Draw the curved arrow mechanism for the formation of an acetal from acidic methanol and 4-methylpentan-2-one in the fewest steps. When given the choice, draw the arrows that lead to the resonance structures with full octets around each atom other than hydrogen. Do not show any inorganic byproducts or counterions. Reagents needed for each step are provided in the boxes.

Answer:

empirical formula: [tex]H_2SO_4[/tex]

2 g H

32.7 g S

65.3 g O

Explanation:

Like the problem said, the first thing we can do is calculate the mass of each of the 3 elements in a 100-gram sample:

- 2.00% * 100g = 2 g H

- 32.7% * 100g = 32.7 g S

- 65.3% * 100g = 65.3 g O

Now we need to find the empirical formula from these. To do so, convert all of those masses into moles by using the molar mass for each element:

- the molar mass of H is 1.01 g/mol

- the molar mass of S is 32.06 g/mol

- the molar mass of O is 16 g/mol

2 g H ÷ 1.01 g/mol = 1.98 mol H

32.7 g S ÷ 32.06 g/mol = 1.02 mol S

65.3 g O ÷ 16 g/mol = 4.08 mol O

Our ratio of H : S : O is now:

1.98 mol : 1.02 mol : 4.08 mol

Divide them all by the smallest number, which is 1.02:

1.98/1.02  :  1.02/1.02  :  4.08/1.02

1.94 : 1 : 4

1.94 ≈ 2

So:

2 : 1 : 4

Thus, the empirical formula is: [tex]H_2SO_4[/tex].

Answer:

Naming ionic compounds with transition metals isn't too hard either. They are named like the binary compounds, with the cation first, then the anion with -ide added to it, but you have to take into account the variations of the metal ions.

Explanation:

Answer:

Explanation:

The fact that each isotope has one proton makes them all variants of hydrogen: the identity of the isotope is given by the number of protons and neutrons. From left to right, the isotopes are protium (1H) with zero neutrons, deuterium (2H) with one neutron, and tritium (3H) with two neutrons.

The enthalpy change when 1.6 kg of methane gas is heated from a temperature of 280 K to 320 K is 140.6 kJ.

The subject this question pertains to is heat capacity and enthalpy change in the field of Chemistry. Simply put, when the temperature of an amount of substance changes, the change in enthalpy (∆H), can be calculated using the formula, ∆H = nCp∆T, where n is the number of moles, Cp is the heat capacity at constant pressure, and ∆T is the change in temperature.

To calculate the answer to your question, we need to convert the mass of methane gas to moles because the molar heat capacity is given. The molar mass of methane (CH4) is approximately 16.04 g/mol. Thus, we have: n (number of moles) = mass (in kg) / molar mass (in kg/mol) = 1.6 kg / 0.01604 kg/mol = 99.75 mol.

The next step is to plug our numbers into the formula: ∆H = nCp∆T = 99.75 mol * 35.31 JK-1mol-1 * (320 K - 280 K) = 140.6 kJ.

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1. **Alkylation**: Diethyl malonate reacts with 1,5-dibromopentane to form the alkylated product.

2. **Hydrolysis**: The alkylated product undergoes acidic hydrolysis to yield a carboxylic acid.

3. **Decarboxylation**: Carboxylic acid formed undergoes decarboxylation to produce the final compound, 1,3-propanediol.

let's break down the reaction step by step:

1. **Alkylation with 1,5-dibromopentane**:

The malonic ester synthesis involves the alkylation of diethyl malonate. In this case, one equivalent of 1,5-dibromopentane is used.

The reaction proceeds via an S_N2 mechanism, where the nucleophilic oxygen attacks the electrophilic carbon of the alkyl halide. The bromine atom is displaced, forming a new carbon-carbon bond.

The reaction can be represented as:

[tex]\[ \text{CH}_2(\text{CO}_2\text{Et})_2 + \text{Br(CH}_2\text{)_4Br} \rightarrow \text{CH}_2(\text{CO}_2\text{Et})_3\text{CH}_2\text{Br} + \text{Br}^-\][/tex]

2. **Hydrolysis with acidic workup**:

The product from the alkylation step undergoes hydrolysis under acidic conditions. This step cleaves the ester groups, yielding carboxylic acids.

[tex]\[ \text{CH}_2(\text{CO}_2\text{Et})_3\text{CH}_2\text{Br} + 2\text{H}_3\text{O}^+ \rightarrow \text{CH}_2(\text{CO}_2\text{H})_3\text{CH}_2\text{OH} + 2\text{EtOH}\][/tex]

3. **Decarboxylation**:

Decarboxylation involves the removal of a carboxyl group from a molecule as carbon dioxide. In this case, since there are three carboxyl groups in the product, three equivalents of carbon dioxide will be released.

[tex]\[ \text{CH}_2(\text{CO}_2\text{H})_3\text{CH}_2\text{OH} \rightarrow \text{CH}_2(\text{CO}_2\text{H})_2\text{CH}_2\text{OH} + \text{CO}_2\][/tex]

The resulting compound is 1,3-propanediol.

So, the carboxylic acid formed in the reaction is glyceric acid (1,3-propanediol).

Diagram is attached

Answer:

The equivalent two-dimensional point lattice for the area-centered hexagon is a rhombus.

Explanation:

The area centered hexagon is illustrated with a centered figure that has dotted center and many other dots around it that connect each other.

In this case, we need to draw the area centered hexagon first.

After drawing, we then connect the centered atoms of the hexagon. This connected centered atoms now form a rhombus like shape

Note: The shape of a rhombus is said to be flat, and it has 4 equal sides.

The pressure of the nitrogen gas confined to a volume of 0.118 L is 13.16 atm.

The pressure of nitrogen gas can be calculated by using the ideal gas equation.

PV =nRT

P = pressure

V = volume = 0.118 L

n =  moles of nitrogen gas

moles = [tex]\rm\dfrac{weight}{molecular\;weight}[/tex]

moles of nitrogen gas = [tex]\rm \dfrac{1.78}{28}[/tex]

Moles of nitrogen gas = 0.0635 moles

R = constant = 0.0821

T = temperature = 25[tex]\rm ^\circ C[/tex] = 298 K

Substituting the values:

P [tex]\times[/tex] 0.118 = 0.0635 [tex]\times[/tex] 0.0821 [tex]\times[/tex] 298

P [tex]\times[/tex] 0.118 = 1.5535

P = 13.16 atm.

The pressure of the nitrogen gas confined to a volume of 0.118 L is 13.16 atm.

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The pressure of the nitrogen gas confined to a volume of 0.118 L at 25°C is approximately 13.16 atmospheres.

To solve this problem, we will use the ideal gas law, which is given by the equation:

[tex]\[ PV = nRT \][/tex]

where:

-  P  is the pressure of the gas,

-  V  is the volume of the gas,

-  n  is the number of moles of the gas,

- R is the ideal gas constant, and

- T  is the temperature in Kelvin.

First, we need to calculate the number of moles of nitrogen gas [tex](\( N_2 \))[/tex]using the molar mass of nitrogen, which is approximately 28.02 g/mol.

[tex]\[ T = 25^\circ \text{C} + 273.15 = 298.15 \text{ K} \][/tex]

Next, we convert the temperature from Celsius to Kelvin by adding 273.15 to the given temperature:

[tex]\[ T = 25^\circ \text{C} + 273.15 = 298.15 \text{ K} \][/tex]

Now, we can rearrange the ideal gas law to solve for pressure  P:

[tex]\[ P = \frac{nRT}{V} \][/tex]

The ideal gas constant R  is approximately 0.0821 L·atm/(mo·K). Plugging in the values we have:

[tex]\[ P = \frac{(0.0635 \text{ mol})(0.0821 \text{ L·atm/(mol·K)})(298.15 \text{ K})}{0.118 \text{ L}} \] \[ P \approx \frac{(0.0635)(0.0821)(298.15)}{0.118} \] \[ P \approx \frac{1.553}{0.118} \] \[ P \approx 13.16 \text{ atm} \][/tex]

Answer:

It opens the can for him

Explanation:

Answer:

Takes the pain off the mans hand.

Explanation:

Saw that you didn't like europa2433, i just kinda wanted to agree he reports to much.

The pH at each volume of added acid is calculated using the Henderson-Hasselbalch equation.

The pH of a solution can be calculated using the equation:

pH = -log[H+]

Using the Henderson-Hasselbalch equation, pH can be calculated at each volume of added acid:

The initial concentration of pyridine is 0.125 M. Since pyridine is a weak base, [H+] can be calculated using the equation:

Kw / Kb = [H+]

where Kw is the equilibrium constant for water (1.0 x 10^-14) and Kb is the base dissociation constant for pyridine (1.7 x 10^-9). Substituting the values, we get:

[H+] = (1.0 x 10^-14) / (1.7 x 10^-9) ≈ 5.88 x 10^-6 M

Now, we can calculate the pH:

pH = -log(5.88 x 10^-6) ≈ 5.23

Using stoichiometry, we can determine the moles of HCI added:

moles of HCI = concentration of HCI x volume of HCI

moles of HCI = (0.100 M) x (0.010 L) = 0.001 mol

Since the acid HCl is strong, all of it will react with the pyridine:

moles of pyridine reacted = moles of HCI added = 0.001 mol

The final volume of the solution is 25.0 mL + 10 mL = 35.0 mL. The concentration of pyridine after the reaction is:

concentration of pyridine after reaction = moles of pyridine / final volume of solution

concentration of pyridine after reaction = (0.125 M)(25.0 mL) / (35.0 mL) ≈ 0.0893 M

Using the Henderson-Hasselbalch equation, we can calculate the pH:

pH = pKa + log ([A-]/[HA])

where pKa is the negative logarithm of the acid dissociation constant for pyridine (9.27) and [A-]/[HA] is the ratio of the conjugate base to the weak acid, which is equal to the ratio of the concentration of pyridine after reaction to the initial concentration of pyridine:

pH = 9.27 + log (0.0893 / 0.125) ≈ 9.07

Using the same approach, we can calculate the concentration of pyridine after the reaction:

concentration of pyridine after reaction = (0.125 M)(25.0 mL) / (45.0 mL) ≈ 0.0694 M

Using the Henderson-Hasselbalch equation, we can calculate the pH:

pH = 9.27 + log (0.0694 / 0.125) ≈ 8.88

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To calculate the pH at each stage, we consider the concentrations of the pyridine and the HCl, and use the formulas for pH and pOH. At 0 mL of HCl, we would calculate pH from the pKa of pyridine. At 10 mL and 20 mL, we need to consider that pyridine is in excess, and calculate pOH first before getting the pH.

The subject of this question is acid-base titration, specifically the titration of pyridine with hydrochloric acid (HCl). To calculate the pH at different volumes of added acid, we'd use the formula for the concentration of the pyridine and HCl.

1. At 0 mL of HCl, the solution is just the pyridine, which is a weak base. We don't have the pKa for pyridine in the question, but assuming we did, we could calculate the pH using the pKa and the formula pH = 14 - pKa. Let's say it's around 5.2.

2. At 10 mL of 0.100 M HCl (1.0 mmol), and 25 mL of 0.125 M pyridine (3.125 mmol), we have more base than acid, so pyridine is in excess. We'd use the equilibrium expression for the reaction of excess pyridine with water to find the pOH, and then calculate the pH.

3. At 20 mL of 0.100 M HCl (2.0 mmol), we still have excess pyridine, and so we'd perform a similar calculation as at 10 mL.

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Answer:

= 14.24g of [tex](NH4)_{2}CO_{3}[/tex] is required.

Explanation:

Reaction equation:

[tex](NH4)_{2}CO_{3}[/tex] → [tex]2NH_{3} + CO_{2} + H_{2} O[/tex]

Mole ratio of ammonium carbonate to carbon dioxide is 1:1

1 mole of CO2 - 44g

?? mole of CO2 - 6.52g

= 6.52/44 = 0.148 moles was produced from this experiment.

Therefore, if 1 mole of [tex](NH4)_{2}CO_{3}[/tex] - 96.09 g

0.148 mol of [tex](NH4)_{2}CO_{3}[/tex] --  ?? g

=0.148 × 96.09

= 14.24g of [tex](NH4)_{2}CO_{3}[/tex] is required.

Answer:  B) Ag(s) is formed at the cathode, and [tex]Mn^{2+}(aq)[/tex] is formed at the anode.

Explanation:

[tex]Ag^{+}(aq)+e^{-1}\rightarrow Ag(s)[/tex]  E=0.80 V

[tex]Mn^{2+}(aq)+2e^{-1}\rightarrow Mn(s)[/tex]  E=-1.18 V

Reduction takes place easily if the standard reduction potential is higher (positive) and oxidation takes place easily if the standard reduction potential is less(more negative).

Thus as reduction potential of Ag is higher , it undergoes reduction and Manganese with lower reduction potential undergoes oxidation. Here Mn undergoes oxidation by loss of electrons, thus act as anode. Ag undergoes reduction by gain of electrons and thus act as cathode.

Cathode : reduction : [tex]Ag^{+}(aq)+e^{-1}\rightarrow Ag(s)[/tex]

Anode : oxidation : [tex]Mn\rightarrow Mn^{2+}(aq)+2e^{-1}[/tex]

Ag(s) is formed at the cathode, and [tex]Mn^{2+}(aq)[/tex] is formed at the anode.

Answer:

1) increases

2) smaller

Explanation:

Generally, as electron- electron repulsion increases and more electrons are added to the atom while Z is held constant, the electron cloud size is increased. The size of the anion formed is usually measured as the size of this extended electron cloud. Hence the larger electron cloud means a larger anion size compared to the size of the neutral atom.

For a cation, the converse is true and the cation is found to be smaller than the neutral atom.

This question is incomplete, I got the complete one from google as below:

I−>I>I+

1. As Z stays constant and the number of electrons increases, the electron-electron repulsions ______ (increases or decreases), and the anion becomes larger.

2. The reverse is true for the cation, which becomes ____ (smaller or larger) than the neutral atom.

Answer:

1. As Z stays constant and the number of electrons increases, the electron-electron repulsion increases, and the anion becomes larger.

2.The reverse is true for the cation, which becomes smaller than the neutral atom.

Explanation:

1. As Z stays constant and the number of electrons increases, the electron-electron repulsion increases, and the anion becomes larger. This is because in anions of the same atoms, the net force of attraction on electrons decreases.

2. The reverse is true for the cation, which becomes smaller than the neutral atom. This is because in cations of the same atoms, the net force of attraction on electrons increases.

Answer:

The following organelles are common to both plant and animal cells:

(N.B: the other options are peculiar to plant cells only)

Explanation:

Plant and animal cells share similarities in components due to the fact that the are both eukaryotic cells ( having DNA as genetic composition).

The organelles common to both plant and animal cells are the cell membrane, nucleus, mitochondria, and cytoplasm.

The organelles that are common to both plant and animal cells are the cell membrane, nucleus, mitochondria, and cytoplasm. Both plant and animal cells have a cell membrane, which controls the movement of substances in and out of the cell. The nucleus is the control center of the cell, responsible for storing and processing genetic information. Mitochondria are the powerhouse of the cell, producing energy through cellular respiration. Cytoplasm is a jelly-like substance that fills the cell and houses the organelles.

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Answer:

An example of engineering material, are plastics, they are derived from organic, natural materials, such as cellulose, coal, natural gas, salt and, of course, oil. Oil is a complex mixture of thousands of compounds and must be processed before being used.

Explanation:

Plastic production begins with distillation at a refinery, where crude oil is separated into groups of lighter components, called fractions. Each fraction is a mixture of hydrocarbon chains (chemical compounds formed by carbon and hydrogen) that differ in terms of the size and structure of their molecules. One of those fractions, naphtha, is the essential compound for the production of plastic.

Two main processes are used to make plastic: polymerization and polycondensation, and both require specific catalysts. In a polymerization reactor, monomers like ethylene and propylene join to form long polymer chains. Each polymer has its own properties, structure and dimensions depending on the type of basic monomer that has been used.

Answer:4.920 X10^-22J

Explanation:

Energy of infrared radiation = hc/λ

where h=plank's constant=6.626X 10^-34js

c= speed of light = 3.0 x 10^8ms⁻¹

wavelength λ= 4.04x 10^-4m

E = 6.626X 10^-34js X 3.0 X 10^8 ms⁻¹/ 4.04 X 10^-4m

=4.920 X10^-22J

A current of 620 A for 90 seconds produces 5.209 g of aluminum.

The mass of pure aluminum produced in the Hall-Heroult process can be calculated by using Faraday's laws of electrolysis and the molar mass of aluminum. First, we determine the moles of electrons transferred using the given current and time. Then, using the stoichiometry of the reaction where 4 moles of Al are produced per 12 moles of electrons:

Calculate the charge (Q) passed: Q = current (I) x time (t) = 620 A x 90.0 s = 55800 C

Determine the moles of electrons: Moles of e- = Q / Faraday's constant (F) Assuming the Faraday constant is approximately 96500 C/mol, this gives Moles of e- = 55800 C / 96500 C/mol = 0.5784 mol

Calculate the moles of aluminum: Since 3 moles of electrons yield 2 moles of Al, 0.5784 mol of e- will yield (4/12) x 0.5784 mol of Al = 0.1931 mol of Al.

Compute the mass of aluminum produced: Mass of Al = moles of Al x molar mass of Al = 0.1931 mol x 26.98 g/mol = 5.209 g

The mass of pure aluminum produced is therefore 5.209 g, assuming 100% efficiency and no other side reactions.

To find the mass of pure aluminum produced, we calculate the charge passed, find the moles of electrons transferred, determine the moles of aluminum from the moles of electrons, and use the molar mass of aluminum to find that approximately 5.20 grams of aluminum is produced.

To calculate the mass of pure aluminum produced in the Hall-Heroult process when 620 A of current is passed through a Hall-Heroult cell for 90.0 seconds, we use Faraday's laws of electrolysis. The reduction of aluminum oxide to aluminum involves the transfer of three moles of electrons per mole of aluminum according to the equation:

Al₂O₃ + 3e⁻ → 2Al + 1.5O₂

Firstly, we calculate the amount of charge passed using the formula Q = It, where Q is the charge in coulombs, I is the current in amperes, and t is time in seconds. Then, using Faraday's constant (96485 C/mol), we determine the moles of electrons transferred. Since we need 3 moles of electrons to produce 1 mole of aluminum, we find the moles of aluminum formed by dividing the moles of electrons by 3. Finally, we calculate the mass using the molar mass of aluminum (26.98 g/mol).

Q = 620 A * 90.0 s = 55800 C
Moles of electrons = Q/F = 55800 C / 96485 C/mol ≈ 0.5785 mol
Moles of aluminum = Moles of electrons / 3 ≈ 0.1928 mol
Mass of aluminum = Moles of aluminum * Molar mass of Al
Mass of aluminum ≈ 0.1928 mol * 26.98 g/mol ≈ 5.20 g

The mass of pure aluminum produced is approximately 5.20 grams.

Answer:

solubility of X in water at 17.0 [tex]^{0}\textrm{C}[/tex] is 0.11 g/mL.

Explanation:

Yes, the solubility of X in water at 17.0 [tex]^{0}\textrm{C}[/tex] can be calculated using the information given.

Let's assume solubility of X in water at 17.0 [tex]^{0}\textrm{C}[/tex] is y g/mL

The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.

So, solubility of X in 1 mL of water = y g

Hence, solubility of X in 36.0 mL of water = 36y g

So, 36y = 3.96

   or, y = [tex]\frac{3.96}{36}[/tex] = 0.11

Hence solubility of X in water at 17.0 [tex]^{0}\textrm{C}[/tex] is 0.11 g/mL.

Answer:

Check the explanation

Explanation:

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

Answer:

It connects the two half‑reactions.

Explanation:

A salt bridge is a laboratory device used to connect the oxidation and reduction of half-cells of a galvanic cell. It prevents the cell from rapidly running its reaction to equilibrium.

Salt bridge or a porous plate connects the solutions of the half-cells that allow ions to pass from one solution to the other. This process balances the charges of the solutions and allows the reaction to continue.

The function of the salt bridge with respect to the electrochemical cell should be that it connected the two -half reactions.

It is the device that should be used in the laboratory for connecting the oxidation and the reduction of half-cells. It also prevents the cell from suddenly run from its reaction to the equilibrium.  Also, it connect the half cell solution where it probide the permission for ions from one solution to another solution. Moreover, there is balancing of the solutions charges due to this it permits the reaction for continuing it.

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Answer:

caffience

Explanation:

Answer:

Check the explanation

Explanation:

As we know the reaction of  EDTA and [tex]Ga^3[/tex]+ and EDTA and [tex]V^3[/tex]+

Let us say that the ratio is 1:1

Therefore, the number of moles of [tex]Ga^3[/tex]+ = molarity * volume

                                    = 0.0400M * 0.011L

                                    = 0.00044 moles

Therefore excess EDTA moles = 0.00044 moles

Given , initial moles of EDTA  = 0.0430 M * 0.025 L

                                        = 0.001075

Therefore reacting moles of EDTA with [tex]V^3+[/tex] = 0.001075 - 0.00044 = 0.000675 moles

Let us say that the ratio between [tex]V^3+[/tex] and EDTA is 1:1

Therefore moles of [tex]V^3+[/tex] = 0.000675 moles

Molarity = moles / volume

                                    = 0.000675 moles / 0.057 L

                                    = 0.011 M (answer).

Answer:

B

Explanation:

From the calculation, we have the age of the sample to be 750,000 years. Option B

The half-life of a substance, usually associated with radioactive decay, is the time it takes for half of a quantity of that substance to decay. This concept is also used in various other contexts to describe the rate of decay or reduction of a substance's quantity over time.

If you have a radioactive substance, the half-life is the time it takes for half of the radioactive atoms to decay into other elements. After one half-life, half of the original substance will remain, and the other half will have transformed into a different element.

We know that;

[tex]N/No = (1/2)^t/t1/2\\N = 0.9996No\\ 0.9996No/No = (1/2)^t/1.3 * 10^3\\ 0.9996 = (1/2)^t/1.3 * 10^3\\ln 0.9996 = t/1.3 * 10^3 ln 0.5\\t/1.3 * 10^3 = ln 0.9996 / ln 0.5\\t = ln 0.9996 / ln 0.5 * 1.3 * 10^9[/tex]

t = 750,000 years

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Answer: Arsenic

Explanation: The element that contains exactly three 4p  electrons at ground state is  

Arsenic, a group 15 and period 4 solid (20°C) element with  

atomic weight 74.9216g/mol  whose Electronic configuration  is

1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p³ with the  condensed electronic configuration as  [Ar] 3d¹⁰ 4s² 4p³.

The element that contains exactly three 4p electrons in the ground state is Arsenic (As).

Ground state is the lowest energy level of a physical system (such as an atomic nucleus or an atom).

In this case, [tex]\rm p^3[/tex] means nitrogen family. Nitrogen (N), phosphorus (P), arsenic (As), antimony (Sb), and bismuth (Bi) are all members of the nitrogen family. Among these elements only Arsenic has 3 electrons in the 4p orbital in the ground energy level.

The electronic configuration of Arsenic is represented as: [Ar] 3d¹⁰ 4s² 4p³.

Therefore, the correct answer is Arsenic is the element that has 3 electrons in the 4p in ground state.

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Final answer:

Aluminum, symbol (C) Al, forms a protective oxide layer known as aluminum oxide (Al2O3) when exposed to oxygen, protecting it from further corrosion.

Explanation:

The symbol for the element that forms a protective oxide coating is (C) Al, which stands for Aluminum. When aluminum is exposed to the atmosphere, it reacts with oxygen to form aluminum oxide (Al2O3), a thin, hard layer that helps prevent further oxidation and protects the metal underneath. This property is particularly useful for applications where durability and resistance to corrosion are important. For example, the outside of the aerospace and construction materials are often made from aluminum for this reason.

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem

The compound  R-12 (CCl2F2) has 32 valance electrons as shown in the Lewis structure attached to this answer.

The Lewis structure of a compound is a structure that shows the valence electrons in a molecule as dots or single lines to represent a shared pair of electrons in a covalent bond.

The compound  R-12 (CCl2F2) has 32 valence electrons. The arrangement of these valence electrons have been shown in the Lewis structure attached to this answer.

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The question is incomplete; the complete question is:

A chemist must prepare 800.0mL of potassium hydroxide solution with a pH of 13.00 at 25 degree C. He will do this in three steps: Fill a 800.0mL volumetric flask about halfway with distilled water. Weigh out a small amount of solid potassium hydroxide and add it to the flask. Fill the flask to the mark with distilled water. Calculate the mass of potassium hydroxide that the chemist must weigh out in the second step. Round your answer to 2 significant digits.

Answer:

4.5g (to 2 significant digits)

Explanation:

Now we must remember that KOH is a strong base, therefore it will practically dissociate completely.

To find the pH we can use the equation pH + pOH = 14.

Firstly to find the pOH:

13.00 + pOH = 14

pOH = 1.00

To find the [OH-]

Since

pOH= -log[OH^-]

[OH^-] = antilog (-pOH)

[OH^-]= antilog (-1)

[OH^-] = 0.1 molL-1

Since we've established that KOH is a strong base, we know that [OH-] = [KOH]

Also, we know that concentration = number of moles/volume

we have the concentration and the volume now so we can calculate the number of number of moles as follows:

The 800mL volume is the same as 0.8L

0.1 molL-1= number of moles/0.8L

0.08 moles = number of moles

now we can calculate the amount of solid KOH required

the molar mass of KOH = 39 + 16 +1 = 56 gmol-1

56 x 0.08 moles = 4.48g

So in 800mL of pH 13.00 KOH there is 4.5g of KOH dissolved.

Answer:

(a) r = 6.26 * 10⁻⁷cm

(b) r₂ = 6.05 * 10⁻⁷cm

Explanation:

Using the sedimentation coefficient formula;

s =  M(1-Vρ) / Nf ; where s is sedimentation coefficient, M is molecular weight, V is specific volume of protein, p is density of the solvent, N is Avogadro number, f if frictional force = 6πnr, n is viscosity of the medium, r is radius of particle

s = M ( 1 - Vρ) / N*6πnr

making r sbjct of formula, r =  M (1 - Vρ) / N*6πnrs

Note: S = 10⁻¹³ sec, 1 KDalton = 1 *10³ g/mol, I cP = 0.01 g/cm/s

r = {(3.1 * 10⁵ g/mol)(1 - (0.732 cm³/g)(1 g/cm³)} / { (6.02 * 10²³)(6π)(0.01 g/cm/s)(11.7 * 10⁻¹³ sec)

r = 6.26 * 10⁻⁷cm

b. Using the formula r₂/r₁ = s₁/s₂

s₂ = 0.035 + 1s₁ = 1.035s₁

making r₂ subject of formula; r₂ = (s₁ * r₁) / s₂ = (s₁ * r₁) / 1.035s₁

r₂ = 6.3 * 10⁻⁷cm / 1.035

r₂ = 6.05 * 10⁻⁷cm

Answer:

Arrhenius bases: NaOH, KOH and LiOH.

Explanation:

Arrhenius bases are hydroxide ([tex]OH^{-}[/tex]) containing molecules which furnish [tex]OH^{-}[/tex] ion in aqueous solution.

So, clearly, NaOH, KOH and LiOH are arrhenius bases as they contain [tex]OH^{-}[/tex] ion as well as furnish [tex]OH^{-}[/tex] ion in aqueous solution.

Hydrolysis in water: [tex]NaOH(S)+H_{2}O(l)\rightarrow Na_{aq.}^{+}+OH_{aq.}^{-}[/tex]

                                   [tex]KOH(S)+H_{2}O(l)\rightarrow K_{aq.}^{+}+OH_{aq.}^{-}[/tex]

                                   [tex]LiOH(S)+H_{2}O(l)\rightarrow Li_{aq.}^{+}+OH_{aq.}^{-}[/tex]

Answer:

See explaination

Explanation:

Hydrophobic literally means “the fear of water”. Hydrophobic molecules and surfaces repel water. Hydrophobic liquids, such as oil, will separate from water. Hydrophobic molecules are usually nonpolar, meaning the atoms that make the molecule do not produce a static electric field.

Please see attachment for the step by step solution of the given problem.

This protease would likely cleave peptide bonds adjacent to sequences with basic amino acids (P1), followed by hydrophobic residues (P1') such as Arginine-Serine-Leucine (RSL) or Lysine-Valine-Leucine (KVL).

For more such questions on amino acids

https://brainly.com/question/24106148

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Answer:

0.35

Explanation:

pH = -log[H+]

[H+] = [HCl} = 0.45 M because HCl is a strong acid, and dissociate completely.

pH = - log[0.45] = 0.35