4-methyl-3-hexanol was prepared by reacting an alkene with either hydroboration-oxidation or oxymercuration-reduction. Draw the structure of the alkene that was used to prepare the alcohol in highest yield.

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Answer 1

Answer:

Answer:

Structure in attachment.

Explanation:

The oxymercuration-demercuration of an asymmetric alkene usually produces the Markovnikov orientation of an addition. The electrophile ⁺Hg(OAc), formed by the electrophile attack of the mercury ion, remains attached to least substituted group at the end of the double bond. This electrophile has a considerable amount of positive charge on its two carbon atoms, but there is more positive charge on the more substituted carbon atom, where it is more stable. The water attack occurs on this more electrophilic carbon, and the Markovnikov orientation occurs.

In hydroboration, borane adds to the double bond in one step. Boron is added to the less hindered and less substituted carbon, and hydrogen is added to the more substituted carbon. The electrophilic boron atom adds to the less substituted end of the double bond, positioning the positive charge (and the hydrogen atom) at the more substituted end. The result is a product with the anti-Markovnikov orientation.

Calculating the pH

a) 0 mL

Initially, the pH of the solution is given by the dissociation of NH₃ in water.

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ (1)

The constant of the above reaction is:

$Kb = ([NH_(4)^(+)][OH^(-)])/([NH_(3)]) = 1.76\cdot 10^(-5)$ (2)

At the equilibrium, we have:

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ (3)

0.030 M - x x x

$1.76\cdot 10^(-5)*(0.030 - x) - x^(2) = 0$

After solving for x and taking the positive value:

x = 7.18x10⁻⁴ = [OH⁻]

Now, we can calculate the pH of the solution as follows:

$pH = 14 - pOH = 14 + log(7.18\cdot 10^(-4)) = 10.86$

Hence, the initial pH is 10.86.

b) 10 mL

After the addition of HCl, the following reaction takes place:

NH₃ + HCl ⇄ NH₄⁺ + Cl⁻ (4)

We can calculate the pH of the solution from the equilibrium reaction (3).

$1.76\cdot 10^(-5)(Cb - x) - (Ca + x)*x = 0$ (5)

Finding the number of moles of NH₃ and NH₄⁺

The number of moles of NH₃ (nb) and NH₄⁺ (na) are given by:

$n_(b) = n_(i) - n_(HCl)$ (6)

$n_(b) = 0.030 mol/L*0.030 L - 0.025 mol/L*0.010 L = 6.5\cdot 10^(-4) moles$

$n_(a) = n_(HCl)$ (7)

$n_(a) = 0.025 mol/L*0.010 L = 2.5 \cdot 10^(-4) moles$

Calculating the concentrations of NH₃ and NH₄⁺

The concentrations are given by:

$Cb = (6.5\cdot 10^(-4) moles)/((0.030 L + 0.010 L)) = 0.0163 M$ (8)

$Ca = (2.5 \cdot 10^(-4) mole)/((0.030 L + 0.010 L)) = 6.25 \cdot 10^(-3) M$ (9)

Calculating the pH

After entering the values of Ca and Cb into equation (5) and solving for x, we have:

$1.76\cdot 10^(-5)(0.0163 - x) - (6.25 \cdot 10^(-3) + x)*x = 0$

x = 4.54x10⁻⁵ = [OH⁻]

Then, the pH is:

$pH = 14 + log(4.54\cdot 10^(-5)) = 9.66$

Hence, the pH is 9.66.

c) 20 mL

We can find the pH of the solution from the reaction of equilibrium (3).

Calculating the concentrations of NH₃ and NH₄⁺

The concentrations are (eq 8 and 9):

$Cb = (0.030 mol/L*0.030 L - 0.025 mol/L*0.020 L)/((0.030 L + 0.020 L)) = 8.0\cdot 10^(-3) M$

$Ca = (0.025 mol/L*0.020 L)/((0.030 L + 0.020 L)) = 0.01 M$

Calculating the pH

After solving the equation (5) for x, we have:

$1.76\cdot 10^(-5)(8.0\cdot 10^(-3) - x) - (0.01 + x)*x = 0$

x = 1.40x10⁻⁵ = [OH⁻]

Then, the pH is:

$pH = 14 + log(1.40\cdot 10^(-5)) = 9.15$

So, the pH is 9.15.

d) 35 mL

We can find the pH of the solution from reaction (3).

Calculating the concentrations of NH₃ and NH₄⁺

$Cb = (0.030 mol/L*0.030 L - 0.025 mol/L*0.035 L)/((0.030 L + 0.035 L)) = 3.85\cdot 10^(-4) M$

$Ca = (0.025 mol/L*0.035 L)/((0.030 L + 0.035 L)) = 0.0135 M$

Calculating the pH

After solving the equation (5) for x, we have:

$1.76\cdot 10^(-5)(3.85\cdot 10^(-4) - x) - (0.0135 + x)*x = 0$

x = 5.013x10⁻⁷ = [OH⁻]

Then, the pH is:

$pH = 14 + log(5.013\cdot 10^(-7)) = 7.70$

So, the pH is 7.70.

e) 36 mL

Finding the number of moles of NH₃ and NH₄⁺

$n_(b) = 0.030 mol/L*0.030 L - 0.025 mol/L*0.036 L = 0$

$n_(a) = 0.025 mol/L*0.036 L = 9.0 \cdot 10^(-4) moles$

Since all the NH₃ reacts with the HCl added, the pH of the solution is given by the dissociation reaction of the NH₄⁺ produced in water.

At the equilibrium, we have:

NH₄⁺ + H₂O ⇄ NH₃ + H₃O⁺

Ca - x x x

$Ka = (x^(2))/(Ca - x)$

$Ka(Ca - x) - x^(2) = 0$ (10)

Calculating the acid constant of NH₄⁺

We can find the acid constant as follows:

$Kw = Ka*Kb$

Where Kw is the constant of water = 10⁻¹⁴

$Ka = (1\cdot 10^(-14))/(1.76 \cdot 10^(-5)) = 5.68 \cdot 10^(-10)$

Calculating the pH

The concentration of NH₄⁺ is:

$Ca = (9.0 \cdot 10^(-4) moles)/((0.030 L + 0.036 L)) = 0.0136 M$

After solving the equation (10) for x, we have:

x = 2.78x10⁻⁶ = [H₃O⁺]

Then, the pH is:

$pH = -log(H_(3)O^(+)) = -log(2.78\cdot 10^(-6)) = 5.56$

Hence, the pH is 5.56.

f) 37 mL

Now, the pH is given by the concentration of HCl that remain in solution after reacting with NH₃ (HCl is in excess).

Calculating the concentration of HCl

$C_(HCl) = (0.025 mol/L*0.037 L - 0.030 mol/L*0.030 L)/((0.030 L + 0.037 L)) = 3.73 \cdot 10^(-4) M = [H_(3)O^(+)]$

Calculating the pH

$pH = -log(H_(3)O^(+)) = -log(3.73 \cdot 10^(-4)) = 3.43$

Therefore, the pH is 3.43.

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Answer:

a)10.87

b)9.66

c)9.15

d)7.71

e) 5.56

f) 3.43

Explanation:

tep 1: Data given

Volume of 0.030 M NH3 solution = 30 mL = 0.030 L

Molarity of the HCl solution = 0.025 M

Step 2: Adding 0 mL of HCl

The reaction: NH3 + H2O ⇔ NH4+ + OH-

The initial concentration:

[NH3] = 0.030M [NH4+] = 0M [OH-] = OM

The concentration at the equilibrium:

[NH3] = 0.030 - XM

[NH4+] = [OH-] = XM

Kb = ([NH4+][OH-])/[NH3]

1.8*10^-5 = x² / 0.030-x

1.8*10^-5 = x² / 0.030

x = 7.35 * 10^-4 = [OH-]

pOH = -log [7.35 * 10^-4]

pOH = 3.13

pH = 14-3.13 = 10.87

Step 3: After adding 10 mL of HCl

The reaction:

NH3 + HCl ⇔ NH4+ + Cl-

NH3 + H3O+ ⇔ NH4+ + H2O

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.010 L = 0.00025 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.00025 =0.00065 moles

Moles HCl = 0

Moles NH4+ = 0.00025 moles

Concentration at the equilibrium:

[NH3]= 0.00065 moles / 0.040 L = 0.01625M

[NH4+] = 0.00625 M

pOH = pKb + log [NH4+]/[NH3]

pOH = 4.75 + log (0.00625/0.01625)

pOH = 4.34

pH = 9.66

Step 3: Adding 20 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.020 L = 0.00050 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.00050 =0.00040 moles

Moles HCl = 0

Moles NH4+ = 0.00050 moles

Concentration at the equilibrium:

[NH3]= 0.00040 moles / 0.050 L = 0.008M

[NH4+] = 0.01 M

pOH = pKb + log [NH4+]/[NH3]

pOH = 4.75 + log (0.01/0.008)

pOH = 4.85

pH = 14 - 4.85 = 9.15

Step 4: Adding 35 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.035 L = 0.000875 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.000875 =0.000025 moles

Moles HCl = 0

Moles NH4+ = 0.000875 moles

Concentration at the equilibrium:

[NH3]= 0.000025 moles / 0.065 L = 3.85*10^-4M

[NH4+] = 0.000875 M / 0.065 L = 0.0135 M

pOH = pKb + log [NH4+]/[NH3]

pOH = 4.75 + log (0.0135/3.85*10^-4)

pOH = 6.29

pH = 14 - 6.29 = 7.71

Step 5: adding 36 mL HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.036 L = 0.0009 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.0009 =0 moles

Moles HCl = 0

Moles NH4+ = 0.0009 moles

[NH4+] = 0.0009 moles / 0.066 L = 0.0136 M

Kw = Ka * Kb

Ka = 10^-14 / 1.8*10^-5

Ka = 5.6 * 10^-10

Ka = [NH3][H3O+] / [NH4+]

Ka =5.6 * 10^-10 = x² / 0.0136

x = 2.76 * 10^-6 = [H3O+]

pH = -log(2.76 * 10^-6)

pH = 5.56

Step 6: Adding 37 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.037 L = 0.000925 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.000925 =0 moles

Moles HCl = 0.000025 moles

Concentration of HCl = 0.000025 moles / 0.067 L = 3.73 * 10^-4 M

pH = -log 3.73*10^-4= 3.43

Whats the difference between china and Europe

Answers

Answer:

the difference is they arnt close

Explanation:

Answer:

China is communist, Europe isn't.

Explanation:

On another planet, the isotopes of titanium have the following natural abundances. a. Isotope 46Ti Abundance 70.900% Mass(amu) 45.95263
b. Isotope 48Ti Abundance 10.000% Mass(amu) 47.94795
c. Isotope 50Ti Abundance 19.100% Mass(amu) 49.94479
d. What is the average atomic mass of titanium on that planet?
e. I got 46.9 amu but it is wrong.

Answers

Answer:

Average atomic mass = 46.91466 amu

Explanation:

Step 1: Data given

Isotopes of titanium

46Ti = 70.900% ⇒ 45.95263 amu

48Ti = 10.000 % ⇒ 47.94795 amu

50Ti = 19.100 % ⇒ 49.94479 amu

Step 2: Calculate the average atomic mass of titanium

Average atomic mass = 0.7090 * 45.95263 + 0.10 * 47.94795 + 0.1910 * 49.94479

Average atomic mass = 46.91466 amu

What isotope has a mass number of 18 and an atomic number of 7

Answers

Answer:

nitrogen 18

nitrogen has an atomic number of 7

Describe the range of radii of most atoms in nanometers (nm)

Answers

Final answer:

The range of radii of most atoms is typically in the nanometer scale (nm) and can be measured using the covalent radius. The size of an atom's nucleus is much smaller than the atom itself. The Bohr model provides a formula to calculate the radius of hydrogen-like atoms.

Explanation:

The range of radii of most atoms is typically in the nanometer scale (nm). The covalent radius, which is defined as half the distance between the nuclei of two identical atoms when they are joined by a covalent bond, provides a practical way to measure the size of atoms. As we move down a group in the periodic table, the covalent radius generally increases, indicating a larger size of the atom. For example, the covalent radius of the halogens increases as we move from fluorine to iodine.

The size of an atom's nucleus, on the other hand, is much smaller than the atom itself. The nucleus has a diameter of about 10-15 meters, while the typical atom has a diameter of the order of 10-10 meters. This difference in size illustrates the emptiness of atoms, with the distance from the nucleus to the electrons being typically 100,000 times the size of the nucleus.

The Bohr model provides a formula to calculate the radius of hydrogen-like atoms, which depends on the principal quantum number (n) and the atomic number (Z). The calculated radii of the orbits of the hydrogen atom have been experimentally verified to have a diameter of a hydrogen atom.

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Final answer:

The range of radii of most atoms is typically measured in nanometers (nm). Covalent radius and hydrogen-like orbits are two methods used to estimate the size of atoms. The size of an atom can vary depending on the element and measurement technique, but most atoms have radii on the order of nanometers (nm).

Explanation:

The range of radii of most atoms is typically measured in nanometers (nm). The size of an atom can be estimated using various techniques. One commonly used measure is the covalent radius, which is defined as one-half the distance between the nuclei of two identical atoms when they are joined by a covalent bond. The covalent radii of different elements can be found in tables and can vary depending on the element and its position in the periodic table.

Another way to estimate the size of atoms is by looking at the sizes of their orbits in hydrogen-like atoms. These orbits are given in terms of their radii by a mathematical expression that includes a constant called the Bohr radius, which is approximately 5.292 × 10-11 m.

Overall, the size of an atom can vary depending on the element and the specific measurement technique used, but most atoms have radii on the order of nanometers (nm).

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A chemist titrates 60.0 mL of a 0.1935 M benzoic acid (HC (H5CO2) solution with 0.2088 M KOH solution at 25 °C. Calculate the pH at equivalence. The pKg of benzoic acid is 4.20.

Answers

Answer:

pH at the equivalence point is 8.6

Explanation:

A titulation between a weak acid and a strong base, gives a basic pH at the equivalence point. In the equivalence point, we need to know the volume of base we added, so:

mmoles acid = mmoles of base

60 mL . 0.1935M = 0.2088 M . volume

(60 mL . 0.1935M) /0.2088 M = 55.6 mL of KOH

The neutralization is:

HBz + KOH ⇄ KBz + H₂O

In the equilibrum:

HBz + OH⁻ ⇄ Bz⁻ + H₂O

mmoles of acid are: 11.61 and mmoles of base are: 11.61

So in the equilibrium we have, 11.61 mmoles of benzoate.

[Bz⁻] = 11.61 mmoles / (volume acid + volume base)

[Bz⁻] = 11.61 mmoles / 60 mL + 55.6 mL = 0.100 M

The conjugate strong base reacts:

Bz⁻ + H₂O ⇄ HBz + OH⁻ Kb

0.1 - x x x

(We don't have pKb, but we can calculate it from pKa)

14 - 4.2 = 9.80 → pKb → 10⁻⁹'⁸ = 1.58×10⁻¹⁰ → Kb

Kb = [HBz] . [OH⁻] / [Bz⁻]

Kb = x² / (0.1 - x)

As Kb is so small, we can avoid the quadratic equation

Kb = x² / 0.1 → Kb . 0.1 = x²

√ 1.58×10⁻¹¹ = [OH⁻] = 3.98 ×10⁻⁶ M

From this value, we calculate pOH and afterwards, pH (14 - pOH)

- log [OH⁻] = pOH → - log 3.98 ×10⁻⁶ = 5.4

pH = 8.6

Final answer:

To calculate the pH at equivalence in a titration, we need to consider the concentration of the excess strong base in the solution. First, we calculate the moles of the acid and the base, then we find the moles of the excess base. Using this information, we can find the concentration of the excess base and subsequently calculate pOH. Finally, we can convert pOH to pH using the pH + pOH = 14 relationship.

Explanation:

pH at the equivalence point in a titration can be determined by considering the concentration of the excess strong base present in the reaction mixture. In this case, the excess strong base is KOH. We can calculate [OH-] using the stoichiometry of the reaction and the given concentrations. Then, we can find the pOH using the formula -log[OH-]. Finally, we can convert pOH to pH using the pH + pOH = 14 relationship.

Given:

Volume of benzoic acid solution (HC (H5CO2)): 60.0 mL
Concentration of benzoic acid solution: 0.1935 M
Concentration of KOH solution: 0.2088 M

Step 1: Determine the amount of benzoic acid (HC (H5CO2)) in moles:

moles of HC (H5CO2) = volume (L) × concentration (M) = 0.0600 L × 0.1935 M = 0.01161 mol

Step 2: Determine the amount of KOH in moles:

moles of KOH = volume (L) × concentration (M) = 0.0600 L × 0.2088 M = 0.01253 mol

Step 3: Determine the amount of excess KOH in moles:

moles of excess KOH = moles of KOH - moles required for neutralizing HC (H5CO2) = 0.01253 mol - 0.01161 mol = 9.2 × 10-4 mol

Step 4: Determine the concentration of excess KOH:

concentration of excess KOH = moles of excess KOH / volume (L) = 9.2 × 10-4 mol / 0.0600 L = 0.0153 M

Step 5: Determine the pOH of the solution:

pOH = -log[OH-] = -log(0.0153) ≈ 1.82

Step 6: Determine the pH of the solution:

pH = 14 - pOH = 14 - 1.82 ≈ 12.18

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4-methyl-3-hexanol was prepared by reacting an alkene with either hydroboration-oxidation or oxymercuration-reduction. Draw the structure of the alkene that was used to prepare the alcohol in highest yield.

Answers

Related Questions

Answers

Calculating the pH

a) 0 mL

b) 10 mL

Finding the number of moles of NH₃ and NH₄⁺

Calculating the concentrations of NH₃ and NH₄⁺

Calculating the pH

c) 20 mL

Calculating the concentrations of NH₃ and NH₄⁺

Calculating the pH

d) 35 mL

Calculating the concentrations of NH₃ and NH₄⁺

Calculating the pH

e) 36 mL

Finding the number of moles of NH₃ and NH₄⁺

Calculating the acid constant of NH₄⁺

Calculating the pH

f) 37 mL

Calculating the concentration of HCl

Calculating the pH

Answers

Answers

Answers

Answers

Final answer:

Explanation:

Learn more about Atomic Radii here:

Final answer:

Explanation:

Learn more about Sizes of Atoms here:

Answers

Final answer:

Explanation:

Learn more about pH at equivalence point here: