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Answer: The mass of [tex]NH_3[/tex] produced is, 3.03 grams.

Explanation : Given,

Mass of [tex]N_2[/tex] = 2.5 g

Mass of [tex]H_2[/tex] = 2.5 g

Molar mass of [tex]N_2[/tex] = 28 g/mol

Molar mass of [tex]H_2[/tex] = 2 g/mol

First we have to calculate the moles of [tex]N_2[/tex] and [tex]H_2[/tex].

[tex]\text{Moles of }N_2=\frac{\text{Given mass }N_2}{\text{Molar mass }N_2}=\frac{2.5g}{28g/mol}=0.089mol[/tex]

and,

[tex]\text{Moles of }H_2=\frac{\text{Given mass }H_2}{\text{Molar mass }H_2}=\frac{2.5g}{2g/mol}=1.25mol[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

[tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g)[/tex]

From the balanced reaction we conclude that

As, 1 mole of [tex]N_2[/tex] react with 3 mole of [tex]H_2[/tex]

So, 0.089 moles of [tex]N_2[/tex] react with [tex]0.089\times 3=0.267[/tex] moles of [tex]H_2[/tex]

From this we conclude that, [tex]H_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]N_2[/tex] is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of [tex]NH_3[/tex]

From the reaction, we conclude that

As, 1 mole of [tex]N_2[/tex] react to give 2 mole of [tex]NH_3[/tex]

So, 0.089 mole of [tex]N_2[/tex] react to give [tex]0.089\times 2=0.178[/tex] mole of [tex]NH_3[/tex]

Now we have to calculate the mass of [tex]NH_3[/tex]

[tex]\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3[/tex]

Molar mass of [tex]NH_3[/tex] = 17 g/mole

[tex]\text{ Mass of }NH_3=(0.178moles)\times (17g/mole)=3.03g[/tex]

Therefore, the mass of [tex]NH_3[/tex] produced is, 3.03 grams.

Answer:

2-hexanone

Explanation:

First, we'll begin by:

1. Locating the longest continuous chain i.e haxane

2. Determine the functional group in the compound. The functional group in ketone (C=O). This changes the name from hexane to hexanone i.e replacing the -e in at the end in hexane with -one to make it hexanone.

3. Give the functional group the lowest low count. In doing this, we'll start counting from the left. The functional group is at carbon 2.

Note: no substitute group is attached.

Now, in naming the compound, we must indicate the position of functional group as illustrated below:

The name of the compound is:

2-hexanone

The compound is called 2-hexanone. This is because there is a six carbon chain with a ketone functional group attached to the second carbon.

You are dealing with an organic compound formula. Let's identify the compound based on its structure. The compound you have given is CH3 - C - CH2 - CH2 - CH2 - CH3. This is a 6 carbon chain and there is double bond between Oxygen and the second carbon. Now, when naming compounds in chemistry we firstly count the length of the chain, it is hexane. Then, we look for any functional group and where it is attached in the chain. A ketone functional group is CO, so this would make it a hexanone. As the CO is attached at the second carbon, it is 2-hexanone. So the correct answer for the name of the compound would be 2-hexanone.

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Answer:

The total pressure in the container is 0.389 atm

Explanation:

Step 1: Data given

Kp = 0.739

The initial pressure of H2S = 0.215 atm

Step 2: The balanced equation

H2S(g) ⇆ H2(g) + S(g)

Step 3: The initial pressures

pH2S = 0.215 atm

pH2 = 0 atm

pS = 0 atm

Step 4: The pressures at the equilibrium

pH2S = 0.215 - X atm

pH2 = X atm

pS = X atm

Step 5:

Kp = 0.739 = (pS)*(pH2) / (pH2S)

0.739 = X*X / (0.215 - X)

0.739 = X² / (0.215 - X)

X² = 0.739*(0.215-X)

X² = 0.1589 - 0.739X

X² +0.739X - 0.1589 = 0

X = 0.174

pH2S = 0.215 - 0.174 atm = 0.041 atm

pH2 = 0.174 atm

pS = 0.174 atm

Step 6: Calculate the total pressure in the container

Total pressure = 0.041 atm + 0.174 atm + 0.174 atm

Total pressure = 0.389 atm

The total pressure in the container is 0.389 atm

Answer:

(FeSCN⁺²) = 0.11 mM

Explanation:

Fe ( NO3)3 (aq) [0.200M] + KSCN (aq) [ 0.002M] ⇒ FeSCN+2

M (Fe(NO₃)₃  = 0.200 M

V (Fe(NO₃)₃ =  10.63 mL

n (Fe(NO₃)₃ = 0.200*10.63 = 2.126 mmol

M (KSCN) =  0.00200 M

V (KSCN) = 1.42 mL

n (KSCN) =  0.00200 * 1.42 = 0.00284 mmol

Total volume = V (Fe(NO₃)₃  + V (KSCN)

                       = 10.63 + 1.42

                       = 12.05 mL

Limiting reactant = KSCN

So,

FeSCN⁺² = 0.00284 mmol

M (FeSCN⁺²) = 0.00284/12.05

                     = 0.000236 M

Excess reactant = (Fe(NO₃)₃

n(Fe(NO₃)₃ =  2.126 mmol -  0.00284 mmol

                  =2.123 mmol

For standard 2:

n (FeSCN⁺²) = 0.000236 * 4.63

                    =0.00109

V(standard 2) = 4.63 + 5.17

                       = 9.8 mL

M (FeSCN⁺²)  = 0.00109/9.8

                      = 0.000111 M = 0.11 mM

Therefore, (FeSCN⁺²) = 0.11 mM

The value of the (F-e-S-C-N⁺²) = 0.11 m-M when the student made additional dilutions.

Since

M (F-e(N-O₃)₃  = 0.200 M

V (F-e(N-O₃)₃ =  10.63 mL

n (F-e(N-O₃)₃ = 0.200*10.63

= 2.126 mmol

M (K-S-C-N) =  0.00200 M

V (K-S-C-N) = 1.42 mL

And,

n (KS-C-N) =  0.00200 * 1.42 = 0.00284 mmol

Now

Total volume = V (F-e(N-O₃)₃  + V (K-S-C-N)

                      = 10.63 + 1.42

                      = 12.05 mL

Also, Limiting reactant = K-S-C-N

So,

F-e-S-C-N⁺² = 0.00284 mmol

M (F-e-S-C-N⁺²) = 0.00284/12.05

                    = 0.000236 M

Now

Excess reactant = (F-e(N-O₃)₃

n(F-e(N-O₃)₃ =  2.126 mmol -  0.00284 mmol

                 =2.123 mmol

Now For standard 2:

n (F-e-S-C-N⁺²) = 0.000236 * 4.63

                   =0.00109

V(standard 2) = 4.63 + 5.17

                      = 9.8 mL

M (F-e-S-C-N⁺²)  = 0.00109/9.8

                     = 0.000111 M = 0.11 mM

Therefore, (F-e-S-C-N⁺²) = 0.11 mM

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For the following reaction, 76.0 grams of barium chloride are allowed to react with 67.0 grams of potassium sulfate.

The reaction consumes _____ moles of barium chloride. The reaction produces _____ moles of barium sulfate and _____ moles of potassium chloride.

Answer: a) The reaction consumes 0.365 moles of barium chloride.

b) The reaction produces 0.365 moles of barium sulfate and 0.730 moles of potassium chloride.

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of barium chloride}=\frac{76.0}{208g/mol}=0.365moles[/tex]

[tex]\text{Moles of potassium sulphate}=\frac{67.0}{174g/mol}=0.385moles[/tex]

[tex]BaCl_2(aq)+K_2SO_4(aq)\rightarrow BaSO_4(s)+2KCl(aq)[/tex]

According to stoichiometry :

1 mole of [tex]BaCl_2[/tex] require 1 mole of [tex]K_2SO_4[/tex]

Thus 0.365 moles of [tex]BaCl_2[/tex] will require=[tex]\frac{1}{1}\times 0.365=0.365moles[/tex]  of [tex]K_2SO_4[/tex]

Thus [tex]BaCl_2[/tex] is the limiting reagent as it limits the formation of product and [tex]K_2SO_4[/tex] is the excess reagent.

As 1 moles of [tex]BaCl_2[/tex] give = 1 moles of [tex]BaSO_4[/tex]

Thus 0.365 moles of [tex]BaCl_2[/tex] give =[tex]\frac{1}{1}\times 0.365=0.365moles[/tex]  of [tex]BaSO_4[/tex]

As 1 moles of [tex]BaCl_2[/tex] give = 2 moles of [tex]KCl[/tex]

Thus 0.365 moles of [tex]BaCl_2[/tex] give =[tex]\frac{2}{1}\times 0.365=0.730moles[/tex]  of [tex]KCl[/tex]

Thus the reaction consumes 0.365 moles of barium chloride. The reaction produces 0.365 moles of barium sulfate and 0.730 moles of potassium chloride.

The standard heat of formation of [tex]\( \text{CuO(s)} \) is \( -149.6 \text{ kJ/mol} \).[/tex]

To find the standard heat of formation of [tex]\( \text{CuO(s)} \)[/tex], we can use the given thermochemical equations and apply Hess's Law.

Given Equations:

1. [tex]\( 2 \text{Cu}_2\text{O(s)} + \text{O}_2\text{(g)} \rightarrow 4 \text{CuO(s)} \quad \Delta H^\circ = -287.9 \text{ kJ} \)[/tex]

2. [tex]\( \text{Cu}_2\text{O(s)} \rightarrow \text{CuO(s)} + \text{Cu(s)} \quad \Delta H^\circ = 11.3 \text{ kJ} \)[/tex]

Goal:

Find the standard heat of formation of [tex]\( \text{CuO(s)} \), \( \Delta H_f^\circ \text{(CuO(s))} \).[/tex]

Steps:

1. Write the formation reaction for [tex]\( \text{CuO(s)} \)[/tex]

[tex]\[ 2 \text{Cu(s)} + \text{O}_2\text{(g)} \rightarrow 2 \text{CuO(s)} \][/tex]

2. Manipulate the given equations to match the formation reaction:

a. Reverse Equation 2 to get [tex]\( \text{CuO(s)} \) and \( \text{Cu(s)} \)[/tex] on the reactant side:  

[tex]\[ \text{CuO(s)} + \text{Cu(s)} \rightarrow \text{Cu}_2\text{O(s)} \quad \Delta H^\circ = -11.3 \text{ kJ} \][/tex]

b. Add this to Equation 1 (which needs no change) to eliminate[tex]\( \text{Cu}_2\text{O(s)} \): \[ \begin{align*} \text{CuO(s)} + \text{Cu(s)} &\rightarrow \text{Cu}_2\text{O(s)} \quad \Delta H^\circ = -11.3 \text{ kJ} \\ 2 \text{Cu}_2\text{O(s)} + \text{O}_2\text{(g)} &\rightarrow 4 \text{CuO(s)} \quad \Delta H^\circ = -287.9 \text{ kJ} \\ \end{align*} \][/tex]

Adding these equations:

[tex]\[ \text{CuO(s)} + \text{Cu(s)} + 2 \text{Cu}_2\text{O(s)} + \text{O}_2\text{(g)} \rightarrow \text{Cu}_2\text{O(s)} + 4 \text{CuO(s)} \][/tex]

Simplify:

[tex]\[ 2 \text{Cu(s)} + \text{O}_2\text{(g)} \rightarrow 2 \text{CuO(s)} \][/tex]

3. Combine the enthalpy changes:

[tex]\[ \Delta H^\circ = -287.9 \text{ kJ} + (-11.3 \text{ kJ}) = -299.2 \text{ kJ} \][/tex]

Since the above reaction is for 2 moles of [tex]\( \text{CuO(s)} \)[/tex], the enthalpy change for the formation of 1 mole of [tex]\( \text{CuO(s)} \)[/tex] is:

[tex]\[ \Delta H_f^\circ \text{(CuO(s))} = \frac{-299.2 \text{ kJ}}{2} = -149.6 \text{ kJ/mol} \][/tex]

Answer:

[NaOH] = 0.107M

Explanation:

25.21ml of NaOH(aq) soln + 0.550g KHPh => NaKPh + H₂O

The reaction is 1 to 1 NaOH to KHPh => moles NaOH neutralized = moles KHPh used. (KHPh => Potassium Hydrogen Phthalate f.wt. = 204.22 g/mol)

moles KHPh = 0.550g / 204.22g·mol⁻¹ = 0.0027 mole

moles NaOH neutralized = moles KHPh used = 0.0027 mole NaOH in 25.21ml aqueous solution ...

Molar concentration of NaOH solution = moles NaOH/Liters solution = 0.0027mol NaOH/0.0251L = 0.107M NaOH solution.

Answer:

[NaOH] = 0.107M

Explanation:

25.21ml of NaOH(aq) soln + 0.550g KHPh => NaKPh + H₂O

The reaction is 1 to ratio 1 NaOH to KHPh => moles NaOH neutralized = moles KHPh used. (KHPh => Potassium Hydrogen Phthalate f.wt. = 204.22 g/mol)

moles KHPh = 0.550g / 204.22g·mol⁻¹ = 0.0027 mole

moles NaOH neutralized = moles KHPh used = 0.0027 mole NaOH in 25.21ml aqueous solution ...

Molar concentration of NaOH solution = moles NaOH/Liters solution = 0.0027mol NaOH/0.0251L = 0.107M NaOH solution.

Answer:

1. Number of gas particles (atoms or molecules)

2. Number of moles of gas

3. Average kinetic energy

Explanation:

Since the two gas has the same volume and are under the same conditions of temperature and pressure,

Then:

1. They have the same number of mole because 1 mole of any gas at stp occupies 22.4L. Now both gas will occupy the same volume because they have the same number of mole

2. Since they have the same number of mole, then they both contain the same number of molecules as explained by Avogadro's hypothesis which states that at the same temperature and pressure, 1 mole of any substance contains 6.02x10^23 molecules or atoms.

3. Being under the same conditions of temperature and pressure, they both have the same average kinetic energy. The kinetic energy of gas is directly proportional to the temperature. Now that both gas are under same temperature, their average kinetic energy are the same.

The answer choices which are the SAME for these two gas samples are

This refers to the type  of energy which has to do with the motion of an object.

Hence, we can see that since the two gas has the same volume and are under the same conditions of temperature and pressure,

We would infer that:

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Answer:

39.3%

Explanation:

Our guide in solving the problem must be the reaction equation hence it is pertinent to put down first:

CaF2 + H2SO4 --> CaSO4 + 2HF

We have a very important information in the question, sulphuric acid is present in excess. This implies that calcium fluoride is the limiting reactant.

Number of moles of calcium fluoride reacted= mass of calcium fluoride reacted/ molar mass of calcium fluoride

Molar mass of calcium fluoride= 78.07 g/mol

Number of moles of calcium fluoride= 11g/78.07 g/mol = 0.14 moles of Calcium flouride

Since 1 mole of calcium fluoride yields two moles of 2 moles hydrogen fluoride

0.14 moles of calcium fluoride will yield 0.14×2= 0.28 moles of hydrogen fluoride

Mass of 0.28 moles of hydrogen fluoride = number of moles× molar mass

Molar mass of hydrogen flouride= 20.01 g/mol

Mass of HF= 0.28 moles × 20.01 g/mol= 5.6 g this is the theoretical yield of HF

Actual yield of HF was given in the question as 2.2g

% yield of HF= actual yield/ theoretical yield ×100

%yield of HF= 2.2/5.6 ×100

% yield of HF= 39.3%

Answer:

A. The theoretical yield of HF is 5.64g

B. The percentage yield of HF is 39%

Explanation:

Step 1:

The balanced equation for the reaction:

CaF2 + H2SO4 --> CaSO4 + 2HF

Step 2:

Determination of the mass of CaF2 that reacted and the mass of HF produced from the balanced equation. This is illustrated below:

Molar Mass of CaF2 = 40 + (19x2) = 40 + 38 = 78g/mol

Molar Mass of HF = 1 + 19 = 20g/mol

Mass of HF from the balanced equation = 2 x 20 = 40g.

From the balanced equation above,

78g of CaF2 reacted and 40g of HF were produced.

A. Determination of the theoretical yield of HF.

This is illustrated below:

From the balanced equation above,

78g of CaF2 reacted to produce 40g of HF.

Therefore, 11g of CaF2 will react to produce = (11 x 40)/78 = 5.64g of HF.

The theoretical yield of HF is 5.64g

B. Determination of the percentage yield.

The percentage yield of HF can be obtained as follow:

Actual yield = 2.2g

Theoretical yield = 5.64g

Percentage yield =?

Percentage yield = Actual yield/Theoretical yield x100

Percentage yield = 2.2/5.64 x 100

Percentage yield = 39%

The percentage yield of HF is 39%

Answer:

The Ksp at 25°C is 1.6 * 10^-5

Explanation:

Step 1: Data given

Temperature = 25°C

Solubility of lead(II) chloride = 1.59 * 10^-2 mol/L

Step 2: The balanced equation

PbCl2(s) <===> Pb2+(aq) + 2Cl-(aq)

Step 3: Calculate the Ksp

Ksp = [Pb2+][Cl-]²

Ksp = [Pb2+][Cl-]²

[Pb2+] = 1.59 *10-2  = 0.0159 M

[Br-] = 2 x 1.59*10-2 = 3.18 *10-2 M

Ksp = (1.59*10-2)(3.18*10-2)²

Ksp =1.6 * 10^-5

The Ksp at 25°C is 1.6 * 10^-5

The value of solubility constant (Ksp) at 25°C in scientific notation is 1.6 × 10-⁵.

According to this question, the following parameters are given:

The balanced equation is as follows:

PbCl2(s) <==> Pb2+(aq) + 2Cl-(aq)

The solubility constant can be calculated as follows:

Ksp = [Pb2+][Cl-]²

Ksp = [Pb2+][Cl-]²

[Pb2+] = 1.59 × 10-² = 0.0159 M

[Br-] = 2 x 1.59*10-² = 3.18 × 10-² M

Ksp = (1.59 × 10-²)(3.18 × 10-²)²

Ksp =1.6 × 10-⁵

Therefore, the solubility constant (Ksp) at 25°C is 1.6 × 10-⁵.

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Answer:

True:

Explanation:

Hybrid orbitals do not exist in isolated atoms. They form only in covalently bonded atoms.

Hybridization happens when several atomic orbitals combine to form other orbitals with the same energy and greater stability.

A set of hybrid orbitals is generated by combining atomic orbitals. The number of hybrid orbitals in a set is equal to the number of atomic orbitals that combined to produce the set.

Hybrid orbitals overlap to form σ bonds. Unhybridized orbitals overlap to form π bonds, and both can appear in an atom at the same time.

The statements about hybridization that are true include:

B. Hybrid orbitals within the same atom have the same energy and shape.

C. Hybrid orbitals are described mathematically as a linear combination of atomic orbitals.

D. An atom can have both hybridized and unhybridized orbitals at the same time.

A sublevel is an energy level that is typically associated with the valence electrons found outside an atomic nucleus.

In Chemistry, there are four (4) types of sublevel and these includes:

I. s orbital (sublevel): it has one (1) orbital i.e 1s.

II. p orbital (sublevel): it has three (3) orbitals.

III. d orbital (sublevel): it has five (5) orbitals.

IV. f orbital (sublevel): it has seven (7) orbitals.

Hybridization can be defined as a phenomenon which involves the linear combination of two or more atomic orbitals of a molecule, so as to form the same number of hybrid orbitals, with each of the orbital having the same energy and shape.

Generally, the two types of hybridization an atom can have include:

1. Hybridized orbitals.

2. Unhybridized orbitals

Hence, we can deduce the following points from the above:

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The formation of 98.7 grams of Fe from the reaction Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s) with an enthalpy change of -852 kJ per 112 grams of Fe formed, would result in approximately -753 kJ of energy being released. The reaction is exothermic, meaning it releases energy as heat.

The question asks about the energy evolved during the formation of 98.7 grams of iron (Fe) from the reaction given: Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s) with a enthalpy change (ΔH°rxn) of -852 kJ. The reaction is exothermic, meaning it releases energy as heat. The ΔH for the reaction is -852 kJ per 112 grams of Fe formed (the molar mass of Fe is 56 g/mol, so 2 mol of Fe is 112 g).

To find out how much energy is released in forming 98.7 g of Fe, you can use simple proportion: (98.7 g Fe / 112 g Fe) x -852 kJ = approx. -753 kJ. So, the formation of 98.7 grams of Fe would result in approximately -753 kJ of energy being released, pointing to answer 753 kJ among the options given.

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The energy evolved during the formation of 98.7 g of Fe is 753 kJ.

The given reaction is:

Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s)

The enthalpy change for the reaction is -852 kJ.

To calculate the energy evolved during the formation of 98.7 g of Fe, we need to use the molar mass of Fe.

The molar mass of Fe is 55.845 g/mol.

By converting the given mass of Fe to moles and using the stoichiometry of the balanced equation, we can calculate the energy evolved.

98.7 g Fe * (1 mol Fe / 55.845 g Fe) * (-852 kJ / 2 mol Fe) = -753 kJ

Therefore, the energy evolved during the formation of 98.7 g of Fe is 753 kJ.

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Answer:

Kb =  7.1 x 10⁻¹³

Explanation:

Ka x Kb = Kw => Kb = 1 x 10⁻¹⁴/1.4 x 10⁻² = 7.1 x 10⁻¹³

The Kb for SO32- at 25ºC is approximately 1.5 x 10^-7.

therefore correct option (b).

To find the Kb for SO32-, we can use the relationship between Ka and Kb. Since the diprotic acid, H2SO3, forms SO32- in the second dissociation step, we can use the formula Kb = Kw/Ka. Kb is the equilibrium constant for the reaction of a base with water to form the hydroxide ion. Kw is the ion product of water, which is 1.0 x 10^-14 at 25ºC. So, using the given Ka2 value of 6.7 x 10^-8, we can calculate the Kb for SO32-:

Kb = Kw/Ka = (1.0 x 10^-14) / (6.7 x 10^-8)

Kb ≈ 1.5 x 10^-7

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Answer:

calories burned = 202 calories

Explanation:

22 minutes = 22/60th of an hour => calories burned in 22 min = (22/60)550 calories = 202 calories.

Answer:

202 calories is the estimate of calories burned

Explanation:

22 minutes = 22/60th of an hour => calories burned in 22 min = (22/60)550 calories = 202 calories.

Switching one substance for another is called displacement.

Chemical reactions are those reactions in which reactants are react with each other for the formation of product.

Hence, given reaction is the displacement reaction.

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Answer:

44.63g

Explanation:

First, let us calculate the number of mole of KBr in 1.50M KBr solution.

This is illustrated below:

Data obtained from the question include:

Volume of solution = 250mL = 250/1000 = 0.25L

Molarity of solution = 1.50M

Mole of solute (KBr) =.?

Molarity is simply mole of solute per unit litre of solution

Molarity = mole /Volume

Mole = Molarity x Volume

Mole of solute (KBr) = 1.50 x 0.25

Mole of solute (KBr) = 0.375 mole

Now, we calculate the mass of KBr needed to make the solution as follow:

Molar Mass of KBr = 39 + 80 = 119g/mol

Mole of KBr = 0.375 mole

Mass of KBr =?

Mass = number of mole x molar Mass

Mass of KBr = 0.375 x 119

Mass of KBr = 44.63g

Therefore, 44.63g of KBr is needed to make 250.0mL of 1.50 M potassium bromide (KBr) solution

Final answer:

To make a 1.50 M potassium bromide solution with a volume of 250.0 mL, you would need approximately 44.625 grams of potassium bromide.

Explanation:

To calculate the mass of potassium bromide required, we can use the formula:

mass (g) = concentration (M) x volume (L) x molar mass (g/mol)

The concentration is given as 1.50 M, the volume is 250.0 mL (which can be converted to 0.250 L), and the molar mass of potassium bromide (KBr) is 119.0 g/mol. Plugging in these values, we get:

mass (g) = 1.50 M x 0.250 L x 119.0 g/mol = 44.625 g

Therefore, you would need approximately 44.625 grams of potassium bromide to make 250.0 mL of a 1.50 M potassium bromide solution.

Answer:

Explanation:

Attach is the solution

Final answer:

The formation of an imine involves nucleophilic addition, carbinolamine formation, and subsequent deprotonations leading to the imine, which is then reduced to an amine by sodium borohydride through nucleophilic attack.

Explanation:

Arrow-pushing Mechanism for Imine and Amine Formation

The question involves describing the arrow-pushing mechanisms for the formation of an imine from an aldehyde and ammonia, followed by the conversion of the imine to an amine using sodium borohydride. Let's address this step by step.

Step 1: Formation of Imine

The formation of an imine involves several key steps:

Nucleophilic addition of the amine to the carbonyl carbon of the aldehyde.Formation of carbinolamine via proton transfer.Protonation of carbinolamine oxygen turns it into a better leaving group, facilitating its elimination as water and resulting in the formation of an iminium ion.Finally, deprotonation of the nitrogen atom yields the imine.

Step 2: Conversion of Imine to Amine

In the next step, sodium borohydride (NaBH4) is used as a reducing agent to convert the imine into an amine. This involves the nucleophilic attack of hydride (H:-) from NaBH4 on the carbon atom of the iminium ion, leading to the formation of an amine.

Answer:

3:1 M ratio at pH 3-4

Explanation:

Answer: The half-cell reaction occurring at cathode is [tex]PdCl_4^{2-}(aq)+2e^-\rightarrow Pd(s)+4Cl^-(aq)[/tex]

Explanation:

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

Reduction reaction is defined as the reaction in which a chemical specie accepts electrons. The oxidation state of the chemical specie reduces.

The given balanced chemical equation is:

[tex]PdCl_4^{2-}(aq)+Cd(s)\rightarrow Pd(s)+4Cl^-(aq)+Cd^{2+}(aq)[/tex]

The half cell reaction occurring at cathode follows:

[tex]PdCl_4^{2-}(aq)+2e^-\rightarrow Pd(s)+4Cl^-(aq)[/tex]

Hence, the half-cell reaction occurring at cathode is given above.

Question:

A) 12

B) 29

C) 2.1 × 10⁻²

D) 8.7 × 10⁻²

E) 47

Answer:

The correct option is;

E) 47

Explanation:

Kc, which is the equilibrium constant of a chemical reaction is derived by finding the ratio between the product of the equilibrium concentration of the product raised to their respective coefficients to the product of the equilibrium concentration of the reactants also raised to their respective coefficients.

Here we have;

[H₂] = 0.14 M

[Cl₂] = 0.39 M

[HCl] = 1.6

The reaction is given as follows;

H₂ (g) + Cl₂ (g) ⇄ 2HCl (g)

The formula for Kc is given as follows;

[tex]Kc = \frac{[HCl]^2}{[H_2][Cl_2]} = \frac{1.6^2}{0.14 \times 0.39} = 46.886[/tex]

Therefore, the Kc for the reaction is approximately equal to 47.

Answer:

The correct answer is B. 734 g

Explanation:

The chemical formulae of lithium sulfate is Li₂SO₄. With this, we can calculate the molecular weight (MM) of lithium sulfate as follows:

MM(Li₂SO₄) = (2 x molar mass Li) + molar mass S + (4 x molar mass O)

                   = (2 x 6.9 g/mol) + 32 g/mol + (4 x 16 g/mol)

                   = 109.9 g/mol

We need to prepare a solution with a molarity of 2.67 M. That means that the solution has to have 2.67 moles of Li₂SO₄ per liter of solution. We can convert from mol to grams with the calculated molecular weight and then we have to multiply by the volume of solution (2500 ml= 2.5 L), as follows:

Mass = 2.67 mol/L x 109.9 g/mol x 2.5 L ≅ 734 g

Answer:

1.04g of iron III carbonate

Explanation:

First, we must put down the equation of reaction because it must guide our work.

2Fe(NO3)3(aq) + 3Na2CO3(aq)→Fe2(CO3)3(s) + 6NaNO3(aq)

From the question, we can see that sodium carbonate is in excess while sodium nitrate is the limiting reactant.

Number of moles of iron III nitrate= mass of iron III nitrate reacted/ molar mass of iron III nitrate

Mass of iron III nitrate reacted= 1.72g

Molar mass of iron III nitrate= 241.88 g∙mol–1

Number of moles of iron III nitrate= 1.72g/241.88 g∙mol–1= 7.11×10^-3 moles

From the equation of the reaction;

2 moles of iron III nitrate yields 1 mole of iron III carbonate

7.11×10^-3 moles moles of iron III nitrate yields 7.11×10^-3 × 1/ 2= 3.56×10^-3 moles of iron III carbonate

Theoretical mass yield of iron III carbonate = number of moles of iron III carbonate × molar mass

Theoretical mass yield of iron III carbonate = 3.56×10^-3 moles ×291.73 g∙mol–1 = 1.04g of iron III carbonate

The theoretical yield of Fe₂(CO₃)₃ obtained from the reaction is 1.04 g

We'll begin by calculating the mass of Fe(NO₃)₃ that reacted and the mass of Fe₂(CO₃)₃ produced from the balanced equation.

2Fe(NO₃)₃ + 3Na₂CO₃ —> Fe₂(CO₃)₃ + 6NaNO₃

Molar mass of Fe(NO₃)₃ = 241.88 g/mol

Mass of Fe(NO₃)₃ from the balanced equation = 2 × 241.88 = 483.4 g

Molar mass of Fe₂(CO₃)₃ = 291.73 g/mol

Mass of Fe₂(CO₃)₃ from the balanced equation = 1 × 291.73 = 291.73 g

From the balanced equation above,

483.4 g of Fe(NO₃)₃ reacted to produce 291.73 g of Fe₂(CO₃)₃

From the balanced equation above,

483.4 g of Fe(NO₃)₃ reacted to produce 291.73 g of Fe₂(CO₃)₃

Therefore,

1.72 g of Fe(NO₃)₃ will react to produce = (1.72 × 291.73) / 483.4 = 1.04 g of Fe₂(CO₃)₃

Thus, the theoretical yield of Fe₂(CO₃)₃ is 1.04 g

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Answer:

3.88 × 10²⁴ molecules

Explanation:

In order to solve this question, we need to consider the Avogadro's number. know that 1 mole of particles contains 6.02 × 10²³ particles. This applies to different kinds of particles: atoms, molecules, electrons.

In this case, 1 mole of molecules of oxygen gas contains 6.02 × 10²³ molecules of oxygen gas. We will use this relation to find the number of molecules of oxygen gas in 6.44 moles of oxygen gas.

[tex]6.44mol \times \frac{6.02 \times 10^{23}molecules }{1mol} = 3.88 \times 10^{24}molecules[/tex]

Answer:

Ka = 1.4 x 10⁻⁴

Explanation:

HC₆H₁₁O₂ ⇄ H⁺ + C₆H₁₁O₂⁻

At equilibrium 1.1M HC₆H₁₁O₂ => [H⁺] = [C₆H₁₁O₂⁻] = 10^-pH = 10⁻²°⁴ = 4.0 x 10⁻³M

Ka = [H⁺][C₆H₁₁O₂⁻]/[HC₆H₁₁O₂] = (4.0 x 10⁻³)²/1.1 = 1.4 x 10⁻⁵    

Answer:

Ka = 1.4 x 10⁻⁴

Explanation:

HC₆H₁₁O₂ ⇄ H⁺ + C₆H₁₁O₂⁻

At equilibrium

1.1M HC₆H₁₁O₂ => [H⁺] = [C₆H₁₁O₂⁻] = 10^-pH = 10⁻²°⁴ = 4.0 x 10⁻³M

Ka = [H⁺][C₆H₁₁O₂⁻]/[HC₆H₁₁O₂] = (4.0 x 10⁻³)²/1.1 = 1.4 x 10⁻⁵  

Answer:

132.93 L

Explanation:

Step 1:

Data obtained from the question:

Volume (V) =?

Number of mole (n) = 2.79 moles

Pressure (P) = 439 mmHg

Temperature (T) = 64°C

Step 2:

Conversion to appropriate unit

For pressure:

760mmHg = 1atm

Therefore, 439 mmHg = 439/760 = 0.58 atm

For temperature:

Temperature (Kelvin) = temperature (celsius) + 273

temperature (celsius) = 64°C

Temperature (Kelvin) = 64°C + 273 = 337 K

Step 3:

Determination of the volume.

The volume occupied by N2 can be obtained by using the ideal gas equation as follow:

Volume (V) =?

Number of mole (n) = 2.79 moles

Pressure (P) = 0.58 atm

Temperature (T) = 337 K

Gas constant (R) = 0.082atm.L/Kmol

PV = nRT

0.58 x V = 2.79 x 0.082 x 337

Divide both side by 0.58

V = (2.79 x 0.082 x 337)/0.58

V = 132.93 L

Therefore, the volume occupied by N2 is 132.93 L

Answer: The theoretical yield of 4-nitrochalcone is, [tex]2.13\times 10^2[/tex]

Explanation : Given,

Volume of acetophenone = 135  microliters = 135 × 10⁻⁶ L = 0.135 mL

conversion used : (1 microliter = 10⁻⁶ L) and (1 L = 1000 mL)

Density of acetophenone = 1.03 g/mL

Mass of acetophenone = Density × Volume = 1.03 g/mL × 0.135 mL = 0.139 g

Mass of 4-nitrobenzaldehyde = 127 mg  = 0.127 g

Conversion used : (1 mg = 0.001 g)

First we have to calculate the moles of acetophenone and 4-nitrobenzaldehyde

[tex]\text{Moles of acetophenone}=\frac{\text{Given mass acetophenone}}{\text{Molar mass acetophenone}}[/tex]

[tex]\text{Moles of acetophenone}=\frac{0.139g}{120.15g/mol}=0.00116mol[/tex]

and,

[tex]\text{Moles of 4-nitrobenzaldehyde}=\frac{\text{Given mass 4-nitrobenzaldehyde}}{\text{Molar mass 4-nitrobenzaldehyde}}[/tex]

[tex]\text{Moles of 4-nitrobenzaldehyde}=\frac{0.127g}{151.12g/mol}=0.000840mol[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

[tex]C_8H_8O+C_7H_5NO_3\rightarrow C_{15}H_{11}NO_3[/tex]

From the balanced reaction we conclude that

As, 1 mole of 4-nitrobenzaldehyde react with 1 mole of acetophenone

So, 0.000840 mole of 4-nitrobenzaldehyde react with 0.000840 mole of acetophenone

From this we conclude that, acetophenone is an excess reagent because the given moles are greater than the required moles and 4-nitrobenzaldehyde is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of 4-nitrochalcone

From the reaction, we conclude that

As, 1 mole of 4-nitrobenzaldehyde react to give 1 mole of 4-nitrochalcone

So, 0.000840 mole of 4-nitrobenzaldehyde react to give 0.000840 mole of 4-nitrochalcone

Now we have to calculate the mass of 4-nitrochalcone

[tex]\text{ Mass of 4-nitrochalcone}=\text{ Moles of 4-nitrochalcone}\times \text{ Molar mass of 4-nitrochalcone}[/tex]

Molar mass of 4-nitrochalcone = 253.25 g/mole

[tex]\text{ Mass of 4-nitrochalcone}=(0.000840moles)\times (253.25g/mole)=0.21273g=212.73mg=2.13\times 10^2mg[/tex]

(1 g = 1000 g)

Therefore, the theoretical yield of 4-nitrochalcone is, [tex]2.13\times 10^2mg[/tex]

Answer : The theoretical yield and percent yield for this reaction is, 3.78 grams and 31.7 % respectively.

Explanation : Given,

Mass of [tex]C_6H_5COOH[/tex] = 3.4 g

Molar mass of [tex]C_6H_5COOH[/tex] = 122.12 g/mol

First we have to calculate the moles of [tex]C_6H_5COOH[/tex]

[tex]\text{Moles of }C_6H_5COOH=\frac{\text{Given mass }C_6H_5COOH}{\text{Molar mass }C_6H_5COOH}[/tex]

[tex]\text{Moles of }C_6H_5COOH=\frac{3.4g}{122.12g/mol}=0.0278mol[/tex]

Now we have to calculate the moles of [tex]C_6H_5COOCH_3[/tex]

The balanced chemical equation is:

[tex]C_6H_5COOH+CH_3OH\rightarrow C_6H_5COOCH_3[/tex]

From the reaction, we conclude that

As, 1 mole of [tex]C_6H_5COOH[/tex] react to give 1 mole of [tex]C_6H_5COOCH_3[/tex]

So, 0.0278 mole of [tex]C_6H_5COOH[/tex] react to give 0.0278 mole of [tex]C_6H_5COOCH_3[/tex]

Now we have to calculate the mass of [tex]C_6H_5COOCH_3[/tex]

[tex]\text{ Mass of }C_6H_5COOCH_3=\text{ Moles of }C_6H_5COOCH_3\times \text{ Molar mass of }C_6H_5COOCH_3[/tex]

Molar mass of  = 136.14 g/mole

[tex]\text{ Mass of }C_6H_5COOCH_3=(0.0278moles)\times (136.14g/mole)=3.78g[/tex]

The theoretical yield of [tex]C_6H_5COOCH_3[/tex] produced is, 3.78 grams.

Now we have to calculate the percent yield of the reaction.

Theoretical yield of the reaction = 3.78 g

Experimental yield of the reaction = 1.2 g

The formula used for the percent yield will be :

[tex]\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]

Now put all the given values in this formula, we get:

[tex]\text{Percent yield}=\frac{1.2g}{3.78g}\times 100=31.7\%[/tex]

The percent yield of the reaction is, 31.7 %

Answer:

W = - 10.5943 KJ, work is negative because it is carried out by the system towards the surroundings.

Explanation:

heat engine:

∴ Th = 485°C

∴ Tc = 42°C

∴ Qh = 9.75 KJ......heat from hot source

first law:

∴ ΔU = Q + W = 0 .........cyclic process

⇒ Q = - W

∴ Q = Qh + Qc = Qh - (- Qc)

∴ Qc: heat from the cold source ( - )

⇒ Qh - ( - Qc) = - W..............(1)

⇒ Qc/Qh = - Tc/Th...........(2)

from (2):

⇒ Qc = - (Tc/Th)(Qh) = - (42°C/485°C)(9.75 KJ)

⇒ Qc = - 0.8443 KJ

replacing in (1):

⇒ - W = 9.75 KJ - ( - 0.8443 KJ)

⇒ - W = 10.5943 KJ

Answer:

16.36A

Explanation:

We'll begin by writing a balanced dissociation equation of aqueous Cr2(SO4)3. This is illustrated below:

Cr2(SO4)3 —> 2Cr^3+ 3(SO4)^2-

From the above, we can see that Cr is trivalent.

Next, let us determine the number of faraday needed to deposit metallic Cr. This is illustrated below:

Cr^3+ 3e- —> Cr

From the above equation, 3 faradays are needed to deposit metallic Cr

1 faraday = 96500C

Therefore, 3 faraday = 3 x 96500C = 289500C.

Molar Mass of Cr = 52g/mol

Now let us determine the quantity of electricity needed for 2.68g of Cr metal

This is shown below:

52g of Cr required 289500C.

Therefore, 2.68g of Cr will require = (2.68 x 289500)/52 = 14920.38C

Now, with this quantity of electricity (i.e 14920.38C), we can easily calculate the current needed for the process. This is illustrated below:

Q (quantity of electricity) = 14920.38C

t (time) = 15.2mins = 15.2 x 60 = 912secs

I (current) =?

Apply the equation Q = It

Q = It

14920.38 = I x 912

Divide both side by 912

I = 14920.38/912

I = 16.36A

Therefore, a current of 16.36A is needed for the process.

The current needed is 16.36 A. The quantity of electricity (i.e 14920.38C), we can easily calculate the current needed for the process.

[tex]Cr_2(SO_4)_3 ---- > 2Cr^{3+}+ 3SO_4^{2-}[/tex]

The number of faradays needed to deposit metallic Cr. This is illustrated below:

[tex]Cr^{3+}+ 3e^- ---- > Cr[/tex]

Given:

3 faradays are needed to deposit metallic Cr

1 faraday = 96500C

Therefore, 3 faraday = 3 * 96500C = 289500C.

Molar Mass of Cr = 52g/mol

52g of Cr required 289500C.

Therefore, 2.68g of Cr will require = (2.68 * 289500)/52 = 14920.38C

Now, with this quantity of electricity (i.e 14920.38C), we can easily calculate the current needed for the process. This is illustrated below:

Q (quantity of electricity) = 14920.38C

t (time) = 15.2mins = 15.2 x 60 = 912secs

To find:

I (current) =?

Apply the equation,

Q = It

14920.38 = I * 912

I = 14920.38/912

I = 16.36A

Therefore, a current of 16.36A is needed for the process.

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Answer: ΔH for the combustion of propanol to carbon dioxide gas and liquid water is 1980 kJ

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

[tex]Q=C\times \Delta T[/tex]

Q = Heat absorbed by calorimeter =?

C = heat capacity of calorimeter = 13.70 kJ/K

Initial temperature of the calorimeter  = [tex]T_i[/tex] = 298.00 K

Final temperature of the calorimeter  = [tex]T_f[/tex]  = 302.34 K

Change in temperature ,[tex]\Delta T=T_f-T_i=(302.34-298.00)K=4.34K[/tex]

Putting in the values, we get:

[tex]Q=13.70kJ/K\times 4.34K=59.4kJ[/tex]

As heat absorbed by calorimeter is equal to heat released by combustion of propanol

[tex]Q=q[/tex]

[tex]\text{Moles of propanol}=\frac{\text{given mass}}{\text{Molar Mass}}=\frac{1.771g}{60g/mol}=0.030mol[/tex]  

Heat released by 0.030 moles of propanol = 59.4 kJ

Heat released by 1 mole of propanol = [tex]\frac{59.4}{0.030}\times 1=1980kJ[/tex]

ΔH for the combustion of propanol to carbon dioxide gas and liquid water is 1980 kJ/mol