Steam reforming of methane (CH4) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills 1.5L a flask with of methane gas and 2.6 atm of water vapor at 47C, and when the mixture has come to equilibrium measures the partial pressure of carbon monoxide gas to be 1.4 atm.


Calculate the pressure equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.

Answer :  The value of standard free energy is, -152.4 kJ/mol

Explanation :

The given balanced cell reaction is:

[tex]FADH_2+\frac{1}{2}O_2\rightarrow FAD+H_2O[/tex]

The half reaction will be:

Reaction at anode (oxidation) : [tex]FADH_2\rightarrow FAD+2H^++2e^-[/tex]     [tex]E^0_{Anode}=+0.03V[/tex]

Reaction at cathode (reduction) : [tex]\frac{1}{2}O_2+2H^++2e^-\rightarrow H_2O[/tex]     [tex]E^0_{Cathode}=+0.82V[/tex]

First we have to calculate the standard electrode potential of the cell.

[tex]E^o=E^o_{cathode}-E^o_{anode}[/tex]

[tex]E^o=(+0.82V)-(+0.03V)=+0.79V[/tex]

Relationship between standard Gibbs free energy and standard electrode potential follows:

[tex]\Delta G^o=-nFE^o_{cell}[/tex]

where,

[tex]\Delta G^o[/tex] = standard free energy = ?

n = number of electrons transferred = 2

F = Faraday constant = [tex]96.48kJ.mol^{-1}V^{-1}[/tex]

[tex]E^o_{cell}[/tex]  = standard electrode potential of the cell = 0.79 V

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

[tex]\Delta G^o=-(2)\times (96.48kJ.mol^{-1}V^{-1})\times (0.79V)[/tex]

[tex]\Delta G^o=-152.4kJ/mol[/tex]

Therefore, the value of standard free energy is, -152.4 kJ/mol

To calculate the standard free energy change for the reduction of O2 with FADH2, we first calculate the difference in reduction potentials to find ΔE0′. We then substitute the values into the relation ΔG∘′ = −nFΔE0′, with n=2 and Faraday constant F=96.48 kJ⋅mol‾¹⋅V‾¹. The calculated ΔG∘′ is -152.4384 kJmol‾¹.

The standard free energy change, ΔG∘′, is related to the change in reduction potential, ΔE0′, by the equation ΔG∘′ = −nFΔE0′. To find the standard free energy released by the reduction of O2 with FADH2, we first need to find ΔE0′.

ΔE0′ is given by the difference in reduction potentials of the two half reactions involved. In this case, the reduction of O2 to H2O (+0.82 V) and the reduction of FAD to FADH2 (+0.03 V). Therefore, ΔE0′ = E(O2/H2O) - E(FAD/FADH2) = +0.82 V - (+0.03 V) = +0.79 V.

Inserting the values into the equation, and knowing that the number of transferred electrons (n) is 2 and the Faraday constant (F) is 96.48 kJ⋅mol‾¹⋅V‾¹, we get ΔG∘′ = −2 * 96.48 kJ⋅mol‾¹⋅V‾¹ * (+0.79 V) = -152.4384 kJmol‾¹. Thus, the standard free energy released by reducing O2 with FADH2 is -152.4384 kJmol‾¹.

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Answer:1 2 4

Explanation:

Answer: 1.3.5.6.

Explanation:

Answer:

Saturated .

Explanation:

Final answer:

The first intermediate formed during a halohydrin reaction is a halonium ion. This intermediate is key in the process of adding a halogen and an OH group to adjacent carbons in the formation of halohydrins.

Explanation:

The expected first intermediate formed during a halohydrin reaction is c. ahalonium ion. In the presence of a halogen and water, the reaction of an alkene leads to the formation of a halonium ion intermediate. Specifically, the alkene undergoes an addition reaction with a halogen such as bromine or chlorine to form a three-membered ring halonium intermediate. This is followed by the nucleophilic attack by water, which opens the ring and leads to the formation of the halohydrin, with the halogen and an OH group on two adjacent carbons. Halohydrins such as bromohydrins or chlorohydrins are important intermediates in organic synthesis.

Answer:

53.4 gMol-1

Explanation:

Let the mass of the unknown gas be M

Let the molar mass of hydrogen gas be 2×1=2gMol-1

Time for diffusion of unknown gas = 12.5 min

Time for diffusion of hydrogen= 2.42 min

From Graham's law:

t1/t2=√M1/M2

Hence:

2.42/12.5= √2/M

Hence M= 53.4 gMol-1

Answer:

The molar mass of the unknown gas is 40.06 g/mol

Explanation:

Graham's law of effusion states that the rate of effusion of a gaseous substance is inversely proportional to the square root of its molar mass.

[tex]\frac{R_{b} }{R_{a} } = \sqrt{\frac{M_{a} }{M_{b} } }[/tex]  [tex]= \frac{t_{a} }{t_{b} }[/tex] where R = rate of effusion, M = molar mass and t= time of effusion

⇒ [tex]\sqrt{\frac{2 g/mol}{x g/mol} } = \frac{162secs}{725secs}[/tex]

x g/mol = [tex]\frac{2}{0.223448^{2} }[/tex]

= 40.06 g/mol

Answer:

New volume is 271 mL

Explanation:

To determine the volume for a gas, when the pressure remains constant we follow this ratio:

V₁ / T₁ = V₂ / T₂

Remember the Ideal Gases Law → P . V = n . R . T

That's why we propose V / T

We need to determine the Absolute T°

25°C + 273 = 298 K

50°C + 273 = 323 K

We convert the volume from mL to L → 250 mL . 1L / 1000 mL = 0.250L

Now we replace: 0.250L / 298K = V₂ / 323K

V₂ = (0.250L / 298K) . 323K  → 0.271 L

In conclussion volume of a gas will be increased, while the temperature is also increased and the pressure remains constant.

Answer:

The final volume is 271 mL

Explanation:

Step 1: Data given

The initial pressure = 2 atm

The pressure will be kept constant

The initial volume = 250 mL = 0.250 L

The initial temperature = 25°C = 298 K

The final temperature = 50 °C = 323 K

Step 2: Calculate the final volume

V1/T1 = V2/T2

⇒with V1 = the initial volume = 0.250 L

⇒with T1 = the initial temperature = 25 °C = 298 K

⇒with V2 = the final volume = TO BE DETERMINED

⇒with T2 = the final temperature = 50 °C = 323 K

0.250 L / 298 K = V2 / 323 K

V2 = (0.250 /298) *323

V2 = 0.271 L = 271 mL

The final volume is 271 mL

Answer : The  pH of buffer is, 5.17

Explanation : Given,

[tex]pK_a=4.75[/tex]

Concentration of acetic acid = 1.00 M

Concentration of sodium acetate = 1.00 M

Volume of solution = 1.00 L

As, [tex]Moles=Concentration\times Volume[/tex]

So,

Moles of acetic acid = 1.00 mol

Moles of sodium acetate = 1.00 mol

Moles of NaOH added = 0.450 mol

The balanced chemical equilibrium reaction is:

                      [tex]CH_3COO+NaOH\rightleftharpoons CH_3COONa+H_2O[/tex]

Initial mole        1                0.450              1

At eqm.         (1-0.450)            0            (1+0.450)

                       = 0.55                                =1.450

Now we have to calculate the pH of buffer.

Using Henderson Hesselbach equation :

[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]

[tex]pH=pK_a+\log \frac{[CH_3COONa]}{[CH_3COOH]}[/tex]

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

[tex]pH=4.75+\log (\frac{1.450}{0.55})[/tex]

[tex]pH=5.17[/tex]

Therefore, the pH of buffer is, 5.17

Answer:

(B) ethyl-propyl-ether

Explanation:

ethyl-propyl-ether

a step by step explanation

As the compound has ether as functional group the name of the compound is ethyl propyl ether.

Functional group is defined as a substituent or group of toms or an atom which causes chemical reactions.Each functional group will react similarly regardless to the parent carbon chain to which it is attached.This helps in prediction of chemical reactions.

The reactivity of functional group can be enhanced by making modifications in the functional group .Atoms present in functional groups are linked to each other by means of covalent bonds.They are named along with organic compounds according to IUPAC nomenclature.

The compound has ether as functional group,thus  the name of the compound is ethyl propyl ether.

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Answer:

X is calcium with the symbol Ca

Explanation:

From the first statement:

X reacted with H2 to produce Z i.e

X + H2 —> Y

Y is heated to form X and H2 i.e

Y —> X + H2

Let us obtain moles of H2 produced. This is illustrated below:

Volume of H2 produced = 559 mL = 0.559 L.

I mole of H2 occupy 22.4L at stp

Therefore, b mol of H2 will occupy 0.559 L at stp i.e

b mol of H2 = 0.559/22.4

b mol of H2 = 0.025 mole

Next 0.025 mole of H2 to gram..

Molar Mass of H2 = 2x1 = 2g/mol

Mole of H2 = 0.025 mole

Mass = number of mole x molar Mass

Mass of H2 = 0.025 x 2

Mass of H2 = 0.05g

From the question given, we were told that 1g of X reacted. This means that 1g of X reacted with 0.05g of H2. Now let us determine the mass of X that will react with 1 mol ( i.e 2g) of H2. This is illustrated below:

From the reaction,

It was discovered that 1g of X reacted with 0.05g of H2.

Therefore, P g of X will react with 2g of H2 i.e

P g of X = 2/0.05 = 40g/mol

The molar mass of X is 40g/mol.

Now let us consider the second statement to see if we'll obtained the same result as 40g/mol of X

From the second statement:

X also combines with chlorine to form a compound Z, which contains 63.89 percent by mass of chlorine i.e

X + Cl2 —> Z

Molar Mass of Z (X + 2Cl) = X + (35.5 x 2) = X + 71.

Z contains 63.89% by mass of Cl2.

We can obtain the molar mass of X as follow:

Percentage by mass of Cl2 in the compound Z is given by:

Mass of Cl2/Molar Mass x100

63.89/100 = 71/(X + 71)

Cross multiply to express in linear form

63.89(X + 71) = 100 x 71

Clear the bracket

63.89X + 4536.19 = 7100

Collect like terms

63.89X = 7100 - 4536.19

63.89X = 2563.81

Divide both side by 63.89

X = 2563.81/63.89

X = 40g/mol

Now we can see that in both experiments, the molar mass of X is 40g/mol.

Comparing the value of X i.e 40g/mol with that from the periodic table, X is calcium with the symbol Ca

The pH of an aqueous solution at 25.0°C with [H+] = 0.0025 M is calculated using the pH formula and results in a pH of 2.60, which is answer option B.

The pH of an aqueous solution at 25.0°C with a hydrogen ion concentration [H+] of 0.0025 M can be found using the pH formula:

pH = -log [H3O+]

By substituting the given concentration into the formula, the calculation would be:

pH = -log(0.0025) = -log(2.5 x 10-3)

Using a scientific calculator:

pH = 2.60

Therefore, the correct answer is B) 2.60.

Answer : The moles of water present in solution are, 0.8297 moles.

Explanation :

First we have to calculate the mass of water.

[tex]\text{Mass of water}=\text{Density of water}\times \text{Volume of water}[/tex]

[tex]\text{Mass of water}=0.9956g/mL\times 15.00mL=14.934g[/tex]

Now we have to calculate the moles of water.

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

Molar mass of water = (2 × Molecular weight of hydrogen) + Molecular weight of oxygen

Molar mass of water = (2 × 1.00g/mol) + 16.00 g/mol

Molar mass of water = 18.00 g/mol

[tex]\text{Moles of water}=\frac{14.934g}{18.00g/mol}[/tex]

[tex]\text{Moles of water}=0.8297mol[/tex]

Therefore, the moles of water present in solution are, 0.8297 moles.

Answer:

Explanation:

The product formed that contains only carbons and hydrogens after free radical bromination of toluene is 1,2-diphenylethane.....

Please go through the attached file for the diagrams .

Answer:

happy happy happy happy happy happy happy

Explanation:

Answer:

A pH greater than 7 is basic. The pH scale is logarithmic and as a result, each whole pH value below 7 is ten times more acidic than the next higher value. For example, pH 4 is ten times more acidic than pH 5 and 100 times (10 times 10) more acidic than pH 6.

Answer : The partial pressure of nitrogen gas in the mixture is, 9.88 atm

Explanation :

According to the Dalton's Law, the total pressure of the gas is equal to the sum of the partial pressure of individual gases.

Formula used :

[tex]p_T=p_{He}+p_{Ar}+p_{N_2}[/tex]

where,

[tex]p_T[/tex] = total pressure of gas  = 13.6 atm

[tex]p_{He}[/tex] = partial pressure of helium gas  = 1831 torr =  2.41 atm

[tex]p_{Ar}[/tex] = partial pressure of argon gas  = 997 torr =  1.31 atm

Conversion used: (1 atm = 760 torr)

[tex]p_{N_2}[/tex] = partial pressure of nitrogen gas  = ?

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

[tex]13.6=2.41+1.31+p_{N_2}[/tex]

[tex]p_{N_2}=9.88atm[/tex]

Thus, the partial pressure of nitrogen gas in the mixture is, 9.88 atm

Answer:

Ksp FeS = 5.2441 E-18

Explanation:

         S          S           S

∴ Ksp = [Fe2+]*[S2-].....solubility product constant

∴ [S2-] = 2.29 E-9 M = S

⇒ Ksp = (S)(S) = S²

⇒ Ksp = (2.29 E-9)²

⇒ Ksp = 5.2441 E-18

Answer:

0.27 mol

Explanation:

Given data

Step 1: Convert the temperature to the Kelvin scale

When working with gases, we need to convert all temperatures to the absolute scale, using the following expression.

[tex]K = \° + 273.15\\K = 70 + 273.15 = 343 K[/tex]

Step 2: Calculate the moles of gaseous oxygen

We will apply the ideal gas equation.

[tex]P \times V = n \times R \times T\\n = \frac{P \times V}{R \times T} = \frac{5 atm \times 1.5 L}{\frac{0.0821atm.L}{mol.K} \times 343 K} = 0.27 mol[/tex]

Answer:

1) Capillary action

2) Cohesion

3)Miniscus

4) Adhesion

5) Viscousity

6) Surface tension

Explanation:

Intermolecular forces are forces that holds molecules together in liquid, these is possible by inter-molecular interactions that exist within the liquids.these forces includes forces such as Waals forces and hydrogen bonds.

When there is great inter-molecular forces, there will be high freezing point and boiling point . At a lower inter-molecular forces the boiling point becomes low too,which brings about great fluidity of the liquid. The liquid flow reluctantly where greater force exist in the liquid. Some of those factors used in characterizing which are;

1)adhesion,

2) surface tension,

3)capillary action,

4)cohesion,

5) meniscus,

6)viscosity.

The statement with its corresponding option is as follows:

Capillarity is a phenomenon through which liquids have the ability to rise or fall through a capillary tube.

Cohesion is the force of attraction between adjacent particles within the same body.

The meniscus is the up or down curve on the surface of a liquid that occurs in response to the surface of its container.

Adhesion is the interaction between the surfaces of different bodies and they are held together by intermolecular forces.

Viscosity refers to the resistance that some liquids possess during their gradual flow and deformation as a result of shear stresses or tensile stresses.

Surface tension refers to the amount of energy required to increase the surface of a liquid per unit area.

Therefore, we can conclude that fluids have elemental properties that define and differentiate them from other forms of matter, such as viscosity, surface tension, among others.

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Answer:

Weak Acid solution

Explanation:

Acidic substances usually have a pH of 0.1-6.9 or it is usually less than 7. Acidic substances too have a very unique sour taste.

However it is understood that it conducts electricity poorly which means the acid doesn’t readily dissociate in water. This makes it a weak acid and an example is Acetic acid.

Answer:

Partial pressure of CO₂ is 406.9 mmHg

Explanation:

To solve the question we should apply the concept of the mole fraction.

Mole fraction = Moles of gas / Total moles

We have the total moles of the mixture, if we have the moles for each gas inside. (3.63 moles of O₂, 1.49 moles of N₂ and 4.49 moles of CO₂)

Total moles = 3.63 mol O₂ + 1.49 mol N₂ + 4.49 mol CO₂ = 9.61 moles

To determiine the partial pressure of CO₂ we apply

Mole fraction of CO₂ → mol of CO₂ / Total moles = P. pressure CO₂ / Total P

Partial pressure of CO₂ = (mol of CO₂ / Total moles) . Total pressure

We replace values: (4.49 moles / 9.61 moles) . 871 mmHg = 406.9 mmHg

Answer:

Partial pressure O2 = 329 mmHg

Partial pressure N2 = 135 mmHg

Partial pressure CO2 = 407 mmHg

Explanation:

Step 1: Data given

Number of moles oxygen (O2) = 3.63 moles

Number of moles nitrogen (N2) = 1.49 moles

Number of moles carbon dioxide (CO2) = 4.49 moles

Total pressure = 871 mmHg

Step 2: Calculate total number of moles

Total moles = moles O2 + moles N2 + moles CO2

Total moles = 3.63 + 1.49 + 4.49

Total moles =  9.61 moles

Step 3: Calculate the mol ratio

Mol ratio number of moles compound / total moles

Mol ratio O2 = 3.63 moles / 9.61 moles

Mol ratio O2 = 0.378

Mol ratio N2 = 1.49 moles / 9.61 moles

Mol ratio N2 = 0.155

Mol ratio CO2 = 4.49 moles / 9.61 moles

Mol ratio CO2 = 0.467

Step 4: Calculate partial pressure

Partial pressure = mol ratio * total pressure

Partial pressure O2 = 0.378 * 871 mmHg

Partial pressure O2 = 329 mmHg

Partial pressure N2 = 0.155 * 871mmHg

Partial pressure N2 = 135 mmHg

Partial pressure CO2 = 0.467 * 871 mmHg

Partial pressure CO2 = 407 mmHg

Answer:

1.4

Explanation:

Mass of pure nitric acid = 751mg

Volume of solution  = 290mL

Unknown:

pH of the solution  = ?

Solution:

To solve this problem, we need the concentration of the acid in the aqueous form.

   This is given by molarity;

               Molarity  = [tex]\frac{number of moles }{volume}[/tex]

Since the number of moles of nitric acid is unknown, we can easily solve for it.

      Number of moles of nitric acid  = [tex]\frac{mass}{molar mass}[/tex]

            molar mass of HNO₃   =  1 + 14 + 3(16)  = 63g/mol

             mass of nitric acid  = 751mg  = 0.751g

     Number of moles  = [tex]\frac{0.751}{63}[/tex]    = 0.012mole

Volume of solution = 290mL  = 0.29dm³

Now molarity of the solution  = [tex]\frac{0.012}{0.29}[/tex]   = 0.041moldm⁻³

Since:

     pH  = -log [H₃O⁺]

         HNO₃   +     H₂O  →    H₃O⁺         +         NO₃⁻

      1moldm⁻³                     1moldm⁻³            1moldm⁻³

   0.041moldm⁻³             0.041moldm⁻³    0.041moldm⁻³

  pH  = -log[0.041]   = 1.4

Answer:

Newton's three laws of motion may be stated as follows:

Every object in a state of uniform motion will remain in that state of motion unless an external force acts on it.

Force equals mass times acceleration

For every action there is an equal and opposite reaction

Explanation:

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Among the following substances, NH₄Cl is never a Brønsted-Lowry base in an aqueous solution because it is a salt.

There are different types of theories of acids and bases like Arrhenius theory, Bronsted Lowry theory and Lewis acids and bases theory. According to Bronsted Lowry theory, acids are the substances which release H+ ions in the solution and bases are the substances which accepts H+ ions in the solution.

This also introduce the concept of conjugate acid base pair. A salt can never be an acid or a base because it is made up of both of them.

Given options are sodium hydrogen phosphate, sodium phosphate, ammonium chloride, sodium bicarbonate and Potassium Hydroxide.

Therefore, Among the following substances, NH₄Cl is never a Brønsted-Lowry base in an aqueous solution because it is a salt.

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The Brønsted-Lowry theory, ammonium chloride (NH₄Cl) is never a base in an aqueous solution. It dissociates into NH₄+ and Cl-, where NH₄+ acts as a Brønsted-Lowry acid. This makes NH₄Cl the correct answer.

The Brønsted-Lowry theory defines a base as a substance that can accept a proton (H+) during a chemical reaction. In an aqueous solution, the behavior of different substances can be predicted based on this definition.

Based on the Brønsted-Lowry definition, ammonium chloride (NH₄Cl) is never a base in an aqueous solution.

Answer: The agents of soil erosion are the same as of other types of erosion for example water, ice, wind, and gravity. Soil erosion is more likely where the ground has been disturbed by agriculture, grazing animals, logging, mining, construction, and recreational activities.Basically what I mean is some causes of solid loss is mining, construction

The scaling factor is determined by dividing the compound's actual molar mass (150 g/mol) by the molar mass of its empirical formula CH2O (30 g/mol), which results in a factor of 5. The molecular formula of the compound is C5H10O5.

The student's question is asking about the scaling factor of a compound's molar mass based on its empirical formula. The empirical formula is CH2O, which has a molar mass of 30 g/mol (C = 12, H = 1 x 2 = 2, O = 16). If the compound's actual molar mass is 150 g/mol, we divide the actual molar mass by the empirical formula's molar mass to find the scaling factor.

Scaling factor = Actual molar mass / Empirical formula molar mass

= 150 g/mol / 30 g/mol

= 5

Thus, the actual compound's formula is obtained by multiplying each subscript in the empirical formula by the scaling factor of 5. Therefore, the molecular formula of the compound is C5H10O5.

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Answer: Density = 0.77g/ml

Explanation:

From the question,

Mass = 0.175g

Volume = 0.000225L

Convert the volume in Litres to Millilitres by multiplying it by 1000, hence the volume is 0.225ml.

Therefore,

Density = mass / volume

0.175g / 0.225ml = 0.77g/ml

Answer:

Density = Mass/ Volume,

Mass = 0.000175kg

Volume = 0.000225L,

Density = 0.000175/ 0.000225

Density = 0.78kg/L

Explanation:

Density is an important physical property. Density is the mass of a substance per unit volume. Volume is the amount of space an object occupies. Chemical properties- These are properties that can only be observed by changing the identity of the substance.

Answer:

1. 2NaN₃(s) → 2Na(s) + 3N₂(g)

2. 14.5 g NaN₃

Explanation:

The answer is incomplete, as it is missing the required values to solve the problem. An internet search shows me these values for this question. Keep in mind that if your values are different your result will be different as well, but the solving methodology won't change.

" The airbags that protect people in car crashes are inflated by the extremely rapid decomposition of sodium azide, which produces large volumes of nitrogen gas. 1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide (NaN₃) into solid sodium and gaseous dinitrogen. 2. Suppose 71.0 L of dinitrogen gas are produced by this reaction, at a temperature of 16.0 °C and pressure of exactly 1 atm. Calculate the mass of sodium azide that must have reacted. Round your answer to 3 significant digits. "

1. The reaction that takes place is:

2. We use PV=nRT to calculate the moles of N₂ that were produced.

P = 1 atm

V = 71.0 L

n = ?

T = 16.0 °C ⇒ 16.0 + 273.16 = 289.16 K

Now we convert N₂ moles to NaN₃ moles:

Finally we convert NaN₃ moles to grams, using its molar mass:

Answer:

Explanation:

find the solution below

Answer:

10.8 ml

Explanation:

The BOD is an empirical test to determine the molecular oxygen used during a specified incubation period (usually five days), for the biochemical degradation of organic matter (carbonaceous demand) and the oxygen used to oxidise inorganic matter.

See attached file

The maximum volume of treated wastewater that will be in the bottle is 10.8 mL.

The given parameters;

The dilution factor (P) is calculated as follows;

[tex]200 \ mg/L= \frac{9.2 \ mg/L \ - \ 2\ mg/L}{P} \\\\P = \frac{7.2 \ mg/L}{200 \ mg/L} \\\\P = 0.036[/tex]

The maximum  volume of treated wastewater that will be in the bottle to have at least 2.0 mg/L DO;

[tex]0.036 = \frac{V_w}{300 \ mL} \\\\V_w = 0.036 \times 300 \ mL\\\\V_w = 10.8 \ mL[/tex]

Thus, the maximum volume of treated wastewater that will be in the bottle is 10.8 mL.

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Final answer:

The maximum possible mass of NO2 that could be formed from the given amounts of reactants is 755.20 grams, and the amount of CO2 that would be produced from the combustion of 37.3 grams of C6H6 is 126.10 grams.

Explanation:

Calculating the Maximum Amount of NO2 Formed

For the balanced reaction 2 NO + O2 → 2 NO2, the stoichiometry shows a 1:1 ratio between NO and NO2. With 16.42 mol of NO, if O2 is in excess, 16.42 mol of NO2 could theoretically be formed. The molar mass of NO2 is 46.0055 g/mol, so the maximum mass of NO2 would be 16.42 mol × 46.0055 g/mol = 755.20 grams of NO2.

Amount of CO2 Produced from C6H6 Combustion

The balanced reaction for the combustion of benzene (C6H6) is 2 C6H6 + 15 O2 → 12 CO2 + 6 H2O. One mole of C6H6 (78.1134 g/mol) produces 6 moles of CO2. With 37.3 grams of C6H6, there would be 37.3 g / 78.1134 g/mol = 0.4775 mol C6H6 which would yield 0.4775 mol C6H6 × 6 mol CO2/mol C6H6 = 2.865 mol CO2.

The molar mass of CO2 is 44.0095 g/mol, so the mass of CO2 would be 2.865 mol × 44.0095 g/mol = 126.10 grams of CO2.

Final answer:

The maximum amount of NO2 that can be formed from 16.42 mol of NO and 14.47 mol of O2 is 755.1 grams. When reacting 37.3 grams of C6H6 with 126.1 grams of O2, the amount of CO2 produced is 125.96 grams.

Explanation:

Reaction Stoichiometry and Limiting Reactants:

To determine the maximum amount of NO2 that can be formed from 16.42 mol of NO and 14.47 mol of O2, we use the balanced chemical equation 2 NO + O2 → 2 NO2. This equation shows that 2 moles of NO react with 1 mole of O2 to produce 2 moles of NO2. We can see that NO will be the limiting reactant because 14.47 mol of O2 would theoretically react with only 28.94 mol of NO, which is more than the 16.42 mol of NO present.

Since NO is the limiting reactant, we can form a maximum of 16.42 mol of NO2. Using the molar mass of NO2 (46.01 g/mol), this is equivalent to 755.1 grams of NO2. For the reaction C6H6 + O2 → CO2 + H2O, first, we would calculate the moles of C6H6 and O2, then use stoichiometry to determine the moles of CO2 produced. The molar mass of C6H6 (78.11 g/mol) means we start with 0.477 moles of C6H6.

From the balanced equation, we can see that for every 2 moles of NO, we need 1 mole of O2 to produce 2 moles of NO2. This gives us the mole ratio of NO to O2 as 2:1.

To find the limiting reagent (the reactant that is completely consumed and determines the maximum amount of product formed), we compare the number of moles of each reactant to the mole ratio.

For NO:

16.42 mol NO * (1 mol O2 / 2 mol NO) = 8.21 mol O2

For O2:

14.47 mol O2 * (2 mol NO / 1 mol O2) = 28.94 mol NO

Since the calculated amount of O2 (28.94 mol) is greater than the actual amount of O2 (14.47 mol), O2 is in excess and NO is the limiting reagent.

Now, we need to calculate the maximum amount of NO2 produced from the limiting reagent, which is NO.

From the balanced equation, we know that 2 moles of NO react to form 2 moles of NO2. Therefore, the mole ratio of NO to NO2 is 2:2.

Using the moles of NO (8.21 mol), we can calculate the moles of NO2:

8.21 mol NO * (2 mol NO2 / 2 mol NO) = 8.21 mol NO2

To convert moles of NO2 to grams, we need to use the molar mass of NO2, which is 46.01 g/mol.

8.21 mol NO2 * 46.01 g/mol = 377.27 g NO2

Therefore, the maximum amount of NO2 that could be formed from 16.42 mol of NO and 14.47 mol of O2 is 377.27 grams of NO2.

For the second question, we have the balanced chemical equation:

C6H6 + 15O2 → 6CO2 + 3H2O

From the balanced equation, we can see that for every mole of C6H6, we need 15 moles of O2 to produce 6 moles of CO2.

To determine the amount of CO2 produced, we first need to find the limiting reagent.

For C6H6:

37.3 g C6H6 * (1 mol C6H6 / 78.11 g C6H6) = 0.477 mol C6H6

For O2:

126.1 g O2 * (1 mol O2 / 32 g O2) = 3.94 mol O2

Since the calculated amount of C6H6 (0.477 mol) is less than the actual amount of O2 (3.94 mol), C6H6 is the limiting reagent.

Using the mole ratio from the balanced equation, we find that 0.477 mol of C6H6 will produce 6 moles of CO2.

To convert moles of CO2 to grams, we need to use the molar mass of CO2, which is 44.01 g/mol.

0.477 mol CO2 * 44.01 g/mol = 21.0 g CO2

Therefore, 37.3 grams of C6H6 reacted with 126.1 grams of O2 will produce 21.0 grams of CO2.

Answer:

AsH3(g) + 4H2O(l)------> H3AsO4(aq) +8H^+(aq) + 8e-

Explanation:

Now we need to work through the problem in steps.

Step1: reaction of arsine with water

AsH3(g) + 4H2O(l) -----> H3AsO4(aq) this is the molecular reaction

Step II: oxygen is balanced using hydrogen ions

AsH3(g) + 4H2O(l)------> H3AsO4(aq) +8H^+(aq)

Step III: We specify the number of electrons transferred in the redox reaction

AsH3(g) + 4H2O(l)------> H3AsO4(aq) +8H^+(aq) + 8e-

The balanced half-reaction for the oxidation of gaseous arsine AsH₃ to

aqueous arsenic acid H₃AsO₄ in acidic aqueous solution is

AsH₃(g) + 4H₂O(l)------> H₃AsO₄(aq) +8H+(aq) + 8e-

AsH₃(g) + 4H₂O(l) -----> H₃AsO₄(aq)

AsH₃(g) + 4H₂O(l)------> H₃AsO₄(aq) +8H⁺(aq)

AsH₃(g) + 4H₂O(l)------> H₃AsO4(aq) +8H⁺(aq) + 8e⁻

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