draw the possible e1 product(s) for the following reactions. do not draw the leaving group or counterion. ignore zaitsev's rule.

Answer:

See Below.

Explanation:

Problem 1

Let x be the mass of 15% solution needed and y be the mass of 20% solution needed. Then, we have the following system of equations:

x + y = 200 (total mass of solution)

0.15x + 0.20y = 0.17(200) (total amount of solute)

Solving this system of equations gives:

x = 60 g (mass of 15% solution)

y = 140 g (mass of 20% solution)

Therefore, 60 g of 15% solution and 140 g of 20% solution are needed to prepare 200 g of 17% solution.

Problem 2

Let x be the mass of 18% solution needed and y be the mass of 5% solution needed. Then, we have the following system of equations:

x + y = 300 (total mass of solution)

0.18x + 0.05y = 0.07(300) (total amount of solute)

Solving this system of equations gives:

x = 120 g (mass of 18% solution)

y = 180 g (mass of 5% solution)

Therefore, 120 g of 18% solution and 180 g of 5% solution are needed to prepare 300 g of 7% solution.

Problem 3

The total mass of the final solution is

200 g + 350 g = 550 g

The total amount of solute in the final solution is:

0.15(200 g) + 0.20(350 g) = 95 g + 70 g = 165 g

Therefore, the mass percentage of the final solution is:

(mass of solute / total mass of solution) x 100% = (165 g / 550 g) x 100% = 30%

Therefore, the mass percentage of the final solution is 30%.

Problem 4

The total mass of the final solution is

300 g + 35 g = 335 g

The total amount of solute in the final solution is:

0.15(300 g) + 35 g = 75 g + 35 g = 110 g

Therefore, the mass percentage of the final solution is:

(mass of solute / total mass of solution) x 100% = (110 g / 335 g) x 100% = 32.8%

Therefore, the mass percentage of the final solution is 32.8%.

Problem 5

The total mass of the final solution is

400 g + 150 g = 550 g

The total amount of solute in the final solution is

0.25(400 g) = 100 g

Therefore, the mass percentage of the final solution is

(mass of solute / total mass of solution) x 100% = (100 g / 550 g) x 100% = 18.2%

Therefore, the mass percentage of the final solution is 18.2%.

Answer:

7.68 x 1025 atoms of phosphorous correspond to 1.06 mole of diphosphorous pentoxide. This can also be written as 1.06 mol of P2O5.

At has a higher initial ionisation energy than Br, while Br has a bigger atomic radius. Se has a bigger atomic radius than Br, and Br has a higher initial ionisation energy than Se.

The most loosely bound electron is further from the nucleus and thus easier to remove in bigger atoms. Hence, the ionisation energy should decrease as size (atomic radius) increases.

This is because the outer electrons aren't bound as strongly because they are farther from the nucleus.

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Answer:

when the solution of potassium hydroxide and zinc chloride are mixed,the double-displacement reaction occur ,resulting in precipitation and the reaction forms potassium chloride and zinc hydroxide .

Answer: It is not necessarily advantageous for a predator to prey exclusively on a single prey species, as this can limit their options and make them vulnerable if the population of that prey species declines or becomes extinct. Predators that are more flexible and able to switch between different prey species may be better equipped to survive and thrive in changing environments.

However, there are some advantages to specializing in a single prey species. For example, a predator that is well adapted to hunting a particular prey species may be more efficient and successful at capturing and consuming that prey, which could provide a reliable source of energy. Additionally, if the predator and prey have co-evolved, the predator may have adaptations that specifically allow it to exploit the weaknesses or vulnerabilities of its prey, giving it an advantage over predators that are less specialized.

Answer:

1. Equilibrium expressions:

a. K = [HSO4-][H3O+]/[H2SO4][H2O]

b. K = [NO]^4[H2O]^6/[NH3]^4[O2]^5

c. K = [NH3][HCl]/[NH4Cl]

d. K = [NO2]^2/[N2O4]

2. Since K = 0.050, the concentrations of the reactants (N2 and H2) are larger than the concentrations of the products (NH3).

3. Since K = 6.8 x 10^3, the concentrations of the products (SO2 and O2) are larger than the concentrations of the reactant (SO3).

4. The Ksp expression for each of the reactions is:

a. Ksp = [Na+][Cl-]

b. Ksp = [Ba2+][SO42-]

5. Brønsted-Lowry acids and bases:

a. Acid: HNO3; Conjugate base: NO3-; Base: H2O; Conjugate acid: H3O+

b. Acid: HCN; Conjugate base: CN-; Base: H2O; Conjugate acid: HCN

c. Acid: H2SO4; Conjugate base: HSO4-; Base: Cl-; Conjugate acid: HCl

d. Acid: NH3; Conjugate base: NH2-; Base: H2O; Conjugate acid: NH4+

e. Acid: H2O; Conjugate base: OH-; Base: O2-; Conjugate acid: OH-

6. Conjugate acids and bases:

a. Acid: H2O; Conjugate base: OH-

b. Acid: H3O+; Conjugate base: H2O

c. Acid: H2CO3; Conjugate base: HCO3-

d. Acid: NH4+; Conjugate base: NH3

e. Acid: HSO4-; Conjugate base: SO42-

7. The strongest acid is HIO4 (highest K value), followed by HCN, HC2H3O2, and H2CO3 (lowest K value). The K values represent the degree to which the acids dissociate in solution. HIO4 is a strong acid, meaning it dissociates almost completely in solution, while H2CO3 is a weak acid, meaning it only dissociates partially.

8. pH and pOH calculations:

a. pH = -log[H3O+] = -log(0.200) = 0.699; pOH = -log[OH-] = -log(1.0 x 10^-14/0.200) = 12.301

b. pOH = -log[OH-] = -log(0.0143) = 1.844; pH = 14.000 - pOH = 12.156

c. pH = -log[H3O+] = -log(3.0) = 0.522; pOH = 13.478

d. pOH = -log[OH-] = -log(0.0062) = 2.206; pH = 14.000 - pOH = 11.794

9. Hydronium and hydroxide ion concentrations:

pH = 3.6; hydronium ion concentration = 10^-pH = 3.98 x 10^-4 M; hydro

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The standard change in Gibbs energy at 25 degree Celsius is  490.6 °C. for given equilibrium partial pressure  .

The Gibbs energy is the thermodynamic potential that is minimized when a system reaches chemical equilibrium at constant pressure and temperature when not driven by an applied electrolytic voltage. Its derivative with respect to the reaction coordinate of the system then vanishes at the equilibrium point.

Using the formula

ΔG° =  - R × T ln K

WHERE R=  8.3144598 J⋅mol⁻¹⋅K⁻¹.

T = 298 K

K = 0.82

SOLVING ,

The standard change in Gibbs energy at 25 degree Celsius is  490.6 °C.

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A few statements may be true regarding a biochemical oxidation-reduction (redox) reaction. The statements are as follows: A redox reaction occurs when there is a transfer of electrons between molecules or atoms.

The electron donor becomes oxidized, and the electron acceptor is reduced, causing a transfer of energy. A redox reaction produces ATP, which is the primary energy currency of the cell. Oxidation and reduction are complementary reactions that occur simultaneously in the same reaction, resulting in the release of energy. Redox reactions are vital in metabolic pathways, and the electron carriers NAD+ and FAD+ are essential in these reactions. Oxygen is frequently used as a final electron acceptor in redox reactions. Redox reactions can also occur in non-cellular environments, such as photosynthesis, respiration, and combustion. The significance of redox reactions is enormous, and they play an essential role in sustaining life on earth. They help in generating energy, breaking down complex molecules, synthesizing molecules, and many other cellular processes.

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Answer:

The correctly written thermochemical equation is:

C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (l), ΔH = -2,220 kJ/mol

This equation represents the combustion of propane (C3H8) in the presence of oxygen (O2) to produce carbon dioxide (CO2) and water (H2O), with a heat release of -2,220 kJ/mol. The state symbols (g) for gases and (l) for liquids indicate the physical state of each substance at standard conditions.

Explanation:

ABOVE

If we use the Gizmo to find the freezing, melting, and boiling points of water at 5,000 meters (16,404 feet) then,

Freezing point: 32 ºF (0ºC)

Melting point: 32 ºF (0ºC)

Boiling point: 203°F (95°C)

The freezing point is defined as the temperature at which a liquid becomes a solid. Increased pressure usually raises the freezing point with the melting point of the solid. The boiling point of a pure substance is defined as the temperature at which the substance transitions from a liquid to the gaseous phase. At the boiling point the vapor pressure of the liquid is equal to the applied pressure on the liquid. The melting point of a substance is defined as the temperature at which the substance changes from a solid to a liquid.

Melting occurs at a single temperature for the pure substances. The normal and average melting point and boiling point of water at 1 atmospheric pressure are 0°C and 100°C respectively. Decreasing the pressure under 1 atm. will lower the boiling point since the external pressure will be lower so it will become equal with the vapor pressure at a lower temperature.

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The pH of 0.335 M trimethylammonium chloride, (CH3)3NHCI is approximately 5.676.

What is pH?

To find the pH of the solution, we need to first determine if (CH3)3NHCI acts as an acid or base. Since (CH3)3NHCI is a salt composed of a weak base, trimethylamine, and a strong acid, hydrochloric acid (HCI), it will undergo hydrolysis in water.

The hydrolysis reaction is given by:

(CH3)3NH+ (aq) + H2O (l) ⇌ (CH3)3N (aq) + H3O+ (aq)

The Kb expression for the equilibrium reaction is:

Kb = [ (CH3)3N ] [ H3O+ ] / [ (CH3)3NH+ ]

Since (CH3)3NH+ and HCl dissociate completely in water, the initial concentration of (CH3)3NH+ is equal to the concentration of (CH3)3NHCI, which is 0.335 M.

Using the Kb value given, we can solve for the concentration of H3O+:

Kb = 6.3 x [tex]10^{-5}[/tex] = [ (CH3)3N ] [ H3O+ ] / 0.335

[ H3O+ ] = Kb x (CH3)3NH+ / (CH3)3N

[ H3O+ ] = 6.3 x [tex]10^{-5}[/tex] x 0.335 / 1

[ H3O+ ] = 2.1095 x [tex]10^{-6}[/tex] M

Finally, we can calculate the pH using the expression:

pH = -log [H3O+]

pH = -log (2.1095 x [tex]10^{-6}[/tex])

pH = 5.676

Therefore, the pH of the solution is approximately 5.676.

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Complete question is: The pH of 0.335 M trimethylammonium chloride, (CH3)3NHCI is approximately 5.676.

The  atoms of lithium that  are in 18.7 g is 16 × 10²³ atoms . This is taken out by mole concept .

The mole is a unit of measurement similar to the pair, dozen, gross, and so on. It provides a precise count of the atoms or molecules in a bulk sample of matter. A mole is the amount of substance that contains the same number of discrete entities (atoms, molecules, ions, etc.)

if 7 grams of lithium contain 6 × 10²³ atoms

then 18.7 will contain 16 × 10²³ atoms

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a balanced equation for the redox reaction: 2 Ca(s) + O2(g) → 2 CaO(s)

A redox reaction is a type of chemical reaction that involves the transfer of electrons between species. One species undergoes oxidation (loses electrons) while another species undergoes reduction (gains electrons).

In the reaction between calcium metal and oxygen gas, the calcium metal is being oxidized (loses electrons) and the oxygen gas is being reduced (gains electrons). This can be seen in the balanced equation where the calcium atoms go from having an oxidation state of 0 to +2, while the oxygen atoms go from having an oxidation state of 0 to -2.

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Answer:

First, we need to calculate the change in temperature:

ΔT = final temperature - initial temperature

ΔT = 85 °C - 35 °C

ΔT = 50 °C

Next, we can use the following formula to calculate the heat energy required:

Q = m·C·ΔT

where Q is the heat energy in calories, m is the mass of the sample in grams, C is the specific heat in cal/g °C, and ΔT is the change in temperature in °C.

Plugging in the given values, we get:

Q = 35.0 g · 0.108 cal/g °C · 50 °C

Q = 189.0 calories

Therefore, 189.0 calories are required to raise the temperature of the sample from 35 °C to 85 °C

Answer:

4897 J

Explanation:

The heat transferred to the surroundings, q_surr, can be calculated using the equation:

q_surr = -q_rxn = -CmΔT

where C is the specific heat capacity of the mixture (assumed to be the same as water, 4.184 J/g°C), m is the mass of the mixture (which we can calculate using the density, assuming that the volumes are additive), and ΔT is the change in temperature (in Celsius).

First, let's calculate the mass of the mixture:

density of water = 1.00 g/cm^3

volume of mixture = volume of acid + volume of base = 106 mL + 157 mL = 263 mL = 0.263 L

mass of mixture = density of water x volume of mixture = 1.00 g/cm^3 x 0.263 L = 263 g

Next, let's calculate the change in temperature:

ΔT = final temperature - initial temperature = 27.29°C - 22.94°C = 4.35°C

Now we can calculate the heat transferred to the surroundings:

q_surr = -CmΔT

q_surr = -(4.184 J/g°C) x (263 g) x (4.35°C)

q_surr = -4897 J

Note that the negative sign indicates that heat is lost by the system to the surroundings. Therefore, the heat transferred to the surroundings, q_surr, is 4897 J.

Answer:

Explanation:

2-bromopropane is the major product because the reaction mechanism involves the formation of the most stable carbocation intermediate. When hydrogen bromide reacts with propene, the hydrogen atom from HBr adds to the carbon atom of the double bond that has fewer hydrogen atoms attached, resulting in the formation of a carbocation intermediate. The intermediate can either form 1-bromopropane or 2-bromopropane depending on the position of the carbocation. The 2-bromopropane is the major product because the secondary carbocation formed in this case is more stable than the primary carbocation formed in the case of 1-bromopropane.

To test for an alkene, bromine water can be used. When an alkene reacts with bromine water, the bromine molecule adds across the double bond, forming a colorless dibromoalkane product. The mechanism for the reaction between hex-1-ene and bromine involves the formation of a cyclic bromonium ion intermediate, followed by the attack of water on the intermediate, resulting in the formation of the dibromoalkane product.

a) Addition polymerization is a process in which unsaturated monomers are joined together to form a polymer. The process involves breaking the double bond of the monomer and joining the monomers together to form a long-chain polymer. The process requires a catalyst to initiate the reaction.

b) The repeating unit of the polymer formed from the addition polymerization of chloroethene monomers is -CH2-CHCl-. This polymer is called polyvinyl chloride (PVC), and it has a wide range of uses, including pipes, electrical cables, and vinyl flooring.

122.5 grams of oxalic add dihydrate (MW = 126.07 g/mole) and disodium oxalate (MW = 133.99 g/mole) were required to prepare this buffer if the total oxalate concentration is 0.115 M.

Weak acids are defined as acids that don't completely dissociate in solution. It can be explained as any acid that is not a strong acid. The strength of a weak acid depends on how much it gets dissociates and the more it dissociates, the stronger the acid. The mass of the weak acid in a solution of a certain pH can be determined by calculating the original concentration of the acid after calculating the concentration of the hydrogen ions with the help of the pH value of the solution.

The Concentration of oxalate ion is  0.115 M.

pKa1 is 1.250.

pKa2 is 4.266.

pH is 5.193.

Molarity = (mass / molar mass) / 1 / volume in liter

The molar mass is 126.07g/mole.

Mass = Molarity × molar mass × Volume in liter

Mass=0.972 M × 126.07 g/mole × 1.00 L

        = 122.5 gram

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The complete question is,

A buffer prepared by dissolving oxalic add dihydrate (H2C2O4⋅2H2O) and disodium oxalate (Na2C2O4) in 1.00 L of water has a pH of 5.193. How many grams of oxalic add dihydrate (MW = 126.07 g/mole) and disodium oxalate (MW = 133.99 g/mole) were required to prepare this buffer if the total oxalate concentration is 0.115 M? Oxalic acid has pKa values of 1.250 (pKa1) and 4.266 (pKa2).

Answer:

lusterous metal

Explanation:

ex gold, iron etc hope it helps

Its atomic radius increases form top to bottom inside a group, then decreases from left and right across a period. As a result, francium is indeed the largest element while helium is the smallest.

Atomic radii inside the periodic table decrease across a row form left to right and increase across a column from top to bottom. Due to these two patterns, the periodic table's lower left and upper right corners, respectively, contain the largest and smallest atoms.

The atomic radius grows form top to bottom inside a group and decreases form left to right during a period, as seen in the images below. As a result, francium is indeed the largest element while helium is the smallest.

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a. The atom with the smaller atomic size is: Nitrogen

a. The atom with the smaller atomic size is: Arsenic.

Atomic size, also known as atomic radius, is the distance between the nucleus of an atom and its outermost electrons. It is typically measured in picometers (pm) or angstroms (Å). The atomic size of an element can be calculated by finding the distance between the nucleus and the outermost electron shell of an atom of that element. This distance can be determined using various methods, including X-ray diffraction and spectroscopy. The atomic size of elements generally decreases from left to right across a period and increases from top to bottom down a group in the periodic table.

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The major organic product of the condensation of ammonia or a primary amine with the carbonyl group of an aldehyde or ketone is an imine.

A functional group or organic substance with a carbon-nitrogen double bond (C=N) is known as an imine. A hydrogen atom or an organic group may be joined to the nitrogen atom. (R). The carbon atom is connected to two more single bonds. Imines are present in numerous processes and are frequently found in manufactured and naturally occurring chemicals.

The five core atoms for ketimines and aldimines, C2C=NX and C(H)C=NX, respectively, are coplanar. The sp2-hybridization of the mutually double-bonded nitrogen and carbon atoms yields planarity. For nonconjugated imines, the C=N distance is 1.29-1.31, whereas for conjugated imines, it is 1.35. The C-N distances in amines and nitriles, on the other hand, are 1.47 and 1.16, respectively. Slow rotation occurs around the C=N bond. E- and Z-isomers were detected using NMR spectroscopy of aldimines have been detected. Owing to steric effects, the E isomer is favored.

An imine is formed when a primary amine reacts with a carbonyl group (C=O) of an aldehyde or ketone to form a new C-N bond. This reaction is known as a condensation reaction, as it involves the loss of a small molecule (e.g. water) to form the product.

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The correct questions is :

What  is the major organic product of the condensation of ammonia or a primary amine with the carbonyl group of an aldehyde or ketone?

The empirical formula of a molecule containing 65.5% carbon, 5.5% hydrogen, and 29% oxygen is CH2O.

To calculate this, you need to first convert the percentage composition into mass composition. This is done by multiplying the percentages by the molecular weight of each element.

Carbon: 65.5% x 12 g/mol = 0.786 g/mol
Hydrogen: 5.5% x 1 g/mol = 0.055 g/mol
Oxygen: 29% x 16 g/mol = 0.464 g/mol
Now that you have the mass composition, you can calculate the moles of each element by dividing the mass of each element by its molar mass.
Carbon: 0.786 g/mol / 12 g/mol = 0.065 mol
Hydrogen: 0.055 g/mol / 1 g/mol = 0.055 mol
Oxygen: 0.464 g/mol / 16 g/mol = 0.029 mol
Finally, divide each element's moles by the smallest moles to get the empirical formula.

Carbon: 0.065 mol / 0.029 mol = 2.24 = 2 mol
Hydrogen: 0.055 mol / 0.029 mol = 1.90 = 1 mol
Oxygen: 0.029 mol / 0.029 mol = 1.00 = 1 mol
Therefore, the empirical formula of the molecule is CH2O.

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Answer:

Carbonothioyl dibromide, also known as CBr2S, has a bond angle of approximately 109.5 degrees, which is the typical tetrahedral bond angle for molecules with sp3 hybridization.

The molecular shape of CBr2S is also tetrahedral, with the two bromine atoms and the sulfur atom arranged at the corners of a tetrahedron, and the carbon atom at the center.

Answer: Manganese

Explanation:

With titanium, it only has two d electrons, so it can't form different high and low spin complexes. It doesn't matter because it will never fill the higher-energy orbitals. The total spin state turns out to be +1 (two unpaired d electrons, no matter what). Therefore, manganese will form both a high and low spin complex.

The R f values for compounds A, B, and C on a silica gel TLC plate developed in hexanes would be determined by measuring the distance each compound traveled compared to the distance the solvent traveled.

a) There is a 4 cm gap between the origin and the solvent front. The Rf value for spot A is[tex]\frac{1.5}{4}= 0.375[/tex], because it travelled 1.5 cm. Due to the 3.5 cm movement of Spot B, its Rf is[tex]\frac{3.5}{4} = 0.875[/tex]. Spot C shifted 3 cm, making its Rf [tex]\frac{3}{4} = 0.75[/tex].

b)Due to its shorter travel distance than the other two compounds, compound A is the most polar. Recall that polar substances adhere to the adsorbent more readily, move less, and have a lower Rf value.

c)Hexanes is less polar than acetone as a solvent. Each of the three compounds would move more quickly if the same method were employed to elute them.The chemicals can be removed from the polar adsorbent more effectively with a more polar eluting solvent. Each compound would have a higher Rf value if acetone were used to elute the TLC plate as opposed to hexanes because each compound travels more quickly.

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Answer: NaAlF6

Explanation:

1) divide by the smallest number of moles

0.953/.322 =2.96 round to 3 Na

.322/.322 = 1 Al

1.93/.322 =5.99 round to 6 F

2) write in order numbers were give

NaAlF6

The empirical formula, CH2O9(menthol) is multiplied by 5 to get the molecular formula, C10H20O.The empirical and molecular formula for menthol are CH2O and C10H20O, respectively.

The molecular formula for menthol is C5H10O.

This can be determined by dividing the molar mass of the empirical formula (156.26 g/mol) by the molar mass of CO2 (44.01 g/mol). This gives a ratio of 3.55, which is equal to the ratio of C atoms in the empirical formula, C10H20O.

Therefore, the molecular formula is C5H10O.
Given:

Menthol is composed of C, H, and O0.1005g sample of menthol is combusted and produces0.2829g of CO2 0.1159g H2O

1. Find: Empirical and molecular formula for menthol.

Let's first calculate the number of moles of CO2 produced. The balanced equation for combustion of menthol is:

C10H20O(s) + 12O2(g) → 10CO2(g) + 10H2O(l)

From the above equation, we can see that for 10 moles of CO2 produced 1 mole of menthol is required.

2. By taking the number of moles of CO2 produced, we can calculate the number of moles of menthol burned.

Moles of CO2 = 0.2829g / 44.01g/mol= 0.00643 mol

C10H20O(s) + 12O2(g) → 10CO2(g) + 10H2O(l)

From the balanced equation,1 mole of C10H20O requires 10 moles of CO2

Moles of C10H20O burned = 10 * 0.00643= 0.0643 mol

Next, we can calculate the number of moles of H2O produced.

Moles of H2O = 0.1159g / 18.015g/mol= 0.00643 mol

C10H20O(s) + 12O2(g) → 10CO2(g) + 10H2O(l)

From the balanced equation,1 mole of C10H20O requires 10 moles of H2O

Moles of C10H20O burned = 10 * 0.00643= 0.0643 mol

3. Now we can calculate the empirical formula of menthol. The empirical formula can be calculated as follows:

Empirical formula = CH2O (Divide all moles by smallest moles) The molecular weight of CH2O = 30 g/mol

The empirical formula mass of the compound is:

mass = (12.011 + 2*1.008 + 15.999) = 30.026

Empirical formula mass of CH2O is 30.026g/mol, and the given sample weighs 0.1005 g.
The number of empirical formula units in the sample is 0.1005 g / 30.026 g/mol = 0.003348Units.

Empirical formula = CH2OThe empirical formula weight of menthol is CH2O, which is equal to 30.026g/mol.

4. To find the molecular formula, we need to know the molecular weight of the menthol. We can calculate it as follows:

Molecular formula mass = Empirical formula mass x n

Where n = integer Molecular formula mass of menthol is 156 g/mol, and the empirical formula mass is 30.026 g/mol.

So, n = 156 g/mol ÷ 30.026 g/mol = 5.192

Thus the empirical formula, CH2O is multiplied by 5 to get the molecular formula, C10H20O.The empirical and molecular formula for menthol are CH2O and C10H20O, respectively.

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(S)-2-butanol will undergo an SN2 reaction with HBr to produce a racemic mixture of alkyl bromides. Here option B is the correct answer.

When optically active alcohol is treated with HBr, the reaction follows an SN1 or SN2 mechanism. In the case of SN1, a carbocation intermediate is formed, and in SN2, a backside attack by the nucleophile occurs. The stereochemistry of the product depends on the configuration of the intermediate and the direction of attack.

In the case of (S)-2-butanol, the hydroxyl group is attached to the second carbon atom, which makes it a primary alcohol. When treated with HBr, it undergoes an SN2 reaction, where the hydroxyl group is replaced by the bromine atom. The nucleophile attacks from the backside of the molecule, leading to an inversion of configuration.

This results in the formation of a racemic mixture of alkyl bromides, as both enantiomers have an equal chance of being attacked from either side. On the other hand, (R)-2-butanol, being the enantiomer of (S)-2-butanol, will also undergo the same reaction and produce the same racemic mixture of alkyl bromides.

In the case of (R)-1-phenyl ethanol and (S)-1-phenyl ethanol, they are secondary alcohols and can undergo either SN1 or SN2 reactions depending on the reaction conditions. However, the reaction mechanism will lead to the formation of a mixture of diastereomers, rather than a racemic mixture of enantiomers.

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Complete question:

Which of the following optically active alcohols, when treated with HBr, results in a racemic mixture of alkyl bromides?

a) (R)-2-butanol

b) (S)-2-butanol

c) (R)-1-phenyl ethanol

d) (S)-1-phenyl ethanol