calculate the molar solubility of barium fluoride in 0.25 m naf at 25°c. ksp for barium fluoride at 25°c is 1.7 × 10-6.

The Reaction is p-acetaminophenol + bromoethane -> phenacetin + HBr. The mechanism involves the deprotonation of p-acetamminophenol followed by the nucleophilic attack by bromoethane and finally the acidic workup to get the desired product.

How does Williamson ester synthesis reaction proceed?

Step 1: Deprotonation of p-acetaminophenol

p-acetaminophenol is treated with a strong base, such as sodium hydride (NaH) or potassium hydroxide (KOH), to form its corresponding phenoxide ion. This deprotonation step is necessary to allow for the subsequent nucleophilic attack by the bromoethane.

p-acetaminophenol + NaH -> p-acetaminophenoxide + Na+ + H2

Step 2: Nucleophilic attack by bromoethane

The deprotonated p-acetaminophenoxide acts as a nucleophile and attacks the electrophilic carbon atom of bromoethane, displacing the bromine atom to form an ether linkage.

p-acetaminophenoxide + CH3CH2Br -> p-ethoxyacetaminophenol + Br-

Step 3: Acidic workup

The resulting p-ethoxyacetaminophenol is then treated with an acidic solution, such as hydrochloric acid (HCl) or sulfuric acid (H2SO4), to protonate the oxygen atom and restore the neutral phenol structure. This step also releases the bromide ion as hydrobromic acid (HBr).

p-ethoxyacetaminophenol + HCl -> phenacetin + CH3CH2OH + H2O

Overall, the Williamson ether synthesis allows for the synthesis of phenacetin from p-acetaminophenol and bromoethane by forming an ether linkage between the oxygen atom of p-acetaminophenol and the carbon atom of bromoethane.

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Digoxin is used in systolic heart failure because it helps to increase the strength and efficiency of the heart's contractions, particularly in cases where the systolic function of the heart is impaired.

Digoxin works by inhibiting the sodium-potassium ATPase pump, which leads to an increase in intracellular calcium concentrations and subsequently improves the contractility of the heart. This makes it an effective treatment option for patients with systolic heart failure, as it can help to improve cardiac output and reduce symptoms such as shortness of breath and fatigue.

However, due to its narrow therapeutic index, careful monitoring is necessary to ensure that digoxin levels remain within a safe and effective range.

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Hence, 1.14 moles of Sulfur dioxide in 73 g and 44 grams of Carbon dioxide are in 1 mole.

The first question asks us to find the number of moles of sulfur dioxide in 73.0 grams. We can use the formula:

Moles = mass / molar mass,

where molar mass is the sum of the atomic masses of all the elements in the compound. For sulfur dioxide, the molar mass is 32.07 g/mol for sulfur and 2*16.00 g/mol for oxygen, giving a total of 64.07 g/mol.

Plugging in the given mass of 73.0 g and the molar mass, we get the number of moles to be 1.14 mol, which corresponds to option (c).

The second question asks us to find the mass of carbon dioxide in 1.00 mole. We can use the formula:

Mass = moles * molar mass,

where the molar mass is 12.01 g/mol for carbon and 2*16.00 g/mol for oxygen, giving a total of 44.01 g/mol for carbon dioxide. Plugging in the given number of moles of 1.00 mol and the molar mass, we get the mass to be 44.0 g, which corresponds to option (c).

Therefore, the answer to both questions is (c) 1.14 mol for the first question and 44.0 g for the second question.

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The change in enthalpy associated with the combustion of 23.00 g of methanol is approximately -521.2 kJ.

To calculate the change in enthalpy associated with the combustion of 23.00 g of methanol, we need to use the stoichiometry and the given enthalpy change per mole (ΔΗ_c = -726 kJ/mol).

First, determine the number of moles of methanol (CH₃OH) by dividing the mass (23.00 g) by its molar mass (32.04 g/mol):

23.00 g / 32.04 g/mol ≈ 0.7178 mol

Now, multiply the moles of methanol by the given enthalpy change per mole:

0.7178 mol * -726 kJ/mol ≈ -521.2 kJ

So, the change in enthalpy associated with the combustion of 23.00 g of methanol is approximately -521.2 kJ (to four significant figures).

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Diethylamine is a primary aliphatic amine, purine is not categorized as any of the amine classifications.

Diethylamin: This is a compound with two ethyl groups attached to a primary amine (-NH2) functional group.

purine: Purine is a heterocyclic aromatic compound that contains two nitrogen atoms in its ring structure. However, it is not classified as an amine because it does not have an -NH2 or -NR2 functional group.

Diethylamine (C4H11N) is a colorless liquid with a fishy odor. It is a common organic compound and is used as a precursor to a variety of chemicals, including pharmaceuticals, insecticides, and rubber chemicals. Diethylamine is a strong base and forms salts with acids. It is also flammable and can react violently with oxidizing agents.

Purine is a heterocyclic aromatic compound with the chemical formula C5H4N4. It is a building block of DNA and RNA, and is found in many foods, including meat, fish, and beans.

Purine is also used in the synthesis of pharmaceuticals, including drugs used to treat gout and leukemia. Its structure consists of a fused pyrimidine and imidazole ring, and its aromaticity arises from the delocalization of π-electrons over the two rings.

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The question cannot be answered without more information. The configuration of each chiral center needs to be specified as either R or S, as they have opposite configurations.

Chirality refers to the property of a molecule or ion that is not superimposable on its mirror image. Chiral centers are atoms in a molecule that are bonded to four different groups, and their configuration can be described using the R/S nomenclature system. The R and S designations are based on the priority of the four substituent groups around the chiral center, which is determined by the atomic number of the attached atoms. Without knowing the specific configuration of each chiral center, it is impossible to determine the chirality of the (1,2) chiral centers.

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The initial rate at [A] = 0.850 M is approximately 0.321 M/s. The correct answer is (a) 0.321 M/s.

Here, the initial rate of the reaction is 0.0100 M/s when the initial concentration of A is 0.150 M, we need to determine the initial rate at an initial concentration of 0.850 M.
For a second-order reaction, the rate law can be written as: rate = k[A]^2
where k is the rate constant and [A] is the concentration of A.
First, we need to find the value of k using the given initial rate and initial concentration: 0.0100 M/s = k(0.150 M)^2
Now, we can solve for k: k = (0.0100 M/s) / (0.150 M)^2
k ≈ 0.4444 M⁻¹s⁻¹
Next, we can use the value of k and the new initial concentration [A] = 0.850 M to find the new initial rate:
rate = k[A]^2rate = (0.4444 M⁻¹s⁻¹)(0.850 M)^2
Calculating the rate, we get:
rate ≈ 0.321 M/s
So, the initial rate at [A] = 0.850 M is approximately 0.321 M/s. The correct answer is (a) 0.321 M/s.

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Specific gravity is the ratio of the density of a substance to the density of a reference substance, usually water.

It is a dimensionless quantity and is often used in the context of fluids and solids to describe their relative densities. The specific gravity of a substance can be calculated by dividing its density by the density of the reference substance.

Specific gravity is commonly used in industries such as oil and gas, where it is used to measure the density of drilling fluids and to determine the concentration of minerals in ores. It is also used in the construction industry to measure the density of construction materials.

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Complete Question:

What is specific gravity and how is it defined?

To calculate the Ksp value for silver sulfide (Ag2S) using its solubility, follow these steps:

1. Convert solubility to molar solubility:
Solubility = 7.37 × 10^-15 g/L
Molar mass of Ag2S = (2 × 107.87 g/mol Ag) + 32.07 g/mol S = 247.81 g/mol
Molar solubility = (7.37 × 10^-15 g/L) / (247.81 g/mol) = 2.97 × 10^-17 mol/L

2. Write the dissolution equilibrium reaction:
Ag2S (s) ⇌ 2Ag+ (aq) + S2- (aq)

3. Set up the Ksp expression:
Ksp = [Ag+]^2 × [S2-]

4. Find the molar concentrations of Ag+ and S2-:
Since 1 mol of Ag2S produces 2 mol of Ag+, the concentration of Ag+ is 2 × 2.97 × 10^-17 mol/L = 5.94 × 10^-17 mol/L.
The concentration of S2- is equal to the molar solubility, 2.97 × 10^-17 mol/L.

5. Plug the concentrations into the Ksp expression and solve:
Ksp = (5.94 × 10^-17)^2 × (2.97 × 10^-17) = 1.05 × 10^-50

So, the Ksp value for silver sulfide (Ag2S) is approximately 1.05 × 10^-50.

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There are 2.04 × 10⁻⁴ moles of hydrogen in 3.06 × 10⁻³ g of glycine.

To determine the quantity of moles of hydrogen in 3.06 × 10⁻³ g of glycine, we first need to find the molar mass of glycine, which is the sum of the atomic masses of all the atoms in one molecule of glycine. The molecular formula for glycine is C₂H₅NO₂, so the molar mass of glycine is:

Molar mass of glycine = 2 × molar mass of carbon + 5 × molar mass of hydrogen + molar mass of nitrogen + 2 × molar mass of oxygen

= 2(12.01 g/mol) + 5(1.01 g/mol) + 14.01 g/mol + 2(16.00 g/mol)

= 75.07 g/mol

Next, we need to determine the number of moles of glycine in 3.06 × 10⁻³ g of glycine by dividing the mass of glycine by its molar mass:

moles of glycine = mass of glycine / molar mass of glycine

= 3.06 × 10⁻³ g / 75.07 g/mol

= 4.08 × 10⁻⁵ mol

Since the molecular formula for glycine contains five hydrogen atoms, the quantity of moles of hydrogen in 3.06 × 10⁻³ g of glycine can be found by multiplying the number of moles of glycine by the number of hydrogen atoms per molecule of glycine:

moles of hydrogen = moles of glycine × 5

= 4.08 × 10⁻⁵ mol × 5

= 2.04 × 10⁻⁴ mol

Therefore, there are 2.04 × 10⁻⁴ moles of hydrogen in 3.06 × 10⁻³ g of glycine.

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a. The concentration of [tex]H_2SO_3[/tex] in the solution is 0.177 MM, the concentration of [tex]HSO^{3-}[/tex] is 0.293 MM, the concentration of [tex]SO_3^{2-}[/tex] is 3.44x[tex]10^{-5}[/tex] MM.

b. The concentration of [tex]H_3O^+[/tex] is 0.000360 MM, and the concentration of OH- is 2.78x[tex]10^{-11}[/tex] MM.

Part A: We are given a 0.470 M solution of [tex]H_2SO_3[/tex] with two dissociation constants, Ka1=1.6×[tex]10^{-2}[/tex] and Ka2 = 6.4×[tex]10^{-8}[/tex]. We can use these dissociation constants to calculate the concentrations of [tex]H_2SO_3[/tex] and [tex]HSO^{3-}[/tex] using the following equations:

Ka1 = [[tex]H_3O^+[/tex]][[tex]HSO^{3-}[/tex]]/[[tex]H_2SO_3[/tex]]

Ka2 = [[tex]H_3O^+[/tex]][[tex]SO_3^{2-}[/tex]]/[[tex]HSO^{3-}[/tex]]

Simplifying these equations, we get:

[[tex]HSO^{3-}[/tex]] = Ka1[[tex]H_2SO_3[/tex]]/[[tex]H_3O^+[/tex]]

[[tex]SO_3^{2-}[/tex]] = Ka2[[tex]HSO^{3-}[/tex]]/[[tex]H_3O^+[/tex]]

[[tex]H_2SO_3[/tex]] = [[tex]H_3O^+[/tex]][[tex]HSO^{3-}[/tex]]/Ka1

Substituting the given values and simplifying, we get:

[[tex]HSO^{3-}[/tex]] = 0.34 M

[[tex]H_2SO_3[/tex]] = 0.13 M

Therefore, the concentration of [tex]H_2SO_3[/tex] is 0.13 M and the concentration of [tex]HSO^{3-}[/tex] is 0.34 M in the given solution.

Part B: We are given the same solution as in Part A, and we need to calculate the concentrations of [tex]SO_3^{2-}[/tex], [tex]H_3O^+[/tex], and OH- using the dissociation constants given.

We can use the following equations to calculate the concentrations:

[[tex]H_3O^+[/tex]] = (Ka1Ka2[C])/([[tex]H_2SO_3[/tex]]+Ka1[C]+Ka1Ka2[C])

[[tex]SO_3^{2-}[/tex]] = Ka2[[tex]H_2SO_3[/tex]]/([[tex]H_2SO_3[/tex]]+Ka1[C]+Ka1Ka2[C])

[OH-] = Kw/[[tex]H_3O^+[/tex]]

Substituting the given values and simplifying, we get:

[[tex]H_3O^+[/tex]] = 1.7 x [tex]10^{-2}[/tex] M

[[tex]SO_3^{2-}[/tex]] = 3.3 x [tex]10^{-9}[/tex] M

[OH-] = 5.9 x [tex]10^{-13}[/tex] M

Therefore, the concentrations of [tex]SO_3^{2-}[/tex], [tex]H_3O^+[/tex], and OH- are 3.3 x [tex]10^{-9}[/tex] M, 1.7 x [tex]10^{-2}[/tex] M, and 5.9 x [tex]10^{-13}[/tex] M, respectively, in the given solution.

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The volume of added base beyond which the pH is calculated by focusing on the amount of excess strong base added is referred to as the equivalence point.

At this point, the amount of added strong base is equal to the amount of acid in the solution. The pH at the equivalence point depends on the strength of the acid and base being titrated.

When a strong base is added to a weak acid, the pH of the solution increases gradually until it reaches the equivalence point. However, if a strong acid is added to a weak base, the pH decreases until it reaches the equivalence point.

In general, the pH changes rapidly near the equivalence point, so it is important to add the base slowly near the end of the titration to accurately determine the equivalence point. Once the equivalence point is reached, the pH is calculated based on the amount of excess strong base added. The pH at the equivalence point can be used to determine the concentration of the acid or base being titrated.

In conclusion, the volume of added base beyond which the pH is calculated by focusing on the amount of excess strong base added is the equivalence point. Accurately determining the equivalence point is crucial in determining the concentration of the acid or base being titrated.

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The solubility product constant (Ksp) for a sparingly soluble salt like barium fluoride ([tex]BaF_{2}[/tex]) can be calculated using the molar solubility of the salt. In this case, the molar solubility of [tex]BaF_{2}[/tex]  is 7.53 mmol/L. the value of Ksp for barium fluoride is 4.3x[tex]10^{-7} .[/tex]

The balanced equation for the dissolution of [tex]BaF_{2}[/tex] is:

[tex]BaF_{2}[/tex] (s) ⇌ [tex]Ba_{2}[/tex]+(aq) + [tex]2F^{-}[/tex](aq)

The Ksp expression is:

Ksp = [tex][Ba_{2}^{+} ][F^{-} ]^2[/tex]

Substituting the molar solubility of [tex]BaF_{2}[/tex] in the expression, we get:

Ksp = [tex](7.53[/tex]x[tex]10^_{-3} )^3[/tex] = 4.3x[tex]10^{-7}[/tex]

Therefore, the value of Ksp for barium fluoride is 4.3x[tex]10^{-7} .[/tex]

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The expected major kinetic product formed from addition of one mole of HBr to the diene is the 1,2-dibromide.

This is because the reaction occurs through a Markovnikov addition mechanism, where the H+ adds to the diene at the carbon with the most hydrogens, and the Br- adds to the carbon with the least hydrogens. This results in the formation of the 1,2-dibromide as the major product.

The reaction occurs in a kinetically controlled manner, meaning that the product formed is the one with the lowest activation energy and therefore forms the fastest.

In summary, the expected major kinetic product formed from the addition of one mole of HBr to the diene is the 1,2-dibromide, formed through a Markovnikov addition mechanism where the H+ adds to the carbon with the most hydrogens and the Br- adds to the carbon with the least hydrogens.

This reaction occurs in a kinetically controlled manner, where the product formed is the one with the lowest activation energy and forms the fastest.

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Answer: we need three distinct calculations

Explanation: The first one is that the heat energy released when  the water is cooled to 0"c.

                       The second one is that the heat energy released when water changes its state to solid at constant temprature.

                        The third one is the heat energy released when the ice at 0'c is changed to -2'c.

a. The solubility of AgIO₃ in a 0.285 M solution of NaIO₃ is 1.06 x 10⁻⁸ M. b. The solubility of La(IO₃)₃ in a 0.285 M solution of NaIO₃ is 4.31 x 10⁻¹² M. c. AgIO₃ is more soluble than La(IO₃)₃.

a. To calculate the solubility of AgIO₃, we need to first write the balanced chemical equation for the dissolution of AgIO₃ in water: AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq). The Ksp expression for this reaction is: Ksp = [Ag⁺][IO₃⁻]. Let x be the solubility of AgIO₃ in a 0.285 M solution of NaIO₃. Then, the equilibrium concentrations of Ag⁺ and IO₃⁻ are both equal to x. The initial concentration of IO₃⁻ is 0.285 M. Substituting the values into the Ksp expression and solving for x gives: x² = Ksp/[IO₃⁻] = (3.0 x 10⁻⁸)/(0.285) = 1.06 x 10⁻⁸ M.

b. The balanced chemical equation for the dissolution of La(IO₃)₃ in water is: La(IO₃)₃(s) ⇌ La³⁺(aq) + 3IO₃⁻(aq). The Ksp expression for this reaction is: Ksp = [La³⁺][IO₃⁻]³. Let x be the solubility of La(IO₃)₃ in a 0.285 M solution of NaIO₃. Then, the equilibrium concentrations of La³⁺ and IO₃⁻ are both equal to x. The initial concentration of IO₃⁻ is 0.285 M. Substituting the values into the Ksp expression and solving for x gives: x⁴ = Ksp/[IO₃⁻]³ = (7.5 x 10⁻¹²)/(0.285)³ = 4.31 x 10⁻¹² M.

c. Since the solubility of AgIO₃ is greater than the solubility of La(IO₃)₃ in a 0.285 M solution of NaIO₃, AgIO₃ is more soluble than La(IO₃)₃.

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The mass of the Pa-234 isotope that remains after 40.1 hours is 0.003 mg.

To solve this problem, we need to use the formula for radioactive decay:

N = N0(1/2)^(t/T)

Where N is the remaining amount of the isotope, N0 is the initial amount, t is the time that has elapsed, T is the half-life of the isotope.

First, we need to find the initial amount of Pa-234. Since the sample is of Pa-239, we need to assume that it decays into Pa-234. The atomic mass of Pa-239 is 239, and it decays into U-235 with a half-life of 23.5 minutes. U-235 decays into Pa-231, which then decays into Pa-234. The decay chain looks like this:

Pa-239 --> U-235 --> Pa-231 --> Pa-234

So, the initial amount of Pa-234 can be calculated from the initial amount of Pa-239 using the decay chain:

N0(Pa-234) = N0(Pa-239) x (1/2)^(40.1/6.69)

N0(Pa-239) = 0.812 mg

N0(Pa-234) = N0(Pa-239) x (1/2)^(40.1/6.69) = 0.812 mg x 0.0243 = 0.0197 mg

Now, we can use the formula for radioactive decay to find the remaining amount of Pa-234 after 40.1 hours:

N(Pa-234) = N0(Pa-234) x (1/2)^(40.1/6.69)

N(Pa-234) = 0.0197 mg x (1/2)^(40.1/6.69) = 0.003 mg

Therefore, the mass of the Pa-234 isotope that remains after 40.1 hours is 0.003 mg.

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The solubility of [tex]Au(OH)_{3}[/tex] in water and 1.0 M nitric acid can be calculated using the solubility product constant (Ksp) expression: the solubility of [tex]Au(OH)_{3}[/tex]  in 1.0 M nitric acid solution is 5.5 x [tex]10^{-18}[/tex] M.

Ksp = [tex][Au_{3} ^{+} ][OH^{-} ]^3[/tex]

where [[tex][Au_{3} ^{+} ][/tex]] is the concentration of the [tex][Au_{3} ^{+} ][/tex] ions and [tex][OH^{-} ]^3[/tex] is the concentration of hydroxide ions.

(a) Solubility of [tex]Au(OH)_{3}[/tex] in water:

Ksp = [tex][Au_{3} ^{+} ][OH^{-} ]^3[/tex]

Ksp = x * [tex](x)^3[/tex] = [tex]x^4[/tex]

x = [tex](Ksp)^(1/4)[/tex] = [tex](5.510^{-46} )^(1/4)[/tex] = [tex]1.110^{-12}[/tex] M

Solubility of [tex]Au(OH)_{3}[/tex] in water is [tex]1.110^{-12}[/tex] M.

(b) Solubility of [tex]Au(OH)_{3}[/tex] in 1.0 M nitric acid solution:

[[tex][Au_{3} ^{+} ][/tex]] = [tex](Ksp / [OH^{-} ]^3)^1/4[/tex]

[[tex]OH^{-}[/tex]] = 1.0 x [tex]10^{-14}[/tex] M (from Kw expression)

[[tex][Au_{3} ^{+} ][/tex]] = (5.5 x [tex]10^{-46}[/tex] / [tex](1.0 x 10^{-14} M)^3)^1/4[/tex]

[[tex][Au_{3} ^{+} ][/tex]] = [tex]5.5 x 10^{-18} M[/tex]

Therefore, the solubility of [tex]Au(OH)_{3}[/tex]  in 1.0 M nitric acid solution is 5.5 x [tex]10^{-18}[/tex] M.

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The product obtained after thermal decarboxylation of 2-pentylmalonic acid is 2-pentylpropanoic acid.

2-pentylmalonic acid has the following structure:

HOOC-CH(COOR)-CH2-CH2-CH2-CH3

Upon heating, the carboxylic acid group (-COOH) undergoes decarboxylation and is removed as carbon dioxide (CO2), leaving behind a ketone group (-C=O) at the alpha position. The remaining molecule is then the corresponding alkyl acid.

Thus, in the given compound, after thermal decarboxylation, the resulting molecule will have the structure:

CH3-CH2-CH2-CH2-CO-CH2-CH2-CH3

which is 2-pentylpropanoic acid.

Therefore, the product obtained after thermal decarboxylation of 2-pentylmalonic acid is 2-pentylpropanoic acid.

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Answer:

31.5 L

Explanation:

Simply use PV=nRT

Convert kPa  to atm by using 101.3 kPa = 1 atm

78.5 kPa x (1 atm/101.3 kPa) = .775 atm

Then find moles of O2 where MM = 32 so we have 1.0 moles

Find T in Kelvin = C +273 = 25 + 273 = 298

(.775 atm)(V) = 1.0 moles(0.082 atm x L / mol x K)(298 K)

V = 31.5 L

In this case, we have 2 moles of Na and 3 moles of H2O, which means that H2O is the limiting reagent. Na is in excess, because we have more than enough to react with all of the H2O.

a. The balanced chemical equation for the reaction is:
2Na(s) + 6H2O(l) → 2NaOH(aq) + 3H2(g)

b. The molecular representations of the reaction can be shown as follows:

After reaction:
2Na + 3H2O → 2NaOH + 3H2

Before reaction:
Na + Na + 3H2O → NaOH + NaOH + 3H2

c. To determine which reagent is limiting, we need to calculate the amount of product that can be formed from each reactant. The balanced equation tells us that 2 moles of Na react with 6 moles of H2O to produce 2 moles of NaOH and 3 moles of H2. Therefore, if we have 2 moles of Na and 6 moles of H2O, we can produce 2 moles of NaOH and 3 moles of H2.

However, if we have less than 6 moles of H2O, then H2O is the limiting reagent, because we will run out of it before all of the Na is used up. If we have less than 2 moles of Na, then Na is the limiting reagent, because we will run out of it before all of the H2O is used up.

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When KBr was added to the copper sulphate solution, the concentration of bromide ions (Br-) increases, the concentration of Cu(H2O)6+2 ions decreases and the concentration of CuBr4-2 ions increases.

When potassium bromide (KBr) was added to the copper sulphate solution, the following changes in the concentration of ions occurred:
1. The concentration of bromide ions (Br-) increased due to the addition of KBr.
2. The equilibrium shifted to the right i.e, forward reaction , as more Br- ions reacted with Cu(H2O)6+2 ions to form CuBr4-2 ions.
3. As a result, the concentration of Cu(H2O)6+2 ions decreased, and the concentration of CuBr4-2 ions increased.
This shift in equilibrium led to a change in colour from blue (due to Cu(H2O)6+2 ions) to green (due to CuBr4-2 ions).

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The expected products of the solvolysis process are methanol (MeOH) and bromide ion (Br-) being the leaving group. Therefore, the products should be methoxide ion (OCH3) and a molecule of HBr. The reaction can be represented as follows:

Br- + MeOH (heat) → OCH3- + HBr

So, the expected product(s) are OCH3- and HBr.
The expected product(s) for this reaction include:

1. OCH3-substituted compound: The bromine atom is replaced by a methoxy group (OCH3) due to nucleophilic substitution by methanol.
2. OH-substituted compound: The bromine atom is replaced by a hydroxyl group (OH) if a small amount of water is present, which is also a common nucleophile.

Solvolysis refers to a type of chemical reaction where a molecule is cleaved or transformed in the presence of a solvent. The solvent can be water or any other polar or nonpolar substance that has the ability to dissolve the reactant. In a solvolysis reaction, the solvent acts as a nucleophile and can replace or modify certain chemical groups in the molecule, resulting in a new product. Solvolysis is a common reaction in organic chemistry and is often used to synthesize new compounds or to break down complex molecules into simpler ones. Examples of solvolysis reactions include hydrolysis, alcoholysis, and ammonolysis.

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In the [tex]C_{13}[/tex] NMR of phenylethylamine, we would expect to see four peaks.

This is because phenylethylamine contains four unique carbon environments: the carbon attached to the amino group, the carbon alpha to the amino group, the two carbons in the phenyl ring, and the carbon beta to the phenyl ring.

Each of these carbon environments will give rise to a distinct peak in the C13 NMR spectrum.

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In the [tex]C_{13}[/tex] NMR of phenylethylamine, we would expect to see four peaks.

This is because phenylethylamine contains four unique carbon environments: the carbon attached to the amino group, the carbon alpha to the amino group, the two carbons in the phenyl ring, and the carbon beta to the phenyl ring.

Each of these carbon environments will give rise to a distinct peak in the C13 NMR spectrum.

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Answer:

I may be incorrect but C? Isnt it positron

Explanation:

The concentration of OH- in a 0.724 M solution of BrO- is 4.0 x 10^-6 M.

To determine the concentration of OH- in a 0.724 M solution of BrO-, we first need to find the concentration of the corresponding BrO- ion. Since BrO- is a weak base, we can use the Kb value to calculate the concentration of OH- ions produced when it dissociates.

First, we need to write the balanced equation for the dissociation of BrO-:
BrO- + H2O ⇌ OH- + HBrO

The Kb expression for this reaction is:
Kb = [OH-][HBrO]/[BrO-]

Since we are given the Kb value and the concentration of BrO-, we can solve for [OH-]:
Kb = [OH-][HBrO]/[BrO-]
4.0 x 10^-6 = [OH-][0.724]/[BrO-]
[OH-] = (4.0 x 10^-6)(0.724)/[BrO-]

To solve for [BrO-], we need to use the fact that it dissociates according to the equation:
BrO- + H2O ⇌ OH- + HBrO

This means that the concentration of BrO- will be equal to the initial concentration of the solution, which is 0.724 M.

Plugging in the values, we get:
[OH-] = (4.0 x 10^-6)(0.724)/0.724
[OH-] = 4.0 x 10^-6 M

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A chemical bond refers to the linking of atoms together to form molecules. Some key characteristics of chemical bonds:

1. They hold atoms together in a molecule. Chemical bonds keep the atoms together rather than having them float apart.

2. They involve the sharing or transfer of electrons between atoms. The bonds are formed due to the electrostatic attraction between positive and negative charges. For example, in an ionic bond, electrons are transferred from one atom to another. In a covalent bond, electrons are shared between atoms.

3. They determine many of the properties of a compound. The strength, polarity, directionality of bonds have a strong influence on properties such as melting point, solubility, conductivity, etc.

4. They can be made and broken. Chemical bonds can form during chemical reactions and break apart during other reactions.

5. They involve the sharing or redistribution of orbital density between atoms. Electrons in atomic orbitals redistribute to form molecular orbitals that surround the nuclei.

6. They align atoms into geometric arrangements. Chemical bonds orient atoms in specific spatial configurations, which determines the molecular geometry and polarity.

7. They influence chemical reactivity. The strength and stability of chemical bonds determine whether a molecule will readily react with other compounds. Weaker bonds are more reactive.

That covers the basic highlights of a chemical bond. Let me know if you have any other questions!

The pH of the solution after the addition of 1.11 mol HCl is 7.85.

First, we need to find the initial concentration of the weak base:

0.85 M = [B]/0.849 L

[B] = 0.85 * 0.849 = 0.72165 mol/L

Next, we can use the pKb value to find the Kb value:

pkb = -log(Kb)

7.85 = -log(Kb)

Kb = 1.74 x 10⁻⁸

Now we can set up the equilibrium expression for the weak base:

B + H2O ⇌ BH+ + OH-

Kb = [BH+][OH-]/[B]

At equilibrium, we can assume that [OH-] is negligible compared to [B] and [BH+]. This allows us to simplify the expression to:

Kb = [BH+][OH-]/[B] ≈ [BH+][OH-]/([B] + [BH+])

Since we are dealing with a weak base, we can also assume that [BH+] is much less than [B]. This allows us to simplify the expression further to:

Kb = [BH+][OH-]/[B] ≈ [BH+][OH-]/[B]

Now we can use the initial concentration of the weak base and the Kb value to find [BH+]:

Kb = [BH+][OH-]/[B]

1.74 x 10⁻⁸= [BH+]/0.72165

[BH+] = (1.74 x 10⁻⁸ * 0.72165) = 1.0138 x 10⁻⁴ M

Next, we can use the balanced chemical equation for the reaction between HCl and the weak base:

HCl + B ⇌ BH+ + Cl-

Since we are adding 1.11 mol of HCl and the weak base is the limiting reactant, all of the weak base will react with the HCl. This means that the final concentration of BH+ will be equal to the initial concentration of the weak base:

[BH+] = 0.72165 mol/L

Now we can use the Henderson-Hasselbalch equation to find the pH of the solution:

pH = pKb + log([BH+]/[B])

pH = 7.85 + log(0.72165/0.72165)

pH = 7.85

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According to the American Chemistry Society, cloth-topped "tennis" or "running" shoes are recommended for general laboratory work. Sandals, woven leather shoes, and high heels are not recommended as they do not provide adequate protection for the feet  against spills or dropped objects in a laboratory setting.
The American Chemical Society is a scientific society based in the United States that supports scientific research in the field of chemistry.

Other safety precautions include: Wearing appropriate gloves, masks, lab coat and shoes. Being careful while using pipette. Washing the glassware properly before using etc.

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