Aniline gives or does not give Bauer’s test

Answer:

rmm doesn't need to include unit and molar mass need to include unit

Yes, If all the coefficients in a balanced chemical equation are multiplied by a common factor, such as 2 in this case, the equation will still be balanced.

This is because the law of conservation of mass still applies, and the number of atoms of each element on the reactant side must be equal to the number of atoms of that element on the product side.

What is law of conservation ?

However, multiplying the coefficients by a factor will affect the equilibrium constant of the reaction. The equilibrium constant is a measure of the ratio of products to reactants at equilibrium, and it is affected by the stoichiometry of the reaction. When the coefficients of the balanced equation are multiplied by a factor, the equilibrium constant is raised to the power of that factor.

For example, consider the following balanced chemical equation:

2A + B → 3C

The equilibrium constant for this reaction is given by:

Kc = [C]³ / ([A]²[B])

If we multiply all the coefficients by 2, the new balanced equation is:

4A + 2B → 6C

The equilibrium constant for this reaction is now:

Kc' =[tex][C]^{6}[/tex] / ([tex][A]^{4} [B]^{2}[/tex])

We can see that Kc' is raised to the power of 2, which means that it is much larger than Kc. This indicates that the equilibrium position of the reaction has shifted towards the products, as predicted by Le Chatelier's principle.

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Specific heat is the amount of heat energy required to raise the temperature of one unit of mass of a substance by one degree Celsius (or one Kelvin) without causing a change in its physical state.

Given information,

Mass of the metal, m = 55 g

Specific heat capacity of the metal, C = 0.662 J/g⁰C

The initial temperature of the metal = 109 ⁰C

Mass of water = 300 g

The initial temperature of water = 30⁰ C

We know that,

Heat loss of metal = Heat gain of water

MCT = MCT

55 × 0.622 × (109-t) = 300 × 4.184 × (t-30)

t = 27.78C

Thus, the final temperature of the water = 27.78C.

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Order of the reaction with respect to N2 =  1

Order of the reaction with respect to H2 =  2

The value of rate constant,  k  =  133 and the unit of rate constant = M-2.sec-1

The sequence of a chemical reaction points to the scaling or potency with which the effluence of a reactant is enhanced in the rate law mathematical statement for the reaction.

The order of the reaction on a precise reactant is identified experimentally by varying the effluence of that reactant and paying heed to how the speed of the reaction alters.

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The fact that an airplane flying due east from an airport will eventually return to its starting point indicates that the Earth is a sphere due to the curvature of the Earth's surface.

How can an airplane flying due east provide evidence for the shape of the Earth?

By continuing to fly east, and with enough fuel, the airplane will eventually reach the same airport it started from. The longest eastward journey would occur from an airport located at one of the Earth's poles. Very short journeys would occur near the equator.

The fact that the airplane returns to its starting point indicates that the Earth is a sphere, as it would not be possible on a flat plane. This phenomenon is due to the curvature of the Earth's surface, which causes the plane's trajectory to follow a circular path along the Earth's surface.

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Answer:

Heat = -28,875 J

Explanation:

We can solve for heat using this equation.

(heat) = (mass)•(specific heat)•(change in temperature)

Plug in what you know

(heat) = 1000g•(0.385J/g°C)•(25°C-100°C)

Solve

Heat = -28,875 J

To calculate the moles of H+ and OH-, you need to know the concentration of the solution in terms of its pH or pOH value.

When you get the pH of the solution, you can use this formula to calculate the concentration of H+ ions: [H+] = 10^(-pH)

Also, if you know the pOH of the solution, you can use this formula to calculate the concentration of OH- ions: [OH-] = 10^(-pOH)

Having determined the concentration of H+ and OH- ions, the molarity formula can be used to calculate the number of moles of each ion as follows: moles = concentration (in mol/L) x volume (in L)

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Answer:

Reverse Direction

Explanation:

Because I got it right...

If the reaction N₂ (g) + 3H₂ (g) ⇄ 2NH₃ (g) is at equilibrium, if the concentration of NH₃ is increased, a shift in the reverse direction.

A chemical reaction that produces products that, in turn, react with each other to produce the reactants again is known as a reversible reaction. Reversible responses will arrive at a balance point where the groupings of the reactants and items will never again change.

The equilibrium expression that is written in the opposite direction of the forward reaction is the same as the one that is written for the forward reaction. K is the constant for the forward reaction, and K is the constant for the reverse reaction (K' = 1/K).

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The location of zero oil velocity is 12 mm from the moving wall. The force required to maintain this motion on the plate is 4.37 N.

Given:

Width of the plate (w) = 30 cm = 0.3 m

Thickness of the oil layer (h) = 3.6 mm = 0.0036 m

Dynamic viscosity of the oil (μ) = 0.027 Pa-s

Velocity of the fixed wall (V) = 3 m/s

Velocity of the moving wall (V1) = 0.3 m/s

(a) To find the velocity profile, we can use the Navier-Stokes equation for laminar flow in the x-direction:

ρu(dv/dx) = -dp/dx + μ(d²v/dy²)

where:

ρ = density of the oil

u = velocity in the x-direction (i.e., the direction of motion)

v = velocity in the y-direction (i.e., perpendicular to the direction of motion)

p = pressure

Assuming steady-state flow and neglecting the pressure gradient (dp/dx), the above equation simplifies to:

ρu(dv/dx) = μ(d²v/dy²)

Since the velocity varies linearly in each layer of the oil:

v(y) = ay + b

where:

a = velocity gradient (i.e., the slope of the velocity profile)

b = velocity intercept (i.e., the y-intercept of the velocity profile)

Using the no-slip boundary condition at the walls:

v(0) = 0 (at the moving wall)

v(h) = 0 (at the fixed wall)

Substituting these boundary conditions and the velocity profile equation into the Navier-Stokes equation:

a = -V1/h

b = V

Therefore, the velocity profile is given by:

v(y) = -V1 × y/h + V

To find the location where the oil velocity is zero (i.e., the point where the plate experiences the maximum resistance), we can set v(y) = 0 and solve for y:

0 = -V1y/h + V

y = Vh/V1

Substituting the given values:

y = 0.012 m = 12 mm

Therefore, the location where the oil velocity is zero is 12 mm from the moving wall.

(b) To find the force required to maintain this motion, we can use the following equation for the drag force on a flat plate:

F = 0.5 × ρ × u² × Cd × A

where:

Cd = drag coefficient

A = area of the plate

Assuming a drag coefficient of 1.0 (appropriate for laminar flow over a flat plate), the area of the plate is:

A = w × h = 0.3 × 0.0036 = 0.00108 m²

Substituting the given values and using the velocity of the plate relative to the oil (i.e., u = V - V1):

F = 0.5 × ρ × (V - V1)² × Cd × A

Using the density of the oil (ρ = 1000 kg/m³):

F = 0.5 x 1000 x (3 - 0.3)² x 1.0 x 0.00108

F = 4.37 N

Therefore, the force that needs to be applied on the plate to maintain this motion is 4.37 N.

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5. There are 5.67 grams of HBr dissolved in 700 mL of the solution with pH of 2.

6. There are 1000 times more hydrogen ions in a solution with a pH of 3 than in a solution with a pH of 6.; option D.

The grams of HBr dissolved in 700 mL of a solution with pH of 2 is determined as follows;

pH = -log[H+]

2 = -log[H+]

[H+] = 10⁻² mol/L

HBr is a strong acid and dissociates completely in water to give H+ and Br- ions as follows:

HBr → H+ + Br-

The concentration of HBr in the solution will then be equal to the concentration of H+ ions.

[HBr] = 10⁻² mol/L

The mass of HBr is then determined using the formula:

The molar mass of HBr is 80.91 g/mol.

mass = 10⁻² mol/L x 0.7 L x 80.91 g/mol

mass = 5.67 g

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The objects found in our solar system include:

Planets and asteroids are common features of our solar system, with eight planets orbiting around the Sun, along with numerous asteroids, comets, and other objects.

The Sun is the center of the solar system and contains more than 99% of the mass of the entire solar system. Nebulae are clouds of gas and dust that are found in space, but they are not exclusive to our solar system.

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371.4 g of glucose must be metabolized to produce 223 g of water.

The balanced chemical equation for the complete combustion of glucose is:

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

According to the equation, for every 1 mole of glucose consumed, 6 moles of water are produced.

The molar mass of glucose is:

6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol) = 180.18 g/mol

To calculate the mass of glucose required to produce 223 g of water, we need to first convert the mass of water to moles:

223 g / 18.015 g/mol = 12.38 mol H₂O

Now we can use the mole ratio from the balanced equation to calculate the moles of glucose required:

1 mol glucose / 6 mol H₂O = x mol glucose / 12.38 mol H₂O

x = 2.06 mol glucose

Finally, we can calculate the mass of glucose:

mass = moles × molar mass

mass = 2.06 mol × 180.18 g/mol = 371.4 g

Therefore, 371.4 g of glucose must be metabolized to produce 223 g of water.

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Secretion of enzymes is an example of chemical digestion.

Enzymes play a crucial role in facilitating chemical reactions in living organisms and belong to a class of proteins. Their function entails lowering activation energy needed to start these biological processes, thus accelerating their pace considerably within our body systems.

Produced from our digestive system's breakdown activities mainly, enzymes facilitate further breakdown processes from undigested food particles into absorbable units routinely taken up by our organs.

Mechanically; movement processes like peristalsis and segmentation trigged along this process complement these enzymatic actions with some chemical aspects noted during deglutition (swallowing) .

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Complete question:

Which of the following is an example of chemical digestion?

A Peristalsis

Segmentation

Deglutition

D Secretion of enzymes

The concept combined gas equation is used here to determine the final temperature of the gas. These laws relate one thermodynamic variable to another holding everything else constant.

The combination of Boyle's law, Charles's law and Gay - Lussac's law give rise to the combined gas law. The state of a gas is determined by the macroscopic and microscopic parameters like pressure, volume, temperature, etc.

The combined gas equation is:

P₁V₁ /T₁ = P₂V₂ /T₂

T₂ = P₂V₂T₁ / P₁V₁

T₁ = 45 + 273 = 318 K

T₂ = 49.5 × 6.50 × 318 / 89.5 × 3.5 = 326.62 K

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The standard reaction Gibbs free energy (ΔGrxn°) using the given partial pressures is -60,200 J/mol.

Energy is the ability to do work, whether it be in the form of heat, light, electrical, chemical, nuclear, or mechanical energy. It is the capacity to move, change, or transform matter, and is essential for all processes of life.

ΔGrxn = -RT ln Kp
Where R is the universal gas constant (8.314 J/mol K) and T is the temperature (25°C).
Therefore, ΔGrxn = -(8.314)(298) ln (2.26 x 10⁴)
= -20,067 J/mol
Now we need to calculate the standard reaction Gibbs free energy (ΔGrxn°) using the given partial pressures:
ΔGrxn° = ∑nΔGrxn°products - ∑nΔGrxn°reactants
ΔGrxn°products = -RT ln (PCH₃OH)
= -(8.314)(298) ln (1.0)
= 0 J/mol
ΔGrxn°reactants = -RT ln (PCO) - 2(RT ln (PH2))
= -(8.314)(298) ln (0.010) - 2(8.314)(298) ln (0.010)
= -60,200 J/mol
Therefore, ΔGrxn° = ∑nΔGrxn°products - ∑nΔGrxn°reactants
= 0 - (-60,200)
= -60,200 J/mol
Finally, ΔGrxn = ΔGrxn° + RT ln Q
= -60,200 + (8.314)(298) ln (1.0)
= -60,200 J/mol.

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The free energy change for the reaction at 25°C and the given partial pressures is -4685 J/mol.  The standard free energy change (ΔG°) for a reaction can be related to the equilibrium constant (Kp) by the following equation:

ΔG° = -RT ln Kp

Where R is the gas constant (8.314 J/K mol), T is the temperature in Kelvin, and ln is the natural logarithm.

To calculate the free energy change (ΔGrxn) for the given reaction at non-standard conditions, we can use the following equation:

ΔGrxn = ΔG° + RT ln Q

Where Q is the reaction quotient, which is the ratio of the product of the partial pressures of the products raised to their stoichiometric coefficients to the product of the partial pressures of the reactants raised to their stoichiometric coefficients.

For the given reaction, the reaction quotient can be expressed as:

Q = (PCH3OH) / (PCO)(PH2)²

Plugging in the given partial pressures, we get:

Q = (1.0 atm) / (0.010 atm)(0.010 atm)² = 1000

Substituting the values into the equation for ΔGrxn, we get:

ΔGrxn = -RT ln Kp + RT ln Q

= -8.314 J/K mol x (298 K) x ln(2.26 x 10⁴) + 8.314 J/K mol x (298 K) x ln(1000)

= -4685 J/mol

Therefore, the free energy change for the reaction at 25°C and the given partial pressures is -4685 J/mol. This negative value indicates that the reaction is spontaneous and will proceed in the forward direction.

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When the reactant concentration is 0.5 M, the reactant's diffusion coefficient is roughly 1.03 x 10-5 cm2/s.

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The diffusion coefficient of the reactant is approximately 1.03 x 10^-5 cm²/s when the concentration of the reactant is 0.5 M.

The mass transfer current density (i_m) for a 2-electron reduction reaction can be expressed as:

i_m = (nFAcD)/δ

where n is the number of electrons involved in the reaction (2 in this case), F is Faraday's constant (96,485 C/mol), A is the surface area of the electrode, c is the bulk concentration of the reactant, D is the diffusion coefficient of the reactant, and δ is the thickness of the diffusion layer.

Assuming the diffusion layer thickness is much smaller than 0.1 mm and that the reactant concentration remains constant beyond this distance, we can assume that the bulk concentration of the reactant is the same as the concentration at the electrode surface.

Rearranging the equation, we can solve for D:

D = (i_m δ)/(nFAc)

Plugging in the given values, we get:

D = (10 mA/cm² * 0.0001 cm) / (2 * 96,485 C/mol * 0.5 mol/cm³ * 1 cm²)

D = 1.03 x 10^-5 cm²/s

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Answer:

To investigate the presence of lithium ions and carbonate ions in the tablet, the following steps can be taken:

Materials needed:

Lithium carbonate tablet

Metal wire

Dilute hydrochloric acid

Limewater

Bunsen burner

Test tubes

Procedure:

Take a small piece of the lithium carbonate tablet and place it on a metal wire.

Hold the wire over a Bunsen burner flame until the sample turns red.

Allow the sample to cool and place it in a test tube.

Add dilute hydrochloric acid to the test tube and observe any gas that is produced.

Pass the gas produced through limewater in a separate test tube and observe if there is any change in the color of the limewater.

Results:

If a red flame is observed in step 2, it indicates the presence of lithium ions in the sample.

If gas is produced in step 4, it indicates the presence of carbonate ions in the sample.

If the limewater turns cloudy or milky in step 5, it indicates the presence of carbonate ions in the sample.

Therefore, the student can conclude that the lithium carbonate tablet contains lithium ions and carbonate ions.

Out of the options given, the expression that represents the ideal gas behavior is:

The ideal gas law is expressed by the formula PV = nRT,

where

P is the pressure,

V is the volume,

n is the number of moles of gas,

R is the gas constant, and

T is the temperature.

By rearranging this equation, we can derive different expressions for the ideal gas behavior in terms of the pressure, volume, and temperature.

This can be obtained by rearranging the ideal gas law equation as follows:

PV = nRT

Dividing both sides by nT, we get:

P × V / (nT) = R

Since R is a constant for a given gas, the left-hand side of the equation must also be constant for an ideal gas. Thus, the expression P × V / T represents the ideal gas behavior.

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The enthalpy change for the reaction is 7520.3 kJ.

What is enthalpy?

Enthalpy is a thermodynamic property of a system that describes the total heat content of the system. It is denoted by the symbol "H" and is defined as the sum of the internal energy of a system and the product of its pressure and volume.

Enthalpy is a state function, meaning that its value depends only on the initial and final states of the system, not on how the system got from one state to the other.To determine the enthalpy change for the reaction:P4O6(s) + 2 O2(g) → P4H10(s)We can use Hess's law, which states that the enthalpy change for a reaction is independent of the pathway taken to get from reactants to products, and depends only on the initial and final states.

In other words, the total enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps of the reaction.

We can write the reaction in terms of the given reactions as follows:P4(s) + 3 O2(g) → P4O6(s) ΔH = -1640.1 kJP4(s) + 5 O2(g) → P4H10(s) ΔH = -2940.1 kJP4O6(s) + O2(g) → P4(s) + 3 O2(g) ΔH = 1640.1 kJ (reverse of reaction 1)To obtain the desired reaction, we need to reverse reaction 2 and multiply it by 2, then add reaction 3 to it.

This gives:2(P4H10(s) → P4(s) + 5 O2(g)) ΔH = 2(2940.1 kJ) = 5880.2 kJP4O6(s) + O2(g) → P4(s) + 3 O2(g) ΔH = 1640.1 kJP4O6(s) + 2 O2(g) → P4H10(s) ΔH = 7520.3 kJ

Therefore, the enthalpy change for the reaction is 7520.3 kJ.

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Answer:

2.09 grams

Explanation:

From this equation, for every 4 moles of water produced, we need 1 mole of propane.

The molar mass of water is 18.015 g/mol, and the molar mass of propane is 44.1 g/mol.

First, we need to calculate the number of moles of water produced:

moles of water = mass of water / molar mass of water

moles of water = 3.43 g / 18.015 g/mol

moles of water = 0.1906 mol

From the balanced chemical equation, we know that 4 moles of water are produced for every 1 mole of propane. Therefore, the number of moles of propane required to produce 0.1906 mol of water is:

moles of propane = 0.1906 mol / 4

moles of propane = 0.04765 mol

Finally, we can calculate the mass of propane required:

mass of propane = moles of propane x molar mass of propane

mass of propane = 0.04765 mol x 44.1 g/mol

mass of propane = 2.10 g

Therefore, the mass of propane reacting to produce 3.43 grams of water is 2.10 grams.

A. Th number of mole of oxygen needed is 75 moles

B. The number of mole of oxygen consumed is 34.5 moles

C. The mass (in grams) of carbon dioxide created is 105.22 g

The number of mole of oxygen needed to burn 25 moles of C₂H₅OH cab be obtain as follow:

C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O

From the balanced equation above,

1 mole of C₂H₅OH needed 3 moles of O₂

Therefore,

25 moles of C₂H₅OH will neede = 25 × 3 = 75 moles of O₂

Thus, the number of mole of oxygen,O₂ needed is 75 moles

The number of mole of oxygen consumed when 23 moles of carbon dioxide are produced can be obtain as follow:

C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O

From the balanced equation above,

2 moles of CO₂ were obtained from 3 moles of O₂

Therefore,

23 moles of CO₂ will be obtain from = (23 × 3) / 2 = 34.5 moles of O₂

Thus, the number of mole of oxygen, O₂ consumed is 34.5 moles

The mass of carbon dioxide, CO₂ created can be obtain as follow:

C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O

From the balanced equation above,

46 g of C₂H₅OH reacted to create 88 g of CO₂

Therefore,

55 g of C₂H₅OH will react to create = (55 × 88) / 46 = 105.22 g of CO₂

Thus, the mass of CO₂ created is 105.22 g

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Answer: 3.46 grams of water is produced.

Explanation:

The chemical formula for aluminum hydroxide is Al(OH)3. When it decomposes, it forms aluminum oxide (Al2O3) and water (H2O).

The balanced chemical equation for this reaction is:

2 Al(OH)3 → Al2O3 + 3 H2O

molar mass of Al(OH)3 = 78.0 g/mol

moles of Al(OH)3 = mass / molar mass = 10.0 g / 78.0 g/mol = 0.1282 mol

mole ratio from the balanced equation to calculate the number of moles of water produced:

moles of H2O = (3/2) * moles of Al(OH)3

moles of H2O = (3/2) * 0.1282 mol = 0.1923 mol

convert the moles of water to grams:

molar mass of H2O = 18.0 g/mol

mass of H2O = moles of H2O * molar mass = 0.1923 mol * 18.0 g/mol = 3.46 g

The kc for the given equilibrium is 12.23. When the observable parameters, such as color, temperature, pressure, concentration, etc. do not vary, the process is said to be in equilibrium.

A chemical reaction is considered to be in a state of chemical equilibrium when both the reactants and the products are in a concentration that does not change over time any more. The forward reaction rate and the backward reaction rate are equal in this state.

The relationship between a reaction's products and reactants with regard to a certain unit is expressed by the equilibrium constant, K.

The equilibrium constant expression is:

Kc = ([SO3]^2 / [SO2]^2 [O2])

Substitute the given equilibrium concentrations,

Kc = ((1.9 M)^2) / ((0.6 M)^2 * (0.82 M))

Now, Simplifying:

Kc ≈ 12.23

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At 298 K, the typical reactions Gibbs energy of

[tex]i).\ 2CH_3CHo(g)+O_2(g)- > 2CH_3COOH(i)\\\\ii).\ 2AgCi(s)+Br_2(i)- > 2AgBr(s)+Cl_2(g)\\\\iii).\ Hg(i)+Cl_2- > HgCl_2(s)[/tex]

are -721.2 kJ/mol, 24.2 kJ/mol, and -92.4 kJ/mol, respectively.

Value is a physical quantity used to indicate the size of a physical system's property in physics. Physical quantities including length, mass, energy, force, velocity, pressure, power, and time are all measured and compared using values. Intensity of a physical quantity, such as temperature, an electric field, or a magnetic field, can also be described by its value. Value is a key cornerstone of physics and is used to calculate a physical system's numerous attributes as well as the interactions between other physical systems.

Using the standard Gibbs energies of formation from Tables 2C.6 and 2C.7 in the thermodynamic data, the following formula is used to compute the

standard Gibbs energy of reaction for the combustion of acetaldehyde to acetic acid at 298 K:

ΔG°rxn = [tex]2[G^of(CH_3COOH)+G^of(O_2)]-[2G^of(CH_3CHO)+G^of(O_2)][/tex]

ΔG°rxn = 2(-382.7) - [2(-170.7) + 0] = -721.2 kJ/mol

Using the values of the standard Gibbs energies of formation from Tables 2C.6 and 2C.7 in the thermodynamic data, the following formula is used

to compute the standard Gibbs energy of reaction for the formation of silver bromide and chlorine gas from silver chloride and bromine at 298 K:

ΔG°rxn = [tex][2G^of(AgBr)+G^of(Cl_2)]-[2G^of(AgCl)+G^of(Br_2)][/tex]

ΔG°rxn = [2(-111.1) + 0] - [2(-127.2) + 0] = 24.2 kJ/mol

Using the values of the standard Gibbs energies of formation from Tables 2C.6 and 2C.7 in the thermodynamic data, the following formula is used

to compute the standard Gibbs energy of reaction for the creation of mercury chloride from mercury and chlorine gas at 298 K:

ΔG°rxn = [tex]G^of (HgCl_2) - [G^of (Hg) + G^of (Cl_2)][/tex][tex]2CH_3CHO(g)+O_2(g)- > 2CH_3COOH(I)[/tex]

ΔG°rxn = -92.4 - [0 + 0] = -92.4 kJ/mol

At 298 K, the typical reactions Gibbs energy of

[tex]i).\ 2CH_3CHo(g)+O_2(g)- > 2CH_3COOH(i)\\\\ii).\ 2AgCi(s)+Br_2(i)- > 2AgBr(s)+Cl_2(g)\\\\iii).\ Hg(i)+Cl_2- > HgCl_2(s)[/tex]

are -721.2 kJ/mol, 24.2 kJ/mol, and -92.4 kJ/mol, respectively.

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The amount of heat energy absorbed by the water when it begins to boil is approximately 77.54 kilojoules.

To calculate the amount of heat energy absorbed by the water when it reaches boiling point, we need to use the specific heat capacity formula:

Q = mcΔT

where:

First, let's convert the volume of water from pint to liters, and then to grams using the density of water, which is 1 g/mL:

0.650 pint x 0.473176 L/pint x 1000 g/L = 307.2 g (rounded to one decimal place)

Next, let's calculate the change in temperature. Since we are heating the water from room temperature to boiling point, assuming room temperature is around 25°C and boiling point is 100°C,

ΔT = 100°C - 25°C = 75°C.

Now, let's use the specific heat capacity of water, which is approximately 4.18 J/g°C.

Plugging in the values:

Q = 307.2 g x 4.18 J/g°C x 75°C

Q = 77542.08 J

Finally, let's convert the heat energy from Joules to kilojoules by dividing by 1000:

Q = 77.54 kJ (rounded to two decimal places)

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Answer:

The cell potential for the given galvanic cell is 0.188 V.

Explanation:

To calculate the cell potential, we can use the Nernst equation:

Ecell = E°cell - (RT/nF)ln(Q)

where E°cell is the standard cell potential, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin (25°C = 298 K), n is the number of moles of electrons transferred (in this case, n = 2), F is the Faraday constant (96,485 C/mol), and Q is the reaction quotient.

First, we need to write the half-reactions and their standard reduction potentials:

Sn4+(aq) + 2e- → Sn2+(aq) E°red = 0.15 V

Fe3+(aq) + e- → Fe2+(aq) E°red = 0.77 V

The overall reaction is the sum of the half-reactions:

Sn2+(aq) + 2Fe3+(aq) → Sn4+(aq) + 2Fe2+(aq)

The reaction quotient Q can be expressed as:

Q = [Sn4+][Fe2+]^2 / [Sn2+][Fe3+]^2

Substituting the given concentrations, we get:

Q = (0.00655)(0.01139)^2 / (0.0624)(0.0437)^2 = 0.209

Now we can calculate the cell potential:

Ecell = 0.15 V + 0.0592 V log([Fe2+]^2/[Fe3+]) + 0.0592 V log([Sn4+]/[Sn2+])

= 0.15 V + 0.0592 V log(0.01139^2/0.0437^2) + 0.0592 V log(0.00655/0.0624)

= 0.188 V

Therefore, the cell potential for the given galvanic cell is 0.188 V.

The cell potential for the given galvanic cell in which the given reaction occurs at 25 °C is 0.188 V.

To find the cell potential, we take the Nernst equation:

Ecell = E°cell - (RT/nF)ln(Q)

In which R is the gas constant (8.314 J/mol·K) and E° cell is the standard cell potential.

T temperature in Kelvin (25°C = 298 K), and n is the number of moles of electrons transferred (n = 2), Q is the reaction quotient and F is the Faraday constant (96,485 C/mol).

Firstly, write the half-reactions and then their standard reduction potentials:

Sn⁴⁺(aq) + 2e⁻  → Sn²⁺(aq) E°red = 0.15 V

Fe³⁺(aq) + e⁻ → Fe²⁺(aq) E°red = 0.77 V

The overall reaction is the sum of the half-reactions:

Sn²⁺(aq) + 2Fe³⁺(aq) → Sn⁴⁺(aq) + 2Fe²⁺(aq)

The Q reaction quotient can be written as:

Q = [Sn⁴⁺][Fe²⁺]² ÷ [Sn²⁺][Fe²⁺]²

Substituting the given concentrations, we observe:

Q = (0.00655)(0.01139)² ÷ (0.0624)(0.0437)² = 0.209

Next, we can find the cell potential:

Ecell = 0.15 V + 0.0592 V log([Fe²⁺]²/[Fe³⁺]) + 0.0592 V log([Sn⁴⁺]/[Sn²⁺])

= 0.15 V + 0.0592 V log(0.01139²÷0.0437²) + 0.0592 V log(0.00655÷0.0624)

= 0.188 V

Thus, the cell potential for the given galvanic cell is 0.188 V.

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