How to prepare ethanoic acid from ethane

Answers

Answer 1

Answer: anonymous
anonymous 5 years ago
First you chlorinate it in presence of light.
C2H6 + Cl2 ---hv -> C2H5Cl + HCl
Then you add aqeuos KOH to get C2H5OH
C2H5Cl+KOH-> C2H5OH+ KCl.
Then you add KMnO4 to get the rquired compound.
C2H5OH ----KMnO4 ---> CH3COOH.

Answer 2

Answer: C2h6 + O2 ----> CH3CooH

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Please help me I’m not sure with that

Answers

What I can’t see the pic ?hhbc639l73773 ghcbhjbch

Answer:

i cant see the picture sorry

Explanation:

For each set of dilutions (bleach and mouthwash), the first tube contained a 1:11 dilution (0.5 mL of agent was added to 5.0 mL of nutrient broth). What was the dilution of the MIC for the mouthwash and the bleach? Hint: You will have to calculate the dilution of each tube 2 – 4.

Answers

The dilution of each tube are as follows;

Tube 2; 8.26 × 10-³
Tube 3; 7.5 × 10-⁴
Tube 4; 6.83 × 10-⁵

For each time a dilution is further diluted;

The dilution ratio is; 1 : 11; In essence, 0.5 mL of agent was added to 5.0 mL of nutrient broth.

For the first tube; dilution factor is; 1 : 11 = 9.1× 10-²

For the second tube; 0.5mL of the first tube was added to 5.0mL of the nutrient broth; The dilution factor is then; (1/11) × (1/11) = 1/121 = 8.26 × 10-³.

For the third tube; 0.5mL of the second tube was added to 5.0mL of the nutrient broth; The dilution factor is then; (1/11) × (1/11) ×(1/11) = 1/121 = 7.5 × 10-⁴.

For the fourth tube; 0.5mL of the third tube was added to 5.0mL of the nutrient broth; The dilution factor is then; (1/11) × (1/11) × (1/11) × (1/11) = 1/14641 = 6.83 × 10-⁵.

Answers

The formula for osmotic pressure is:

$\Pi = iMRT$

where $\Pi$ is osmotic pressure, $i$ is van't Hoff's factor, $M$ molarity, $R$ is Ideal gas constant, and T is Temperature.

$\Pi$ = 132 atm

The van't Hoff's factor for glucose, $i$ = 1

$R = 0.08206 Latmmol^(-1)K^(-1)$

$T = 298 K$

Substituting the values in the above equation we get,

$132 atm = 1* M* 0.08206 Latmmol^(-1)K^(-1)* 298$

$M = (132 atm)/(1* 0.08206 Latmmol^(-1)K^(-1)* 298) = 5.4797 molL^(-1) \simeq 5.48 molL^(-1)$

So, the molarity of the solution is $5.48 molL^(-1)$ .

One gram of a compound requires the following quantities of solvent to dissolve: 47 mL of water, 8.1 mL of chloroform, 370 mL of diethyl ether, or 86 mL of benzene. Calculate the solubility of the compound in these four solvents (as g/100 mL). Estimate the partition coefficient of the compound between chloroform and water, ethyl ether and water, and benzene and water. Which solvent would you choose to extract the compound from an aqueous solution

Answers

Answer:

Chloroform.

Explanation:

Given,

Solvent requires 1g of compound per 100 mL

For water,

= 1g/47ml

= 2.1

For Chloroform,

= 1 g/8.1 mL

= 12.345679

For Diethyl ether,

= 1 g/370 mL

= 0.27

For Benzene,

= 1 g/86 mL

= 1.2

Partition coefficients:

Water = -

chloroform = 5.9

Diethyl = .13

Benzene = .57

The solvent chloroform would be chosen for drawing out the compound out of an aqueous solution as it has the maximum solubility.

Final answer:

The solubility of a compound in different solvents will determine its concentration in each solvent. The partition coefficient represents the relative solubility of a compound in two immiscible solvents. Chloroform would be the best choice to extract the compound from an aqueous solution.

Explanation:

The solubility of a compound is usually expressed as grams of solute per 100 mL of solvent. To calculate the solubility, you can use the following formula:

Solubility (g/100 mL) = (mass of solute / volume of solvent) * 100

Using this formula, the solubility of the compound in water is 47 g/100 mL, in chloroform is 97.53 g/100 mL, in diethyl ether is 2.70 g/100 mL, and in benzene is 1.16 g/100 mL.

The partition coefficient is a measure of the compound's solubility in two immiscible solvents. To calculate it, divide the solubility of the compound in one solvent by its solubility in another solvent. For example, the partition coefficient between chloroform and water would be:

Partition coefficient = Solubility in chloroform / Solubility in water = 97.53 g/100 mL / 47 g/100 mL = 2.07

The larger the partition coefficient, the more soluble the compound is in the first solvent compared to the second solvent. Based on the partition coefficients, chloroform would be the best choice to extract the compound from an aqueous solution.

Learn more about Solubility and partition coefficient here:

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Calculate the molar mass of the following:(a)the anesthetic halothane, C2HBrClF3
(b)the herbicide paraquat, C12H14N2CL2
(c)caffein, C8H10N4O2
(d)urea, CO(NH2)2
(e)a typical soap,C17H35CO2Na

Answers

Answer:

a)C2HBrClF3 = 197.35 g/mol

b)C12H14N2CL2 = 229.06g/mol

c)C8H10N4O2 = 194.22g/mol

d) CO(NH2)2=60.07 g/mol

e)C17H35CO2Na = 306.52 g/mol

Explanation:

Molar mass of a compound is equal to the sum of the atomic masses of the constituent elements.

a) C2HBrClF3

$Molar\ mass = 2(At. mass C)+1(at.mass H) +1(At. mass Br) + 1(At.mass Cl) + 3(At.mass F)\n=2(12.01 g/mol) + 1(1.01g/mol)+1(79.90 g/mol) +1(35.45g/mol)+3(18.99g/mol)=197.35g/mol$

b) C12H14N2CL2

$Molar\ mass = 12(C) + 14(H) + 2(N) + 2(Cl)\n\n=12(12.01) + 14(1.01) + 2(14.01) + 2(35.45) = 229.06g/mol$

c) C8H10N4O2

$Molar\ mass = 8(C) + 10(H) + 4(N) + 2(O)\n\n=8(12.01) + 10(1.01) + 4(14.01) + 2(16.00) =194.22g/mol$

d) CO(NH2)2

$Molar\ mass = 1(C) + 1(O) + 2(N) + 4(H)\n\n=1(12.01) + 1(16.00) + 2(14.01)+4(1.01) =60.07 g/mol$

e) C17H35CO2Na

$Molar Mass = 18(C) + H(35) +2(O) + 1(Na)\n\n=18(12.01) + 35(1.01) + 2(16.00) + 1(22.99) =306.52 g/mol$

calculate the mol fraction of ethanol and water in a sample of rectified spirit which contains 95% of ethanol by mass

Answers

Answer:

We are given that there is 95% ethanol by mass in rectified spirit

so, we can say that in a 100g sample, we have 95 grams of ethanol and 5 grams of water

we will find the number of moles of ethanol and water in 100g solution of rectified spirit and use that to calculate the mole fraction

Moles of Ethanol:

Molar mass of ethanol = 46 grams / mol

Number of moles = Given mass / molar mass

Number of moles = 95 / 46

Moles of Ethanol = 2 moles (approx)

Moles of Water:

Molar mass of water = 18 grams per mol

Number of moles = Given mass / molar mass

Moles of water = 5 / 18

Moles of water = 0.28 moles (approx)

Mole Fractions:

Mole fraction of a specific compound is the number of moles of that compound divided by the total number of moles in the solution

Mole fraction of Ethanol:

Moles of ethanol / (moles of ethanol + moles of water)

2 / (2 + 0.28)

2 / (2.28) = 0.9 (approx)

Mole fraction of Water:

Moles of water / (Moles of ethanol + moles of water)

0.28 / (2 + 0.28)

0.28 / (2.28) = 0.1 (approx)

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