. With Na2Cr2O7/H2SO4, compound B was oxidized to an optically inactive dicarboxylic acid C, C6H10O4. Provide the structures of A, B, and C (ignore specific configuration of any stereocenters).

Chemistry
1 answer:
Daniel [21]3 years ago
8 0

Answer:

Check the explanation

Explanation:

Acidipic Acid <u><em>(which is an essential dicarboxylic acid for manufacturing purposes with about 2. 5 billion kilograms produced per year. It is mainly used for the production of nylon and its related materials.)</em></u>

Going by the question, since the H_{2}  CrO_{4} is comparatively mild oxidizing agent than the CrO_{3}, it only oxidizes the carbon group

Kindly check the attached image below for the full explanation to the question above.

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WORTH 28 points MUST ANSER ALL 3 Easy POINTS
Vadim26 [7]
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3 0
3 years ago
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Can someone please help with chemistry? Lead has a density of 10.5 g/cm^3. What is the diameter of a lead ball that has a mass o
adelina 88 [10]
Density = Mass / Volume

10.5 = 500 / x

10.5x = 500

x = 50 cm^3 (1 sig fig) volume of the sphere

Volume of a sphere = 4/3 pi r^2

pi = 3.14 r = radius

50 = 4/3 pi r^3

50•3/4 = pi r^3

37.5 = pi r^3

37.5/pi = r^3

11.9 = r^3

cube root(11.9) = r

2.3 = r

Diameter = 2•radius

Diameter = 2 • 2.3

Diameter = 5 cm (1 sig fig) diameter



3 0
3 years ago
In Haber’s process, 30 moles of hydrogen and 30 moles of nitrogen react to make ammonia. If the yield of the product is 50%, wha
devlian [24]

Answer:

Therewill be produced 170.6 grams NH3, there will remain 25 moles of N2, this is 700 grams

Explanation:

<u>Step 1:</u> Data given

Number of moles hydrogen = 30 moles

Number of moles nitrogen = 30 moles

Yield = 50 %

Molar mass of N2 = 28 g/mol

Molar mass of H2 = 2.02 g/mol

Molar mass of NH3 = 17.03 g/mol

<u>Step 2:</u> The balanced equation

N2 + 3H2 → 2NH3

<u>Step 3:</u> Calculate limiting reactant

For 1 mol of N2, we need 3 moles of H2 to produce 2 moles of NH3

Hydrogen is the limiting reactant.

The 30 moles will be completely be consumed.

N2 is in excess. There will react 30/3 =10 moles

There will remain 30 -10 = 20 moles (this in the case of a 100% yield)

In a 50 % yield, there will remain 20 + 0,5*10 = 25 moles. there will react 5 moles.

<u>Step 4:</u> Calculate moles of NH3

There will be produced, 30/ (3/2) = 20 moles of NH3 (In case of 100% yield)

For a 50% yield there will be produced, 10 moles of NH3

<u>Step 5</u>: Calculate the mass of NH3

Mass of NH3 = mol NH3 * Molar mass NH3

Mass of NH3 = 20 moles * 17.03

Mass of NH3 = 340.6 grams = theoretical yield ( 100% yield)

<u>Step 6: </u>Calculate actual mass

50% yield = actual mass / theoretical mass

actual mass = 0.5 * 340.6

actual mass = 170.3 grams

<u>Step 7:</u> The mass of nitrogen remaining

There remain 20 moles of nitrogen + 50% of 10 moles = 25 moles remain

Mass of nitrogen = 25 moles * 28 g/mol

Mass of nitrogen = 700 grams

6 0
4 years ago
Read 2 more answers
In hypothetical element, two energy levels are separated by an energy of 282 kj/mol. what wavelength of light (in nm) is involve
scZoUnD [109]

Answer:

4.23*10^2 nm = 423 nm

Explanation:

E={\frac  {hc}{\lambda }}

h=6.62607015*10^{-34}J*s\\c=299 792 458 m / s\\h*c=1.98644568*10^{-25}J*m

282000 j/mol *1 mol/6*10^(23) = 4.7*10^(-19) J for one electron

\lambda={\frac  {hc}{E }}=\frac{1.98644568*10^{-25}J*m}{4.7*10^{-19}J} =\\\\=4.22648017*10^{-7} m = 4.22648017*10^{-7} m*\frac{10^9 nm}{1 m} =\\\\4.23*10^2 nm = 423 nm

6 0
3 years ago
What is the angle of refraction?
Andrej [43]

Answer:

The answer is the angle made by a refracted ray with a perpendicular to the refracting surface.

Explanation:

  • <u><em>A prism uses refraction to form a spectrum of colors from an incident beam of light.</em></u>
6 0
3 years ago
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