A saturated solution of lead(II) chloride, PbCl2, was prepared by dissolving solid PbCl2 in water. The concentration of Pb2+ ion in the solution was found to be 1.62×10−2 M . Calculate Ksp for PbCl2.

The equilibrium constant for the reaction is 32.2

The equilibrium constant (K꜀) for a given reaction is simply defined as the ratio of the concentration of the products raised to their coefficients to the concentration of the reactants raised to their coefficients

With the above information, we can obtain the equilibrium constant for the reaction given above as follow:

8H₂S(g) <=> 8H₂(g) + S₈(g)

Concentration of Hydrogen sulphide [H₂S] = 0.250 M

Concentration of Hydrogen [H₂] = 0.400 M

Concentration of sulphur [S₈] = 0.750 M

[tex]K_{C} = \frac{[H_{2}]^8[S_{8}]}{[H_{2}S]^{8}} \\\\K_{C} = \frac{0.4^{8} * 0.750}{0.25^{8}}\\\\[/tex]

Therefore, the equilibrium constant for the reaction is 32.2

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The equilibrium constant Kc for the given reaction is determined from the ratio of the product concentrations to the reactant concentrations, each raised to the power of their stoichiometric coefficients.

The equilibrium constant Kc for a chemical reaction is calculated from the concentrations of the products divided by the concentrations of the reactants, each raised to the power of their stoichiometric coefficients.

Given the reaction equation: 8H2S(g) ⇌ 8H2(g) + S8(g)

And the equilibrium concentrations: [H2S] = 0.250 M, [H2] = 0.400 M, [S8] = 0.750 M

The equilibrium constant Kc is therefore: Kc = ([H2]^8 * [S8]) / ([H2S]^8) = (0.400^8 * 0.750) / (0.250^8).

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Answer: The boiling point of water is an absolute constant

Explanation:

The boiling point of a compound is NOT an absolute constant. This is because at certain conditions such as a change in altitude, the boiling point of a compound changes.

As temperature increases, evaporation increases and vapour pressure increases. Compounds boils when vapour pressure equal to the atmospheric pressure.

At higher altitudes, atmospheric pressure is decreses.

When atmospheric pressure is decresed, the vapour pressure of the compound is lowered to reach boiling point. Therefore, the temperature needed for vapour pressure to equal atmospheric pressure is lower. The boiling point is lower at higher altitude.

Hence boiling point depends on atmospheric pressure

The boiling point of a compound is not an absolute constant.

The statement "The boiling point of a compound is an absolute constant" is FALSE.

The boiling point of a compound is affected by several factors, including intermolecular forces and molecular structure. Compounds with stronger intermolecular forces, such as hydrogen bonds, will generally have higher boiling points. Viscosity, which is the resistance of a liquid to flow, is not directly related to boiling point. Nonvolatile compounds, which do not easily evaporate, may have lower boiling points depending on their molecular structure.

Therefore, the correct statement is that the boiling point is not an absolute constant.

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Answer:

The false statement is: (C) The average kinetic energy of the molecules is not related to the absolute temperature

Explanation:

According to the Kinetic Theory of Gases, gases are composed of small fast moving particles that are in constant random motion. These gaseous particles are known as molecules.

It is assumed that the mass of all the gaseous molecules is the same and the intermolecular interactions (attractive or repulsive) are negligible.

Also, the total volume of all the gaseous molecules is considered to be negligible as compared to the volume of the container.

According to this theory, average kinetic energy (K) of the gaseous particles is directly proportional to the absolute temperature (T) of the gas, by the equation:

[tex]K = \frac{3}{2}k_{b}T[/tex]

Here, [tex]k_{b}[/tex] is the Boltzmann constant

Therefore, the false statement about the kinetic molecular theory is (C)

Statement C, which suggests that the average kinetic energy of the molecules is not related to the absolute temperature, is incorrect.

The question is focused on the Kinetic Molecular Theory, which primarily has four postulates that describe the behavior of gases. The incorrect statement in the list is the following:

In reality, according to the Kinetic Molecular Theory, the average kinetic energy of gas particles is directly proportional to the absolute temperature. This means that as the temperature increases, the average kinetic energy of the gas particles also increases, and vice versa. Therefore, it's clear that the temperature and average kinetic energy are related, making statement C incorrect.

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The balanced chemical equation for the combustion of butane is correct. Converting the mass of butane to moles and using the mole ratio, the volume of carbon dioxide produced is approximately 596 liters, as calculated using the ideal gas law.

Combustion of Butane and its Volume Calculation:

1. Balanced Chemical Equation:

The provided equation is correct:

C₄H₁₀ + ¹³/₂O₂ → 4CO₂ + 5H₂O

It is balanced because it has the same number of atoms of each element on both sides of the equation.

2. Volume of Carbon Dioxide Produced:

a) Converting Mass of Butane to Moles:

The mass of butane (0.360 kg) to moles (6.19 mol) using the following formula:

moles = mass / molar mass

The molar mass of butane is approximately 58.12 g/mol.

b) Mole Ratio and CO₂ Production:

The balanced equation tells us that 1 mole of butane reacts with 13/2 moles of oxygen to produce 4 moles of carbon dioxide. Therefore, the 6.19 moles of butane will produce:

6.19 mol butane * (4 mol CO₂ / 1 mol butane) = 24.76 mol CO₂

c) Ideal Gas Law and Volume Calculation:

The ideal gas law relates the volume (V), pressure (P), temperature (T), and number of moles (n) of a gas:

V = nRT/P

where:

R is the gas constant (0.08206 L atm/mol K)

T is the temperature in Kelvin (293.15 K)

P is the pressure in atm (1 atm)

Plugging in the values:

V = 24.76 mol * 0.08206 L atm/mol K * 293.15 K / 1 atm ≈ 596 L

Therefore, the volume of carbon dioxide produced is approximately 596 liters.

Answer:

E) 2.38

Explanation:

The pH of any solution , helps to determine the acidic strength of the solution ,

i.e. ,

and ,

pH is given as the negative log of the concentration of H⁺ ions ,

hence ,

pH = - log H⁺

From the question ,

the concentration of the solution is 0.0042 M , and being it a strong acid , dissociates completely to its respective ions ,

Therefore , the concentration of H⁺ = 0.0042 M .

Hence , using the above equation , the value of pH can be calculated as follows -

pH = - log H⁺

pH = - log ( 0.0042 M )

pH =  2.38 .

Final answer:

HCl is a strong acid, and its dissociation yields an equal concentration of hydrogen ions. Thus, the correct options is E) 2.38.

Explanation:

The pH of a 0.0042 M hydrochloric acid solution can be calculated using the formula [tex]pH = -log[H+][/tex], where [tex][H+][/tex] is the concentration of hydrogen ions.

In a solution of hydrochloric acid, which is a strong acid, the concentration of hydrogen ions [tex][H+][/tex] is equal to the concentration of the acid itself because [tex]HCl[/tex]dissociates completely in water.

Thus, the pH of the 0.0042 [tex]M HCl[/tex] solution is calculated as follows:

[tex]pH = -log(0.0042)[/tex]

= 2.38.

Therefore, the correct answer is E) 2.38.

The final temperature of the granite and the water will be 26.6 °C.

In a mixture between two substances of different temperatures, there is heat transfer, in which the one with a lower temperature receives this heat and the one with a higher temperature loses the heat.

In a system, the value of heat lost and received is equal.

This heat is represented by the following expression:

                                         [tex]Q = m \times c \times T[/tex]

The value of heat is given in calories, the value of temperature is given in celsius and the value of mass is given in grams

Thus, to find the value of the final temperature of the mixture, it is enough to equate the heat of the two substances.

                     [tex]1100 \times (\frac{0.803}{4,18}) \times (t_{f} - 83.7) = 2000 \times 1 \times (t_{f} - 20.3)[/tex]

                               [tex]220 (t_{f} - 83.7) = 2000 (t_{f} - 20.3)[/tex]

As the value of the granite final temperature is lower than the initial temperature, the result of the temperature variation would be negative, so the signs of the expression can be changed.

                                 [tex]18414 - 220 t_{f} = 2000 t_{f} - 40600[/tex]

                                         [tex]2220 t_{f} = 59,014[/tex]

                                              [tex]t_{f} = 26.6[/tex]

So, the final temperature of the granite and the water will be 26.6 °C.

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Using conservation of energy, the final equilibrium temperature of the granite and water system can be calculated by setting the heat lost by granite equal to the heat gained by water and solving for the final temperature.

The question is asking for the final equilibrium temperature of a two-part system initially at different temperatures when they are put in contact. This is a problem that can be solved using the concept of heat transfer and the principle of conservation of energy. The heat lost by the hot granite will equal the heat gained by the cooler water until thermal equilibrium is reached.

Using the specific heats of granite and water, and their masses and initial temperatures, we set up the equation based on the principle that heat lost = heat gained:

m1c1(Tfinal - T1(initial)) = m2c2(T2(initial) - Tfinal)

Where m1 and m2 are the masses of granite and water, c1 and c2 are their specific heats, Tfinal is the final equilibrium temperature and T1(initial) and T2(initial) are the initial temperatures of granite and water, respectively. You'll need to convert the mass of the granite to grams, as the specific heat is given in grams, and use the specific heat of water (4.184 J/g°C).

Then you can solve for Tfinal to find the temperature at which the granite and water reach equilibrium.

Answer:

ΔG° = -133,1 kJ

Explanation:

For the reactions:

(1) V₂O₃(s) + 3CO(g) → 2V(s) + 3CO₂(g); ΔH° = 369,8 kJ; ΔS° = 8,3 J/K

(2) V₂O₅(s) + 2CO(g) → V₂O₃(s) + 2CO₂(g); ΔH° = –234,2 kJ; ΔS° = 0,2 J/K

By Hess's law it is possible to obtain the ΔH° and ΔS° of:

2V(s) + 5CO₂(g) → V₂O₅(s) + 5CO(g)

Substracting -(1)-(2), that means:

ΔH° = -369,8 kJ - (-234,2 kJ) = -135,6 kJ

ΔS° = - 8,3 J/K - 0,2 J/K = -8,5 J/K

Using: ΔG° = ΔH° - TΔS° at 298K

ΔG° = -135,6 kJ - 298K×-8,5x10⁻³kJ/K

ΔG° = -133,1 kJ

I hope it helps!

Final answer:

To calculate the standard Gibbs free energy change (ΔG°) for the desired reaction at 298 K, reverse the first reaction, multiply the second by 2, sum their ΔH° and ΔS° values, and use the Gibbs free energy equation.

Explanation:

To calculate the standard Gibbs free energy change (ΔG°) for the reaction 2V(s) + 5CO2(g) → V2O5(s) + 5CO(g) at 298 K, we need to use the given information from two different reactions and apply Hess's law. First, we need to reverse the first reaction and multiply the second reaction by 2 to get the formation of V2O5(s) from V(s) and CO2(g). The ΔH° and ΔS° values of the first reaction would become -369.8 kJ and -8.3 J/K, respectively, after being reversed. For the second reaction, these values are -468.4 kJ and 0.4 J/K after being multiplied by 2. Adding these values provides the ΔH° and ΔS° for the overall reaction.

The ΔH° and ΔS° are then used to calculate ΔG° using the Gibbs free energy equation: ΔG° = ΔH° - TΔS°. Substituting the values, we find ΔG° for the overall reaction. This will tell us whether the reaction is spontaneous at 298 K, with a negative ΔG° indicating a spontaneous process.

Answer:

9.86 × 10²⁴ photons

Explanation:

Let's consider the photosynthesis global reaction.

6 CO₂(g) + 6 H₂O(l) →  C₆H₁₂O₆(s) +  6 O₂(g)

We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.

ΔG° = ∑np . ΔG°f(p) - ∑nr . ΔG°f(r)

where,

ni are the moles of reactants and products

ΔG°f(p) are the standard free Gibbs energies of formation of reactants and products

For the balanced equation for 1.00 mol of glucose, ΔG° is:

ΔG° = [1 mol × ΔG°f(C₆H₁₂O₆(s)) + 6 mol × ΔG°f(O₂(g))] - [6 mol × ΔG°f(CO₂(g)) + 6 mol × ΔG°f(H₂O(l))]

ΔG° = [1 mol × (-910.6 kJ/mol) + 6 mol × 0 kJ/mol] - [6 mol × (-394.4 kJ/mol) + 6 mol × (-237.1 kJ/mol)]

ΔG° = 2878 kJ

The energy of each photon (E) can be calculated using the Planck-Einstein's equation.

E = h . c. λ⁻¹

where,

h is the Planck's constant

c is the speed of light

λ is the wavelength

E = (6.626 × 10⁻³⁴ J.s) × (3.00 × 10⁸ m/s) × (680 × 10⁻⁹ m)⁻¹ = 2.92 × 10⁻¹⁹ J

The minimum number of photons required is:

[tex]2878\times 10^{3} J.\frac{1photon}{2.92\times 10^{-19}J} =9.86 \times 10^{24}photon[/tex]

Final answer:

The net change in photosynthesis involves forming glucose and O2 from CO2 and H2O with the aid of solar energy. The minimum number of photons needed at 680 nm to form 1.00 mol of glucose can be calculated using the energy requirements for the synthesis of glucose in photosynthesis.

Explanation:

The question pertains to the light-dependent reactions of photosynthesis, where solar energy is absorbed by chlorophyll to produce glucose from CO2 and H2O. Calculating the minimum number of photons with a wavelength of 680 nm needed to prepare 1.00 mol of glucose requires understanding the energy associated with photons and the process of photosynthesis. Since solar energy is converted into chemical energy in the form of ATP and NADPH during photosynthesis, and a total of 54 ATP equivalents are required to synthesize one molecule of glucose, the number of photons must provide at least this much energy. Considering that each photon's energy is inversely proportional to its wavelength, we can use the relationship E = hc/λ (where E is energy, h is Planck's constant, c is the speed of light, and λ is wavelength) to calculate the energy of a photon of 680 nm, and subsequently, determine the total number of photons required to provide the 54 ATP equivalents needed for glucose synthesis.

Answer:

The correct option is: B.) M^-1 s^-1

Explanation:

According to the rate law, the rate of a given chemical reaction is equal to the product of the reactant concentrations. Where, the exponents a and b are the partial orders for the concentration of X and Y, respectively.

The rate equation for the chemical reaction is:

Rate = k [X]ᵃ[Y]ᵇ

Here, k is the rate constant.

The concentration of X and Y reactants is given in mol.L⁻¹ or M (molarity).

The unit of rate of a chemical reaction is mol.L⁻¹.s⁻¹ or molarity per second M.s⁻¹

Given: rate = k [X][Y]              

In the given rate equation, the exponents of concentrations of reactants X and Y is 1. Thus the overall order of the reaction is 2.

Therefore, it is a second order reaction.

Since the unit of rate is M.s⁻¹ and units of [X] and [Y] is M.

∴ [tex]Rate (M.s^{-1}) = k [X (M)] [Y (M)][/tex]

⇒ [tex]k = \frac{Rate (M.s^{-1})}{[X (M)] [Y (M)]} [/tex]

⇒ the unit of k in the given rate law equation = M⁻¹.s⁻¹

The units of k in the rate law equation rate=k[X][Y] are M^-1 s^-1, which correspond to a second-order reaction. These units arise because the rate constant unit must counteract the reactant concentrations unit, providing the reaction rate (usually M s^-1).

The given reaction rate law is rate=k[X][Y] where k is the rate constant, and [X] and [Y] are the molar concentrations of substances X and Y. The units of k, the rate constant, are determined by the reaction's overall order, which, in this case, is a sum total of the individual reaction orders of X and Y. In this scenario, the reaction is second-order (one for [X] and one for [Y]). Therefore, the units of k would be M-1 s-1 (Molarity inverse second inverse) as given by option B. It is inferred as M-1 s-1 because the rate constant unit must cancel out the units of the reactant concentrations to give the rate of the reaction (which is typically in M s-1).

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Answer:

The expression will be given as:

[tex][CH_3OH]=K_c\times [CO]\times [H_2]^2[/tex]

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as [tex]K_{c}[/tex]

[tex]aA+bB\rightleftharpoons cC+dD[/tex]

[tex]K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}[/tex]

[tex]CO(g)+2H_2(g)[/tex] ⇌ [tex]CH_3OH(g)[/tex]

The equilibrium-constant expression for the given reaction  is given by:

[tex]K_{c}=\frac{[CH_3OH]}{[CO][H_2]^2}[/tex]

If we are given with equilibrium constant and equilibrium concentration of carbon monoxide and hydrogen gas we can determine the concentration of methanol at equilibrium.

The expression will be given as:

[tex][CH_3OH]=K_c\times [CO]\times [H_2]^2[/tex]

Answer:

Explanation:

To solve this problem we use PV=nRT for both gases in their containers, in order to calculate the moles of each one:

645 Torr ⇒ 645 /760 = 0.85 atm

25°C ⇒ 25 + 273.16 = 298.16 K

0.85 atm * 1.40 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

n = 0.0487 mol O₂

1.13 atm * 0.751 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

n = 0.0347 mol N₂

Now we can calculate the partial pressure for each gas in the new container, because the number of moles did not change:

P(O₂) * 2.00 L = 0.0487 mol O₂ * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

P(O₂) = 0.595 atm

P(N₂) * 2.00 L = 0.0347 mol N₂ * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

P(N₂) = 0.424 atm

Finally we add the partial pressures of all gases to calculate the total pressure:

Final answer:

To find the partial pressures and total pressure in the container, convert pressures to atm, apply the ideal gas law for each gas, and sum the resulting partial pressures. The partial pressures are 0.5959 atm for O₂ and 0.4249 atm for N₂, with a total pressure of 1.0208 atm.

Explanation:

The question asks to calculate the partial pressures of O₂ and N₂ as well as the total pressure in a 2.00 L container at 25 °C, after transferring 1.40 L of O₂ at 645 Torr and 0.751 L of N₂ at 1.13 atm into it. To solve this, the pressure values need to be converted to a common unit (atmospheres), and the ideal gas law used for calculations.

First, convert the O₂ pressure from Torr to atm: 645 Torr / 760 = 0.8487 atm. Then use the ideal gas law (P₁V₁ = P₂V₂) to find the new pressures in the 2.00 L container. For O₂, (0.8487 atm)(1.40 L) / 2.00 L = 0.5959 atm. The pressure of N₂ is already in atm, so, (1.13 atm)(0.751 L) / 2.00 L = 0.4249 atm.

The total pressure is the sum of the partial pressures: 0.5959 atm (O₂) + 0.4249 atm (N₂) = 1.0208 atm. Therefore, the partial pressures are 0.5959 atm for O₂ and 0.4249 atm for N₂, with a total pressure of 1.0208 atm in the container.

Answer:

b C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻

e HC₂H₃O₂ + H₂O ⇄ H₃O⁺ + C₂H₃O₂⁻

f ka<<1

g. Weak acid molecules and water molecules

Explanation:

The water molecule could act as a base and as an acid, a molecule that have this property is called as amphoteric.

b The salt NaC₂H₃O₂ is dissolved in water as Na⁺ and C₂H₃O₂⁻. The reaction of the anion with water is:

C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻

Where the C₂H₃O₂⁻ is the base and water is the acid.

e. The reaction of HC₂H₃O₂ (acid) with water (base), produce:

HC₂H₃O₂ + H₂O ⇄ H₃O⁺ + C₂H₃O₂⁻

f. As the acetic acid (HC₂H₃O₂) is a week acid, the dissociation in C₂H₃O₂⁻ is not complete, that means that ka<<1

g. The ka for this reaction is 1,8x10⁻⁵, that means that there are more weak acid molecules (HC₂H₃O₂) than conjugate base ions. Also, the water molecules will be in higher proportion than hydronium ions.

I hope it helps!

HC2H3O2 is a weak acid so it dissociates to a small extent in solution.

When NaC2H3O2 is dissolved in water, the following reaction occurs;

[tex]NaC2H3O2(aq) ---->Na^+(aq) + C2H3O2-(aq)[/tex]

The ion combines with water as follows;

[tex]C2H3O2-(aq) + H2O(l) -----> HC2H3O2(aq) + OH-(aq)[/tex]

When  HC2H3O2 is added to water, the following reaction occurs;

[tex]HC2H3O2(aq) + H2O(l) -----> H3O+(aq) + C2H3O2-(aq)[/tex]

We must note that HC2H3O2 is a weak acid. Weak acids only dissociate to a small extent in water therefore Ka << 1.

Since Ka << 1, it the follows that weak acid molecules and water molecules are present in the greatest concentration in solution because HC2H3O2 only dissociates to a very small extent.

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Answer:

that results in an alcohol and a carboxylic acid.

Explanation:

Saponification is a chemical reaction process of alkaline hydrolysis of esters(R'COOR group) by which soap is obtained.

For Example, when a base such as sodium hydroxide [NaOH]  is used to hydrolyze an ester, the products are a carboxylate salt and an alcohol. Because soaps are prepared by the alkaline hydrolysis of fats and oils.

In a saponification reaction, alkaline hydrolysis of fats and oils with sodium hydroxide  yields propane-1,2,3-triol and the corresponding sodium salts of the component fatty acids.

i.e  Fat or oil + caustic alkali ⇒ Soap + propane-1,2,3-triol

As a specific example, ethyl acetate and NaOH react to form sodium acetate and ethanol:

The reaction goes to completion in the image below:

Answer : The enthalpy of the reaction is, -2552 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given enthalpy of reaction is,

[tex]4B(s)+3O_2(g)\rightarrow 2B_2O_3(s)[/tex]    [tex]\Delta H=?[/tex]

The intermediate balanced chemical reactions are:

(1) [tex]B_2O_3(s)+3H_2O(g)\rightarrow 3O_2(g)+B_2H_6(g)[/tex]     [tex]\Delta H_A=+2035kJ[/tex]

(2) [tex]2B(s)+3H_2(g)\rightarrow B_2H_6(g)[/tex]    [tex]\Delta H_B=+36kJ[/tex]

(3) [tex]H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)[/tex]    [tex]\Delta H_C=-285kJ[/tex]

(4) [tex]H_2O(l)\rightarrow H_2O(g)[/tex]    [tex]\Delta H_D=+44kJ[/tex]

Now we have to revere the reactions 1 and multiple by 2, revere the reactions 3, 4 and multiple by 2 and multiply the reaction 2 by 2 and then adding all the equations, we get :

(when we are reversing the reaction then the sign of the enthalpy change will be change.)

The expression for enthalpy of the reaction will be,

[tex]\Delta H=-2\times \Delta H_A+2\times \Delta H_B-6\times \Delta H_C-6\times \Delta H_D[/tex]

[tex]\Delta H=-2(+2035kJ)+2(+36kJ)-6(-285kJ)-6(+44)[/tex]

[tex]\Delta H=-2552kJ[/tex]

Therefore, the enthalpy of the reaction is, -2552 kJ/mole

Answer:

We need 247 mL of NaOH

Explanation:

Step 1: Data given

Molarity of NaOH = 0.587 M

Volume of 0.445 M NiBr2 solution = 163 mL = 0.163 L

Step 2: The balanced equation

NiBr2(aq) + 2NaOH(aq) Ni(OH)2(s) + 2NaBr(aq)

Step 3: Calculate moles of NiBr2

Moles NiBR2 = Molarity NiBR2 * volume

Moles NiBR2 = 0.445 M * 0.163 L

Moles NiBR2 = 0.0725 moles

Step 3: Calculate moles of NaOH

For 1 mol NiBr2 consumed, we need 2 moles NaOH

For 0.0725 moles NiBR2, we need 2* 0.0725 = 0.145 moles NaOH

Step 4: Calculate volume of NaOH

Volume = moles NaOH / Molarity NaOH

Volume = 0.145 moles / 0.587 M

volume = 0.247L = 247 mL

We need 247 mL of NaOH

Answer: b. Catalysis occurs at the active site, which usually consists of a crevice on the surface of the enzyme.

c. Generally, an enzyme is specific for a particular substrate. For example, thrombin catalyzes the hydrolysis of the peptide bond between Arg and Gly.

Explanation:

The role of the enzyme is to produce a specific product, whereas the non-biological catalyst produces more than one product. Thus they have a different degree of reaction specificity.

The active site of the enzyme is the catalytic site of the enzyme. It is the region where the substrate molecule bind and undergoes a chemical reaction. The enzyme exhibit a special crevice or opening at the active site which facilitates the binding with the substrate.

Enzymes usually bind to a specific substrate as they have an active site that allows a particular substrate to bind to the active site of the enzyme. This is because of the shape of the active site of the enzyme which has a binding affinity with a particular substrate any other substrate cannot bind to the active site.

To determine the stronger oxidizing agent in each pair of substances, compare the oxidizing strengths using data from Appendix E. The stronger oxidizing agents are O3(g), Cl2(g), Co2+(aq), and BrO3-(aq).

To determine the stronger oxidizing agent in each pair, we need to compare the oxidizing strengths of the substances. From Appendix E in the textbook, we can find the relevant data. Here are the answers:

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To determine the stronger oxidizing agent, compare the standard reduction potentials of the species in each pair. The substance with the higher standard reduction potential is the stronger oxidizing agent.

To determine the stronger oxidizing agent in each pair, we need to compare their standard reduction potentials. The substance with the higher standard reduction potential is the stronger oxidizing agent. From Appendix E, we can find the standard reduction potentials of the species in each pair and compare them to determine the stronger oxidizing agent.

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Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

                         H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium:  0.50M - x          x            x

The equilibrium constant (Ka₁) is:

[tex] K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x} [/tex]

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

[tex] [HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M [/tex]

Similarly, from the second dissociation of sulfurous acid we have:

                              HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium:  7.94x10⁻²M - x          x            x

The equilibrium constant (Ka₂) is:  

[tex] K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x} [/tex]  

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

[tex] [SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M [/tex]

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

When .50 moles of sulfurous acid are dissolved in a 1 L solution, the concentration that will be least at equilibrium is [H2SO3].

In the given equilibrium reactions, when .50 moles of sulfurous acid are dissolved in a 1 L solution, the concentration that will be LEAST at equilibrium is [H2SO3], which represents the original acid concentration.

Answers and Explanation:

a)- The chemical equation for the corresponden equilibrium of Ka1 is:

2. HNO2(aq)⇌H+(aq)+NO−2

Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.

b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:

ΔG= ΔGº + RT ln Q

Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)

At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:

⇒ 0 = ΔGº + RT ln Ka

   ΔGº= - RT ln Ka

   ΔGº= -8.314 J/K.mol x 298 K x ln (4.5 10⁻⁴)

  ΔGº= 19092.8 J/mol

c)- According to the previous demonstation, at equilibrium ΔG= 0.

d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:

ΔG= ΔGº + RT ln Q

Q= ((H⁺) (NO₂⁻))/(HNO₂)

Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)

Q= 1.88 10⁻⁴

We know that   ΔGº= 19092.8 J/mol, so:

ΔG= ΔGº + RT ln Q

ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)

ΔG= -2162.4 J/mol

Notice that ΔG<0, so the process is spontaneous in that direction.

HNO2 is a weak acid whose dissociation occurs as follows;  HNO2 (aq)    ⇋       H+ (aq)  +      NO2- (aq)

The equation for the dissociation of HNO2 is;

           HNO2 (aq)    ⇋       H+ (aq)  +      NO2- (aq)

We can use the value of Ka to find the  ΔG∘ for the dissociation of nitrous acid in aqueous solution as follows;

ΔG∘ = -RTlnKa

Where;

R = Gas constant = 8.314 J/K. mol

Ka = Acid dissociation constant = 4.5×10^−4

T = temperature = 25 ∘C or 298 K

Substituting values;

ΔG∘ = -(8.314 J/K. mol × 298 K × ln 4.5×10^−4)

ΔG∘ = 19.1 KJ/mol

Given that;

Q = [ H+] [NO2- ]/[HNO2]

[ H+] =  5.9×10−2 M

[NO2- ] = 6.7×10−4 M

[HNO2] = 0.21 M

Q = [5.9×10−2 M] [6.7×10−4 M]/[0.21 M]

Q = 1.88 × 10^−4

From the formula;

ΔG = ΔG∘ + RTlnQ

ΔG =  19.1  KJ/mol + (8.314 J/K. mol  ×  298 K ×  ln 1.88 × 10^−4)

ΔG = -2.15 KJ/mol

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Explanation:

-Filter help — delete some big unreacted, undesirable species (norit is probably from what you are sorting through, its only carbon which cleans up things)

— extract with DCM because you are probably in an aqueous phase, and some butanoate is in it

- Anhydrous sodium absorbs excess of  water (dries the material)

-evaporation in the hood to clear the DCM and crystallize the product.

Answer : The value of [tex]\Delta G_{rxn}[/tex] is, -27.0kJ/mole

Explanation :

The formula used for [tex]\Delta G_{rxn}[/tex] is:

[tex]\Delta G_{rxn}=\Delta G^o+RT\ln K_p[/tex]   ............(1)

where,

[tex]\Delta G_{rxn}[/tex] = Gibbs free energy for the reaction

[tex]\Delta G_^o[/tex] =  standard Gibbs free energy  = -32.8 kJ

/mol

R = gas constant = 8.314 J/mole.K

T = temperature = 298 K

[tex]K_p[/tex] = equilibrium constant

First we have to calculate the value of [tex]K_p[/tex].

The given balanced chemical reaction is,

[tex]2NO(g)+O_2(g)\rightarrow 2NO_2(g)[/tex]

The expression for equilibrium constant will be :

[tex]K_p=\frac{(p_{NO_2})^2}{(p_{NO})^2\times (p_{O_2})}[/tex]

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

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

[tex]K_p=\frac{(0.800)^2}{(0.500)^2\times (0.250)}[/tex]

[tex]K_p=10.24[/tex]

Now we have to calculate the value of [tex]\Delta G_{rxn}[/tex] by using relation (1).

[tex]\Delta G_{rxn}=\Delta G^o+RT\ln K_p[/tex]

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

[tex]\Delta G_{rxn}=-32.8kJ/mol+(8.314\times 10^{-3}kJ/mole.K)\times (298K)\ln (10.24)[/tex]

[tex]\Delta G_{rxn}=-27.0kJ/mol[/tex]

Therefore, the value of [tex]\Delta G_{rxn}[/tex] is, -27.0kJ/mole

Answer:

[tex]e(\%)=0.17\%[/tex]

Explanation:

Volume of a drop:

[tex]V_{drop}=\frac{1 mL}{20 drop}[/tex]

[tex]V_{drop}=0.05 mL[/tex]

To estimate the error:

[tex]e(\%)=\frac{V_{real}-V{theorerical}}{V{theoretical}}*100\%[/tex]

[tex]e(\%)=\frac{(30 mL+0.05mL)-30mL}{30mL}*100\%[/tex]

[tex]e(\%)=0.17\%[/tex]

Answer:

A) 6

B) 6

C) 4

E) 1

Explanation:

Coordination number, also called Ligancy, the number of atoms, ions, or molecules bound to the metal atom in a complex ion

A) [Fe(gly)2(H2O)2]

Gly = the aminoacid glycin, and works as a  bidentate ligand,Bidentate ligands bind through two donor sites. Bidentate means "two-toothed".  It can bind to a metal via two donor atoms at once.

This makes the coordination number = 6

B) [Pb(EDTA)]2

EDTA is a hexadentate ligand and forms very stable complexes.

EDTA forms 6 bonds to the Pb atom (2 Pb-N bonds and 4 Pb-O bonds). The coordination number = 6.

C) [Pt(NH3)4]2+

It's a squareplanar complex with coordination number = 4 because 4 bindings with NH3

D) Na[Au(CN)2]

Cyanide binds only to 1 Au- atom. The coordination number = 1

A) [tex][Fe(gly)2(H2O)_2]^+[/tex]: Coordination number is 6.

B)[tex][Pb(EDTA)]_2[/tex]−: Coordination number is 6.

C) [tex][Pt(NH_3)_4]^2+[/tex]: Coordination number is 4.

D) [tex]Na[Au(CN)_2]:[/tex] Coordination number is 2.

The coordination numbers for each of the given complexes are as follows:

A) [tex][Fe(gly)2(H2O)2]+[/tex]: The coordination number of the iron (Fe) ion in this complex is 6. This is because there are two glycine (gly) ligands and two water (H2O) molecules coordinated to the iron ion. Each glycine is bidentate, meaning it binds to the iron ion with two donor atoms (usually nitrogen and oxygen), contributing a total of four donor atoms. The two water molecules each provide one donor atom (oxygen). Thus, the total number of donor atoms around the iron ion is 6, giving it a coordination number of 6.

 B) [tex][Pb(EDTA)]2[/tex]: The coordination number of the lead (Pb) ion in this complex is 6. Ethylenediaminetetraacetic acid (EDTA) is a hexadentate ligand, meaning it can bind to a metal ion with all six of its donor atoms (four oxygens and two nitrogens). Since EDTA is fully deprotonated in this complex, it forms a negatively charged species that can wrap around the lead ion, resulting in a coordination number of 6.

 C) [tex][Pt(NH_3)_4]^2+:[/tex] The coordination number of the platinum (Pt) ion in this complex is 4. There are four ammonia (NH3) molecules coordinated to the platinum ion, each acting as a monodentate ligand and providing one donor atom (nitrogen). Therefore, the platinum ion is surrounded by four donor atoms, giving it a coordination number of 4.

 D) [tex]Na[Au(CN)_2][/tex]: The coordination number of the gold (Au) ion in this complex is 2. There are two cyanide (CN) ligands coordinated to the gold ion, each acting as a monodentate ligand and providing one donor atom (carbon). Thus, the gold ion is surrounded by two donor atoms, giving it a coordination number of 2.

 To summarize:

A) [tex][Fe(gly)2(H2O)_2]^+[/tex]: Coordination number is 6.

B)[tex][Pb(EDTA)]_2[/tex]−: Coordination number is 6.

C) [tex][Pt(NH_3)_4]^2+[/tex]: Coordination number is 4.

D) [tex]Na[Au(CN)_2]:[/tex] Coordination number is 2.

Answer:

50. 5mol

Explanation:

The mass of Zinc required = 3.30 kg = 3300 grams

The relation between moles and mass is:

[tex]moles=\frac{mass}{atomicmass}[/tex]

Atomic mass of Zinc = 65.38g/mol

[tex]molesofZinc=\frac{3300}{65.38}[/tex]

[tex]molesofzinc=50.47mol[/tex]

The answer to three significant figures will be 50.5 mol

To find the number of moles of zinc in 3.30 kg of zinc, divide the mass in grams by the molar mass of zinc. The number of moles of zinc is 50.5 mol.

To find the number of moles of zinc in 3.30 kg of zinc, we can use the molar mass of zinc to convert the mass to moles. The molar mass of zinc (Zn) is 65.38 g/mol. First, we convert the mass from kg to g by multiplying by 1000:

3.30 kg × 1000 g/kg = 3300 g

Then, we divide the mass in grams by the molar mass to get the number of moles:

3300 g / 65.38 g/mol = 50.5 mol

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Answer:

1. 4.32 h

2. 4.12 g/L

Explanation:

1. For a batch culture, the time (tb) can be calculated by:

tb =  ln (X/X0)/μmax

Where X0 is the initial mass concentration of the cells (12 g/100L = 0.12 g/L), X is the mass concentration of the cells at tb, and μmax is the maximum specific growth rate of the cells.

The biomass yield (Y) is:

Y = (X - X0)/ (S0 - S)

Where S is the mass concentration of the substrate at tb and S0 the initial mass concentration of the substrate (glucose in this case).

Reorganizing:

X = Y*(S0 -S) + X0

Let's assume that at the stationary state all substrate was consumed, so S = 0.

tb = ln[(YS0 + X0)/X0)]/μmax

tb = ln[(0.575*10 + 0.12)/0.12]/0.9

tb = 4.32 h

2.  If 70% of the substrate is consumed, S = 10 - 0.7*10 = 3 g/L

tb = ln[(0.575*(10-3) + 0.12)/0.12]/0.9

tb = 3.93 h

The initial concentration is X0 = 0.12 g/L, the X:

tb =  ln (X/X0)/μmax

3.93 = ln(X/0.12)/0.9

ln(X/0.12) = 3.537

X/0.12 = [tex]e^{3.537}[/tex]

X/0.12 = 34.36

X = 4.12 g/L

Answer:

a) to = 4.3 h

b) xf = 4.1 g/L

Explanation:

We have the following data:

So = 10 g*L^-1 = initial concentration

Ys = biomass yield = 0.575 g*g^-1

umax = maximum specific growth rate = 0.9 h^-1

The initial cell concentration is equal to:

xo = 12 g/100 L = 0.12 g/L

a) The batch time it takes to reach the stationary phase equals:

to = (1/umax)*ln(1 + (Ys*(so-sf)/xo))) = (1/0.9)*ln(1+(0.575*(10-0))/0.12)))) = 4.3 h

b) Since the fermentation stops after consuming only 70% of the glucose, we have the following:

sf = (1-07)*so = 0.3*10 = 3 g/L

tb = (1/0.9)*ln(1+(0.575*(10-3))/(0.12)))) = 3.94 h

Finally, the final cell concentration can be found by the following equation:

xf = xo*e^(umax * tb) = 0.12 * e^(0.9 * 3.94) = 4.1 g/L

Answer: The [tex]K_{sp}[/tex] of gold (III) chloride is [tex]3.2\times 10^{-25}[/tex]

Explanation:

We are given:

Solubility of gold (III) chloride = [tex]1.00\times 10^{-4}g/L[/tex]

Molar mass of gold (III) chloride = 303.3 g/mol

To calculate the solubility in mol/L, we divide the given solubility (in g/L) with molar mass, we get:

[tex]\text{Solubility (in mol/L)}=\frac{\text{Solubilty (in g/L)}}{\text{Molar mass}}[/tex]

Putting values in above equation, we get:

[tex]\text{Solubility of gold (III) chloride (in mol/L)}=\frac{1.00\times 10^{-4}g/L}{303.3g/mol}\\\\\text{Solubility of gold (III) chloride (in mol/L)}=3.29\times 10^{-7}mol/L[/tex]

The balanced equilibrium reaction for the ionization of gold (III) chloride follows:

[tex]AuCl_3\rightleftharpoons Au^{3+}+3Cl^-[/tex]

                s       3s

The expression for solubility constant for this reaction will be:

[tex]K_{sp}=[Au^{3+}][Cl^-]^3[/tex]

We are given:

[tex]s=3.29\times 10^{-7}mol/L[/tex]

Putting values in above equation, we get:

[tex]K_{sp}=(s)\times (3s)^3\\\\K_{sp}=9s^4\\\\K_{sp}=9\times (3.29\times 10^{-7})^4[/tex][tex]K_{sp}=(s)\times (3s)^3\\\\K_{sp}=27s^4\\\\K_{sp}=27\times (3.29\times 10^{-7})^4=3.2\times 10^{-25}[/tex]

Hence, the [tex]K_{sp}[/tex] of gold (III) chloride is [tex]3.2\times 10^{-25}[/tex]

The Ksp of AuCl3, calculated from its solubility and molar mass, is 1.2x10^-26.

In order to calculate the Ksp for AuCl3, we first need to convert the solubility from g/L to mol/L. We do this by dividing the given solubility (1.00x10^-4 g/L) by the molar mass of AuCl3 (303.3 g/mol). This gives us a molar solubility of 3.3x10^-7 mol/L. The dissolution reaction of AuCl3 can be represented as follows: AuCl3(s) -> Au3+(aq) + 3Cl-(aq). The solubility product (Ksp) is then obtained by multiplying the concentration of each ion raised to the power of its stoichiometric coefficient in the balanced equation. This gives Ksp=Au3+ (aq) * Cl-(aq)^3 = (3.3x10^-7)^4 = 1.2x10^-26, hence, the answer is option C.) 1.2x10^-26.

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Final answer:

In an isothermal reversible expansion, ΔG and ΔA do not differ and can be calculated using the natural log of the ratio of final to initial volumes. During an isothermal expansion against a constant external pressure, while ΔG can vary due to the external work being done, ΔA remains more reflective of internal work under constant conditions.

Explanation:

The question asks for the calculation of ΔG (Gibbs Free Energy change) and ΔA (Helmholtz Free Energy change) for the isothermal expansion of an ideal gas under two scenarios: (a) a reversible path and (b) an expansion against a constant external pressure. Both ΔG and ΔA can be insightful in understanding the spontaneity and energy changes of a chemical process.

(a) Isothermal Reversible Expansion

For an isothermal reversible expansion, ΔG and ΔA do not differ because both depend on the same variables in an ideal condition. Since the temperature is constant and the process is reversible, the free energy changes can be calculated using the formulas:

ΔG = -nRT ln(Vf/Vi) ΔA = -nRT ln(Vf/Vi)

Given: n = 2.50 mol, T = 298 K, Vi = 10.0 L, and Vf = 50.0 L

This leads to the same value for both ΔG and ΔA, showing that the only driving force behind the reversible isothermal expansion is entropy.

(b) Isothermal Expansion Against a Constant External Pressure

When expanding against a constant external pressure of 0.750 bar, the pressure-volume work becomes different. However, ΔG can still be calculated using the formula for isothermal processes in the presence of external pressure, but would yield a different context compared to ΔA which is not typically affected by external pressure in the same straightforward manner.

ΔG is defined by the system's ability to do non-PV work, and in isothermal conditions against constant external pressures, its calculation might deviate, illustrating differences in real-world versus ideal conditions. ΔA would not generally change because it pertains to the work done in a closed system at constant temperature and volume.

Answer: The standard enthalpy change of the reaction is coming out to be -16.3 kJ

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as [tex]\Delta H[/tex]

The equation used to calculate enthalpy change is of a reaction is:  

[tex]\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]

For the given chemical reaction:

[tex]Mg(OH)_2(s)+2HCl(g)\rightarrow MgCl_2(s)+2H_2O(g)[/tex]

The equation for the enthalpy change of the above reaction is:

[tex]\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})][/tex]

We are given:

[tex]\Delta H_f_{(Mg(OH)_2(s))}=-924.5kJ/mol\\\Delta H_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H_f_{(MgCl_2(s))}=-641.8kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol[/tex]

Putting values in above equation, we get:

[tex]\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ[/tex]

Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ

To calculate the standard enthalpy change for the given reaction, use the standard enthalpy of formation values for the reactants and products. Calculate the enthalpy change for the formation of each compound using their respective standard enthalpy of formation values. Then, calculate the overall enthalpy change for the reaction by subtracting the sum of the enthalpy changes for the reactants from the sum of the enthalpy changes for the products. Round the result to the appropriate number of significant figures.

The standard enthalpy change for a reaction can be calculated using the standard enthalpy of formation values. For the given reaction, Mg(OH)2(s) + 2 HCl(g) ⟶ MgCl2(s) + 2 H2O(g), we need to use the standard enthalpy of formation values for the reactants and products.

Step 1: Calculate the enthalpy change for the formation of Mg(OH)2(s) using its standard enthalpy of formation value.

Step 2: Calculate the enthalpy change for the formation of MgCl2(s) using its standard enthalpy of formation value.

Step 3: Calculate the enthalpy change for the formation of H2O(g) using its standard enthalpy of formation value.

Step 4: Calculate the overall enthalpy change for the reaction by subtracting the sum of the enthalpy changes for the reactants from the sum of the enthalpy changes for the products.

Step 5: Round the result to the appropriate number of significant figures.

Answer: The standard enthalpy change for the given reaction at 25°C can be calculated using the standard enthalpy of formation values for the reactants and products.

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Answer: The balloon will burst as it occupies higher volume than  maximum inflated volume.

Explanation:

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is inversely proportional to the volume of the gas at constant temperature.  

The equation given by this law is:

[tex]P_1V_1=P_2V_2[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas= 1.00 atm = 760.0 mm Hg  

[tex]V_1[/tex] =  initial volume of gas = 2.50 L

[tex]P_2[/tex] = final pressure of gas= 450.0 mm Hg

[tex]V_2[/tex] = final volume of gas = ?

Putting values in above equation, we get:

[tex]760.0\times 2.50=450.0\times V_2[/tex]

[tex]V_2=4.22L[/tex]

Thus the final volume of the gas is 4.22 L and as the maximum inflated volume is 3.00 L, the balloon will burst.

Answer:

The exception here is

C2 and B2

both the bonds are pi bond in the above cases

Explanation:

The fault here is the unforeseen energy order of molecular orbitals. Since if filled, sigma 2px will be firmly pushed away by the electron density of orbitals already filled with sigma 2s (both electron densities lie on the inter-nuclear axis). Because of this, sigma 2px's energy is found shockingly greater than pi 2py and pi 2pz.

The electrons therefore occupy pi 2py and pi 2pz orbitals as opposed to normal sigma 2px (which remains vacant) while filling.

The existence of these 4 electrons in orbitals pi 2py and pi 2pz results in two pi bonds being formed.

While most species follow the general rule that a single bond is a sigma bond and a double bond comprises a sigma and pi bond, B2 and C2 are exceptions, with their sigma and pi bond energy levels differing from the norm.

The generalization that a single bond is a sigma bond, and a double bond is made up of a sigma bond and a pi bond, with a triple bond consisting of one sigma bond and two pi bonds, is questioned in the given species. In most cases, this generalization holds true, with N2 possessing a triple bond that contains one sigma bond and two pi bonds, O2 containing a double bond with one sigma and one pi bond, and F2 having a single bond that, by definition, is a sigma bond.

However, according to the provided information, there is an exception among the listed species where the energy levels of the sigma and pi bonds differ from the norm. For B2, C2, and N2, the sigma bonds (2p) have higher energy than the pi bonds (2p). This anomalous behavior suggests that in these species, the generalization about bonding is violated.

Therefore, among the species listed, B2 and C2 are exceptions to the rule, with B2 and C2 having differences in their bond energies that do not align with the typical characteristics of sigma and pi bonds.