) Predict the product for the following reaction. Assume you have an excess of potassium tert-butoxide. (CH3),COK Br

Answer: The law was applicable only to calcium. It could not include other elements beyond calcium.  With the discovery of rare gases, it was the ninth element and not the eighth element having similar chemical properties.

Explanation:

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Answer:

The balanced chemical equation for the reaction is:

C + 2H2 → CH4

From the equation, we can see that 1 mole of carbon reacts with 2 moles of hydrogen to produce 1 mole of methane. Therefore, to produce 7.8 moles of methane, we would need:

1 mole of carbon = 1 mole of CH4 / 2 moles of H2 = 1/2 mole of CH4

7.8 moles of CH4 = 7.8 × (1/2) moles of C = 3.9 moles of C

Now, we can use the molar mass of carbon to convert moles to grams:

Atomic mass of carbon (C) = 12.01 g/mol

3.9 moles of C × 12.01 g/mol = 46.8 g of C

Therefore, we need 46.8 grams of carbon to produce 7.8 moles of methane (CH4). Rounded to the nearest tenth, the answer is 46.8 grams.

To neutralize 50.0 mL of 0.80 M NaOH, 200 mL of 0.20 M HCl are needed.

When sodium hydroxide (NaOH) and hydrochloric acid (HCl) are mixed, sodium chloride (NaCl) and water (H2O) are the results. The chemical formula for the neutralizing reaction is as follows:NaOH+HClNaCl+H2O.

We must apply the following balanced chemical equation for the neutralization reaction to calculate how much HCl is needed to neutralize 50.0 mL of 0.80 M NaOH:

HCl + NaOH NaCl + H2O

One mole of HCl interacts with one mole of NaOH to form one mole of NaCl and one mole of water, as shown by the equation.

Let's first determine the quantity of NaOH in moles.

Moles of NaOH = volume (in liters) x molarity

Moles of NaOH = 50.0 mL x (1 L/1000 mL) x 0.80 M

Moles of NaOH = 0.040 moles

moles of HCl = volume (in liters) x molarity

0.040 moles = volume (in liters) x 0.20 M

Volume (in liters) = 0.040 moles / 0.20 M

Volume (in liters) = 0.20 L

Finally, we can convert the volume from liters to milliliters:

Volume (in milliliters) = 0.20 L x (1000 mL/1 L)

Volume (in milliliters) = 200 mL

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According to the given Information:

The aqueous solution that has the lowest freezing point is 1.0 M C2H5OH (ethanol).

Because it determines the concentration of solute particles in the solution.

Ionic solutes, such as NaCl, dissociate into multiple ions in water, producing a higher concentration of solute particles per unit concentration than molecular solutes, such as ethanol.

This results in a greater degree of freezing point depression for ionic solutes than molecular solutes.

An aqueous solution is one in which water serves as the solvent.

Aqueous solutions are very common in nature and in laboratory settings. Many substances can dissolve in water to form aqueous solutions, including salts, acids, bases, and gases.

Aqueous solutions are important in many fields of science, including chemistry, biology, and environmental science.

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The gas law that is employed to deflate a football is Boyle's law.

Boyle's law states that for a fixed amount of gas at a constant temperature, the pressure and volume of the gas are inversely proportional to each other. The equation for Boyle's law is:

[tex]P_{1} V_{1} /P_{2} V_{2}[/tex]

Where [tex]P_{1}[/tex] is the initial pressure of the gas, [tex]V_{1}[/tex]  is the initial volume of the gas,[tex]P_{2}[/tex] is the final pressure of the gas, and [tex]V_{2[/tex] is the final volume of the gas. In the case of deflating a football, the pressure of the air inside the football is reduced by letting some of the air out. The volume of the football decreases as the pressure decreases, and this is in accordance with Boyle's law.

Therefore, The Boyle's Law is used to deflate a football. It states that the volume of a gas is inversely proportional to its pressure, when the temperature is constant.

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A compound that is defined by its ability to produce hydroxide ions when dissolved in water is known as a base.

Bases are compounds that dissolve in water to form hydroxide ions (OH-). They are hydroxide ion donors, to be precise. Bases have a pH value greater than 7. The OH- ions are released when bases are dissolved in water. Sodium hydroxide (NaOH) is a good example of a base.

When NaOH is dissolved in water, it produces hydroxide ions (OH-) and sodium ions (Na+). As a result, the solution is more basic, and the pH is greater than 7. The following are some examples of bases:

Bases are commonly utilized in several chemical reactions. They're utilized as pH modifiers, reagents, and buffer solutions, among other things. They are also used in industries like cosmetics, detergents, and food. Furthermore, they are utilized in water treatment plants to control acidity levels and remove impurities.

Therefore, a compound that is defined by its ability to produce hydroxide ions when dissolved in water is known as a base.

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The compound that behaves as an acid when dissolved in water is HI (hydrogen iodide). Thus, the correct option will be D.

HI is an Arrhenius acid, meaning it produces hydrogen ions (H⁺) in aqueous solution. The compound that behaves as an acid when dissolved in the water Hydrogen iodide (HI). HI is a diatomic molecule and a colorless gas at room temperature.

Hydrogen iodide is a strong acid when dissolved in water, with a pKa of −10. Hydrogen iodide is also used as a reducing agent in organic chemistry in the production of iodinated compounds.

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A sense strand is the mRNA strand that is translated from a DNA strand with a same nucleotide sequence. the codons have specific functions when the mRNA sequence is translated into a protein.

The DNA sequence serves as a template for the synthesis of a complementary mRNA molecule during transcription. The nucleotide arrangement of the DNA template strand dictates the sequencing of the mRNA. The mRNA sequence is not identical to the template DNA strand; rather, it is complementary to it. RNA polymerase, which builds the mRNA molecule on the DNA template strand, adds complementary RNA nucleotides to the lengthening mRNA chain. Since RNA nucleotides have uracil (U) as a base instead of thymine (T), the mRNA sequence will have the same nucleotide sequence as the DNA template strand. The mRNA sequence is read in groups of three nucleotides called codons, and the codons have specific functions when the mRNA sequence is translated into a protein.

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Methanol, or CH3OH, is a Brnsted-Lowry base in this reaction because it can receive a proton from the hydroxide ion, or OH-, to generate CH3O- (methoxide ion).

The Brnsted-Lowry base OH- (hydroxide ion), on the other hand, may transfer a proton (H+) to[tex]CH3OH[/tex]to create H2O. (water).So the reactants are CH3OH (base) and OH- (base), and the products are CH3O- (conjugate base of CH3OH) and H2O (conjugate acid of OH-).I apologize for the mistake in my previous response. You are correct that methanol, or CH3OH, is a Brønsted-Lowry acid in this reaction because it donates a proton (H+) to the hydroxide ion (OH-) to form CH3O- (methoxide ion). The hydroxide ion (OH-) is a Brønsted-Lowry base because it accepts a proton (H+) from CH3OH to form H2O (water). Therefore, the reactants are [tex]CH3OH[/tex]  (acid) and OH- (base), and the products are CH3O- (conjugate base of CH3OH) and H2O (conjugate acid of OH-).

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When scientists first discovered that atoms were composed of smaller particles, such as electrons, protons, and neutrons, they did not reject atomic theory because the experimental evidence supported the idea that atoms were still the fundamental building blocks of matter.

Instead of rejecting atomic theory, scientists modified it to incorporate the new information about the structure of atoms. For example, Ernest Rutherford's famous gold foil experiment in 1911 showed that atoms had a small, positively charged nucleus at their center, which was surrounded by negatively charged electrons.

This discovery led to the development of the modern model of the atom, which is still based on the idea that atoms are the smallest units of matter that retain the properties of an element. However, the model has been updated to reflect our current understanding of the subatomic particles that make up atoms.

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1. The balanced chemical equation shows that for every 2 molecules of octane burned, 16 molecules of carbon dioxide are produced. Therefore, the number of molecules of carbon dioxide represented by the equation is 16.

2. The balanced chemical equation shows that for every 2 molecules of octane burned, 25 molecules of oxygen are required. Therefore, the ratio of octane to oxygen is 2:25. From this, we can determine the number of moles of octane represented by the equation by dividing the given amount of oxygen by the ratio:

25 mol O2 × (2 mol C8H18 / 25 mol O2) = 2 mol C8H18

Therefore, the equation represents 2 moles of octane.

3. The simplified mole ratio of octane to carbon dioxide can be determined by dividing both sides of the equation by the coefficient of octane (2):

2C8H18(g) +25O2(g) → 16CO2(g) + 18H2O(l)

Dividing by 2, we get:

C8H18(g) + 12.5O2(g) → 8CO2(g) + 9H2O(l)

The simplified mole ratio of octane to carbon dioxide is therefore 1:8.

4. The simplified mole ratio of oxygen to octane can be determined in the same way, by dividing both sides of the equation by the coefficient of octane:

2C8H18(g) +25O2(g) → 16CO2(g) + 18H2O(l)

Dividing by 2, we get:

C8H18(g) + 12.5O2(g) → 8CO2(g) + 9H2O(l)

The simplified mole ratio of oxygen to octane is therefore 12.5:1.

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The breakdown of 6.54 g of potassium chlorate results in the production of 4.81 x [tex]10^{22}[/tex]oxygen molecules.

The balanced chemical equation for the decomposition of potassium chlorate is:

2 KClO3(s) → 2 KCl(s) + 3 O2(g)

This equation tells us that for every 2 moles of potassium chlorate that decompose, 3 moles of oxygen gas are produced.

To determine the number of molecules of oxygen produced by the decomposition of 6.54 g of potassium chlorate, we first need to convert the mass of potassium chlorate to moles using its molar mass. The molar mass of KCLO₃ is:

K: 39.10 g/mol

Cl: 35.45 g/mol

O: 3(16.00 g/mol) = 48.00 g/mol

Total molar mass of KCLO₃: 39.10 + 3(35.45) + 48.00 = 122.55 g/mol

Number of moles of KCLO₃ = 6.54 g / 122.55 g/mol = 0.0533 mol

Now we can use the mole ratio from the balanced equation to calculate the number of moles of oxygen produced:

3 moles O₂ / 2 moles KCLO₃ = x moles O₂ / 0.0533 moles KCLO₃

x = 3/2 x 0.0533 = 0.0799 moles O₂

Finally, we can convert the number of moles of oxygen to the number of molecules using Avogadro's number:

Number of molecules of O2 = 0.0799 mol x 6.022 x [tex]10^{23}[/tex] molecules/mol = 4.81 x [tex]10^{22}[/tex] molecules

Therefore, 4.81 x [tex]10^{22}[/tex] molecules of oxygen are produced by the decomposition of 6.54 g of potassium chlorate.

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When KCl is dissolved in water, the following ions are produced: K+ and Cl-.

The solution of an ionic compound dissolved in water will be broken into ions, with the positive ions separated from the negative ions. The cation, which is positively charged, is usually a metal, while the anion, which is negatively charged, is usually a non-metallic element or a group of atoms. When a solute dissolves in water, it forms an electrolyte, which is a substance that conducts electricity when dissolved in water.

KCl, or potassium chloride, is an ionic compound. It is a white crystalline powder with a salt-like taste that dissolves in water. It is used in food processing as a sodium replacement, in medicine as a potassium supplement, and in industrial chemical synthesis and manufacturing.

The chemical formula of KCl is K+Cl-. Potassium chloride (KCl) consists of K+ ions and Cl- ions. In water, these ions disassociate (separate) to produce K+ ions and Cl- ions. So, when KCl is dissolved in water, the ions K+ and Cl- are formed. The answer is K+ and Cl-.

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The  molecule that is the main product of the Calvin cycle is glucose. The Calvin cycle is also known as the light-independent reactions.

It is a series of biochemical reactions that occur in the stroma of the chloroplast in photosynthetic organisms to produce glucose. The Calvin cycle is made up of three stages: Carbon fixation, Reduction and regeneration of ribulose bisphosphate. Here's a breakdown of each stage:

Carbon fixation: Carbon dioxide enters the Calvin cycle and is converted to organic molecules. During carbon fixation, Rubisco, which is a crucial enzyme in photosynthesis, catalyzes the reaction between carbon dioxide and ribulose bisphosphate, leading to the formation of a six-carbon molecule that splits into two three-carbon molecules. This three-carbon molecule is the starting material for the reduction process.

Reduction: The ATP and NADPH produced during the light-dependent reactions are used to convert the three-carbon molecule produced during carbon fixation into glyceraldehyde-3-phosphate. This process involves a series of biochemical reactions that require the use of energy from ATP and electrons from NADPH.

Regeneration of ribulose bisphosphate: Glyceraldehyde-3-phosphate, which is the main product of the Calvin cycle, is used to regenerate the starting material for carbon fixation, ribulose bisphosphate. During this stage, ATP is used to convert the remaining glyceraldehyde-3-phosphate molecules into ribulose bisphosphate. The Calvin cycle is an essential process in photosynthesis, as it produces glucose, which is the main source of energy for plants and other photosynthetic organisms.

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Yes, this is a good idea as it is an efficient use of resources. Advantages include reduced costs of purchasing new drying agents and decreased wastage of materials. Disadvantages could include loss of quality of the recycled drying agent, and extra energy used to dry out the salt.

Drying agents can be collected after an experiment and the hydrated salt heated in an oven to drive off the water. The recycled drying agent can then be used again for another experiment.

What are drying agents?

In order to absorb water vapor, drying agents are added to organic solvents to make them anhydrous.

What are the advantages and disadvantages of recycling drying agents?

The recycling of drying agents has a few advantages and disadvantages:

Advantages of recycling drying agents:

Cost-effective: If the solvent used is expensive, recycling drying agents can save money. A drying agent like anhydrous magnesium sulfate is a good example since it can be reused numerous times. No pollution: The disposal of waste is reduced. If every time a new drying agent is employed, it must be disposed of properly, which is both time-consuming and costly. The amount of waste that has to be disposed of is reduced if the same drying agent is used repeatedly. Recyclable waste: Used drying agents are recyclable. It's just a matter of heating the salt to remove any water and returning it to the drying agent stock. This procedure helps to prevent waste.

Disadvantages of recycling drying agents:

Contamination: Even though the recycled drying agent is supposed to be pure, it may still contain minor quantities of impurities, which might result in contamination of the final product. Impurities: If the drying agent is not cleaned properly, impurities will be transferred from one experiment to the next. Excessive heating: Anhydrous drying agents should not be heated excessively because they may lose their effectiveness. If the salt is heated for too long, the surface area exposed to moisture will be decreased. Therefore, while recycling drying agents is a good idea, some precautions should be taken to ensure that the drying agent is pure and effective.

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The volume of the balloon at a temperature of 263 K will be approximately 683.1 mL.

Charles's Law states that the volume of a gas is directly proportional to its absolute temperature at constant pressure.

This means that the volume and temperature of a gas are directly proportional to each other as long as the pressure is constant.

It is expressed as:

V₁/T₁ = V₂/T₂

Where V₁  and T₁ are the initial volume and temperature, V₂ is the final volume, and T₂ is the final temperature.

Given that:

Solving for V₂, we get:

V₂ = V₁T₂ / T₁

V₂ = ( 800 × 263 ) / 308

V₂ = 210400 / 308

V₂ = 683.1 mL

Therefore, the volume is  683.1 mL.

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The change that is useful for the environment and living things is called "positive environmental change."

Positive environmental change refers to any alteration or modification in the environment that improves or benefits living organisms' well-being. Examples of positive environmental changes include reducing pollution, conserving water, using renewable energy sources, and recycling waste products. Positive environmental change is essential to ensure a sustainable future and to maintain the planet's biodiversity.

It can be achieved by implementing new policies, practices, and technologies that promote sustainable development and reduce the negative impact on the environment. Positive environmental change can also help to address climate change and other environmental challenges faced by humanity. By taking positive steps to protect the environment, we can ensure that future generations can also enjoy a healthy, prosperous, and sustainable planet.

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Answer: I think its 111.6

Explanation:

To estimate the system pressure and vapor-phase mole fraction when x1 = 0.8, we can use the 1-parameter Margules equation.

This equation assumes that the vapor-liquid equilibrium is a linear relationship between the mole fraction of each component.

Since the given experiment gives us three points, we can use linear interpolation to estimate the parameters of the Margules equation.

From the given experiment, we know the values for x1, y1, and P when x1 = 0.0, 0.2, and 1.0 respectively. Therefore, we can calculate the slope and y-intercept of the Margules equation as follows:

Slope = (P2 - P1)/(y2 - y1) = (68.23 - 41.02)/(0.51 - 0.0) = 68.23

y-intercept = P1 - (slope * y1) = 41.02 - (68.23 * 0.0) = 41.02

Using these values and the x1 value of 0.8, we can then estimate the system pressure and vapor-phase mole fraction as follows:

System Pressure = (slope * 0.8) + y-intercept = (68.23 * 0.8) + 41.02 = 78.2 kPa

Vapor-phase Mole Fraction = (System Pressure - y-intercept) / slope = (78.2 - 41.02) / 68.23 = 0.80

Therefore, the estimated system pressure and vapor-phase mole fraction when x1 = 0.8 is 78.2 kPa and 0.80 respectively.

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Part A: True.

Part B: False. Rinsing with water may not be enough to clean the buret completely.

Part C: False. Soapy water should not be used to clean a buret since it can leave residue.

Part D: False. After rinsing with water and soapy water solution, the buret should be rinsed with distilled water and dried before adding the titrating solution.

Part E: False. The buret should be rinsed with the titration solution only once before beginning a titration.

Titration is a laboratory procedure used to compare a solution's concentration to that of a reference solution with known concentration. It entails gradually mixing the standard solution into the sample solution up until the reaction is finished, which can be detected by a colour change or another quantifiable signal.

In many disciplines, including chemistry, medicine, and environmental research, titration is used. It can be used to quantify the quantity of a certain component in a sample, examine the concentration of acids and bases, and ascertain the purity of a substance.

Titration calls for exact volume and concentration measurements, as well as safe chemical handling and disposal. There are several different kinds of titration techniques, including complexometric, redox, and acid-base titration.

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Each reaction's Lewis acid and Lewis base are as follows:

AlCl3 is the Lewis acid in the reaction, whereas Cl- is the Lewis base. F- is the Lewis base and BF3 is the Lewis acid in the reaction Cl- + AlCl3 --> AlCl4- BF3 + F- --> BF4-

The Lewis base in this reaction is NH3, and the Lewis acid is H+. NH3 + H+ --> NH4+

Explanation: A Lewis acid acts as an electron pair acceptor in a Lewis acid-base reaction, whereas a Lewis base acts as an electron pair donor. In the initial reaction, Cl- provides AlCl3 with a pair of electrons, which AlCl3 accepts to produce AlCl4-. As a result, AlCl3 is the Lewis acid and Cl- is the Lewis base. In the subsequent response, F- provides two pairs. BF3 takes the electrons and transforms them into BF4-. Hence, the Lewis bases are F- and BF3, respectively. In the third reaction, H+ absorbs a pair of electrons from NH3 and forms NH4+ as a result. As a result, the Lewis bases are NH3 and H+.

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The volume of the ammonia that can be produced from the reaction that has been written is;  155 * 10^7 L

We know that;

PV = nRT

For the hydrogen;

n = PV/RT

n = 155 * 4.0 x 10^7 /0.082 * 776

n = 620 * 10^7/63.63

n = 9.7 * 10^7 moles

Now the reaction equation is;

N₂(g) + 3H₂(g) → 2NH3(g)

3 moles of hydrogen  produced 2 moles of ammonia

9.7 * 10^7 moles will produce  9.7 * 10^7 moles * 2 moles/ 3 moles

x = 6.5 * 10^7 moles

For the volume of the ammonia;

V = nRT/P

V =  6.5 * 10^7 moles * 0.082 * 293.6/1.007

V = 155 * 10^7 L

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Answer:

To prepare a 1.00 mol dm^-3 sodium chloride solution, we need to dissolve one mole of sodium chloride in one liter of solution (1000 cm^3).

However, we only need to prepare 100 cm^3 of the solution, which is 1/10 of a liter. So we need to dissolve:

1/10 * 1.00 mol = 0.100 mol

of sodium chloride in 100 cm^3 of solution.

The molar mass of sodium chloride is 58.5 g/mol. So to calculate the mass of sodium chloride required, we can use:

mass = number of moles x molar mass

mass = 0.100 mol x 58.5 g/mol

mass = 5.85 g

Therefore, we need 5.85 g of sodium chloride to prepare 100 cm^3 of 1.00 mol dm^-3 sodium chloride solution.

The order of increasing reaction rate of alkyl halides with triethylamine is iodoethane, 1-bromopropane, and 2-bromopropane.

When the primary alkyl halide reacts with the triethylamine, it's faster than the secondary alkyl halide. Since triethylamine is a strong, bulky base that tends to perform nucleophilic substitution, it undergoes a reaction with both primary and secondary alkyl halides. When halides react with triethylamine, the bond between nitrogen and carbon is formed.

In this reaction, the rate of reaction will be slower with secondary alkyl halides due to steric hindrance.  Iodoethane will be more reactive than 1-bromopropane because of the higher electronegativity of iodine which makes it more prone to nucleophilic substitution. Based on the above discussion, the order of increasing rate of reaction with triethylamine is 2-bromopropane < 1-bromopropane < iodoethane.

Therefore, iodoethane is the most reactive of the three alkyl halides, 1-bromopropane is more reactive than  2-bromopropane and 2-bromopropane is the least reactive.

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Charged ions such as sodium, potassium, and chloride are called electrolytes.

Ions are atoms or molecules that have a positive or negative charge. They develop an electrical charge when an atom or molecule gains or loses one or more electrons, becoming an ion. Cations are ions with a positive charge, whereas anions are ions with a negative charge. The conductivity of fluids is due to charged ions like electrolytes.

Sodium, potassium, chloride, bicarbonate, calcium, and phosphate are examples of electrolytes that are vital for the body's daily function. Electrolytes play a significant role in maintaining the correct water balance and assisting in the transmission of electric impulses across cells.

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The average mass of chemistrium (Ch) in amu is: 12.5 amu.

Chemistrium is an element with the atomic number 106. It is a transactinide synthetic element with an atomic weight of 268 u. Until 2009, this element was known as unnilhexium (Unh). It was named chemistrium in honor of the chemistry in recognition of the Moscow-based Joint Institute for Nuclear Research's contributions to the synthesis of new elements.

If a sample of the element chemistrium (Ch) contains 100 atoms of Ch-12 and 10 atoms of Ch-13 (for a total of 110 atoms in the sample), the average mass of chemistrium in amu can be calculated as follows:

Average mass of Ch = [(number of atoms of Ch-12 x atomic weight of Ch-12) + (number of atoms of Ch-13 x atomic weight of Ch-13)] / Total number of atoms of Ch= [(100 x 12.000000) + (10 x 13.003355)] / 110= [1200.0000 + 130.03355] / 110= 1330.03355 / 110= 12.18212318 amu, which is rounded off to 12.5 amu.

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The mass of 1 mole of solute dissolved to make the solution will be 40 g/mol (mass of 1 mole of solute).

To determine the mass of 1 mole of solute, we can use the molar mass of the solute. The formula for molar mass is:
Molar Mass = Mass of Solute ÷ Number of Moles

Let's use this formula to solve the problem:
Mass of Solute = 20 grams
Volume of Solution = 500 mL = 0.5 L
Concentration of Solution = 1 M
Number of Moles of Solute = Concentration × Volume = 1 M × 0.5 L = 0.5 mol

Now, we can use the molar mass formula to calculate the mass of 1 mole of solute:
Molar Mass = Mass of Solute ÷ Number of Moles
Molar Mass = 20 grams ÷ 0.5 mol
Molar Mass = 40 grams/mol

Therefore, the mass of 1 mole of solute is 40 grams.

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Answer:

a) Methanoic acid can be prepared from methanol through oxidation using potassium permanganate and sulfuric acid. The reaction proceeds as follows:

CH3OH + 2[O] → HCOOH + H2O

b) Methanal (formaldehyde) can be prepared from methanol through oxidation using potassium dichromate and sulfuric acid. The reaction proceeds as follows:

CH3OH + [O] → CH2O + H2O

c) Butanone can be prepared from 2-butanol through oxidation using Jones reagent (CrO3/H2SO4) or pyridinium chlorochromate. The reaction proceeds as follows:

CH3CH(OH)CH2CH3 + [O] → CH3COCH2CH3 + H2O

d) Pentanal can be prepared from 1-pentanol through oxidation using potassium permanganate and sulfuric acid. The reaction proceeds as follows:

CH3(CH2)3CH2OH + 3[O] → CH3(CH2)3CHO + 3H2O

e) Hexanoic acid can be prepared from 1-hexanol through oxidation using potassium permanganate and sulfuric acid. The reaction proceeds as follows:

CH3(CH2)4CH2OH + 4[O] → CH3(CH2)4COOH + 4H2O

f) Hexanal can be prepared from 1-hexanol through oxidation using pyridinium chlorochromate. The reaction proceeds as follows:

CH3(CH2)4CH2OH + [O] → CH3(CH2)5CHO + H2O

g) Hexan-3-one can be prepared from 3-hexanol through oxidation using Jones reagent (CrO3/H2SO4) or pyridinium chlorochromate. The reaction proceeds as follows:

CH3(CH2)4CH(OH)CH3 + [O] → CH3(CH2)3COCH3 + H2O

(please could you kindly mark my answer as brainliest)

Answer:

2,1,1  

Explanation: