Please help thanks so much!!!!!!!!!!!!

a. Br  will have the negative charge

b. The effective charges on the I and Br atoms are approximately +1.012e and -1.012e, respectively.

a. To identify which atom within an IBr molecule will have a partial negative charge, we must consider each atom's electronegativity.

On the periodic table, iodine (I) has an electronegativity value of 2.66 while bromine (Br) boasts 2.96; since Br has higher electronegativity it will attract electrons more strongly and hence have an even stronger partial negative charge.

B. To calculate the effective charges on the I and Br atoms in IBr, we can use the dipole moment equation:

μ = Q * d

where μ is the dipole moment, Q is the effective charge, and d is the bond length.

We are given the dipole moment (μ) as 1.21 D, and the bond length (d) as 249 pm. However, we need to convert the units to the SI system before proceeding with the calculation.

[tex]1 D (Debye) = 3.336 * 10^-^3^0 cm,\\\\1 pm = 10^-^1^2 m.[/tex]

Now we can solve for the effective charge (Q):[tex]u = 1.21 D * (3.336 × 10^-^3^0 Cm/D)\\ \\= 4.03656 * 10^-^3^0 cm\\d = 249 pm * (10^-12 m/pm) = 2.49 * 10^-^1^0 m[/tex]

Q = μ / d

[tex]Q = (4.03656 * 10^-^3^0 cm) / (2.49 *10^-^1^0 m) \\\\\\=1.62151 * 10^-^2^0 C[/tex]

This effective charge represents the charge difference between the I and Br atoms. To express the charges in units of the elementary charge (e), we need to divide the effective charge by the elementary charge value (e = 1.602 × 10^-19 C):

Q_e =[tex]\frac{(1.62151 * 10^-^2^0 C)}{(1.602 * 10^-^1^9 C)} = 1.012[/tex]

The effective charges on the I and Br atoms are approximately +1.012e and -1.012e, respectively.

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Answer: 0.24 Moles. WARNING: "mm" in the original question seems to be a typo and was assumed to be "M" (moles per liter) for concentration.

Explanation:

To calculate the number of moles of solute needed to prepare a solution, we can use the formula:

moles of solute = concentration (in moles per liter) * volume of solution (in liters)

Given:

Concentration of NaCl solution = 0.8 M (moles per liter)

Volume of solution to be prepared = 300 mL = 300/1000 L (converted to liters)

Plugging in the given values:

Concentration = 0.8 M

Volume of solution = 300/1000 L

moles of NaCl = 0.8 M * 300/1000 L

Calculating:

moles of NaCl = 0.24 moles

So, 0.24 moles of NaCl are needed to prepare 300 mL of a 0.8 M NaCl solution.

The specific heat capacity of the ring, given that the ring gives up 667.5 J of energy to the water is 0.444 J/gºC. The ring is made of nickel.

The specific heat capacity of the ring can be obtain as illustrated below:

Q = MCΔT

-667.5 = 25.5 × C × -59

-667.5 = -1504.5 × C

Divide both sides by -1504.5

C = -667.5 / -1504.5

C = 0.444 J/gºC

Thus, the specific heat capacity of the ring is 0.444 J/gºC. Hence, the ring is made of nickel

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We can use the combined gas law to determine the volume of the balloon at a higher altitude. The combined gas law relates the pressure, volume, and temperature of a gas:

(P1 x V1) / T1 = (P2 x V2) / T2

where P1, V1, and T1 are the pressure, volume, and temperature of the gas at the initial state, and P2, V2, and T2 are the pressure, volume, and temperature of the gas at the final state.

We are given the initial pressure (P1 = 761 mmHg), volume (V1 = 56.0 L), and temperature (T1 = 23.1 °C = 296.25 K) of the gas, and the final pressure (P2 = 0.0772 atm), and temperature (T2 = -6.97 °C = 266.18 K) of the gas. We can solve for V2, the final volume of the gas:

(P1 x V1) / T1 = (P2 x V2) / T2

V2 = (P1 x V1 x T2) / (P2 x T1)

V2 = (761 mmHg x 56.0 L x 266.18 K) / (0.0772 atm x 296.25 K)

V2 = 2,040 L (rounded to three significant figures)

Therefore, the volume of the weather balloon at the higher altitude is approximately 2,040 L.

Answer:its b

Explanation:

jusut bcss

The molarity of the solution that has 2.0 moles of solute in 3.0 L of solution is 0.67 mol/L

Molarity is  described as a measure of the concentration of a chemical species, in particular of a solute in a solution, in terms of amount of substance per unit volume of solution.

Molarity = moles of solute / liters of solution

we then substitute the given values, and have

Molarity = 2.0 moles / 3.0 L

Molarity = 0.67 mol/L

Molarity is  very important because the ration used to express the concentration of  any  solution.

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First, we need to write out the balanced chemical equation for the reaction: 4 KO2 + 2 CO2 → 2 K2CO3 + 3 O2

From the equation, we can see that 4 moles of KO2 react with 2 moles of CO2 to produce 3 moles of O2. Therefore, we need to convert the given masses of KO2 and CO2 into moles:

moles of KO2 = 2.61 g / molar mass of KO2 = 2.61 g / 71.10 g/mol = 0.0367 mol
moles of CO2 = 4.46 g / molar mass of CO2 = 4.46 g / 44.01 g/mol = 0.1013 mol

Next, we need to determine the limiting reagent (the reactant that will be completely consumed in the reaction) by comparing the mole ratios of KO2 and CO2 in the balanced equation. The ratio of moles of KO2 to moles of CO2 is:
0.0367 mol KO2 / 4 mol KO2 per 2 mol CO2 = 0.0184 mol CO2

Since this ratio is less than the actual number of moles of CO2 we have (0.1013 mol), CO2 is in excess and KO2 is the limiting reagent.

Using the mole ratio from the balanced equation, we can calculate the number of moles of O2 produced:

moles of O2 = 3 mol O2 per 4 mol KO2 × 0.0367 mol KO2 = 0.0275 mol O2

Finally, we can convert the moles of O2 to grams:

mass of O2 = moles of O2 × molar mass of O2 = 0.0275 mol × 32.00 g/mol = 0.88 g
Therefore, 2.61 g of KO2 and 4.46 g of CO2 would produce 0.88 g of O2.

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Answer:

solution resists changes in pH

Explanation:

the inherent property of buffers is to resist change to ph even when acids and bases are added. when the base is added, it is quickly neutralized by the conjugate acid, so the ph won't change.

Answer:

B: The solutions resists changes in pH.

Explanation:

Buffer reactions maintain stable pH of solutions.

The rate of appearance of NOBr at the given time is 9.80 × [tex]10^-^4[/tex] M/s which is the second option as the question asks to determine the rate of appearance of NOBr, which is a product of the given chemical reaction, at the given time when the rate of disappearance of Br₂ is known. 

For every 1 mole of Br₂ that disappears, 2 moles of NOBr appear. Thus, we can set up a proportion:

(2 mol NOBr / 1 mol Br₂) = (rate of appearance of NOBr / rate of disappearance of Br₂)

Substituting the given values,

(2 mol NOBr / 1 mol Br₂) = (rate of appearance of NOBr / 4.90 x  [tex]10^-^4[/tex] M/s)

Solving for the rate of appearance of NOBr,

rate of appearance of NOBr = (2 mol NOBr / 1 mol Br₂) x (4.90 x  [tex]10^-^4[/tex] M/s) = 9.80 ×  [tex]10^-^4[/tex]M/s

The rate of appearance of NOBr at the given time is 9.80 × [tex]10^-^4[/tex] M/s.

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Answer:

39.21 psi

Explanation:

According to Dalton's Law, the total pressure in the system is the sum of all the partial pressures, so all you need to do is add the partial pressures :)

The rate constant, k = 5.27 E-2 s⁻¹, determines the rate law for the reaction P → E + Z.

The rate of the reaction P → E + Z can be expressed as:

Rate = - d[P]/dt = d[E]/dt = d[Z]/dt

where d[P], d[E], and d[Z] = changes in the concentrations of P, E, and Z, respectively, over a small time interval dt.

Use the experimental data to determine the rate constant and the order of the reaction.

Calculate the initial rate of the reaction in each trial by dividing the change in concentration of P by the time interval:

rate1 = (d[P]/dt)1 = (0.30 M - 0 M)/(20 s) = 0.015 M/s

rate2 = (d[P]/dt)2 = (0.60 M - 0.30 M)/(20 s) = 0.015 M/s

rate3 = (d[P]/dt)3 = (0.90 M - 0.60 M)/(20 s) = 0.015 M/s

The initial rates are the same in all three trials, which suggests that the reaction is first-order with respect to P.

Now using any of the three trials to determine the value of the rate constant k, trial 1:

Rate1 = k[P]1

k = Rate1/[P]1 = (1.58 E-2 M/s)/(0.30 M) = 5.27 E-2 s⁻¹

Therefore, the rate law for the reaction P → E + Z is:

Rate = k[P]

where k = 5.27 E-2 s⁻¹ is the rate constant.

Use the rate law to calculate the expected rates of the reaction at different concentrations of P. For example:

Rate2 = k[P]2 = (5.27 E-2 s⁻¹)(0.60 M) = 3.16 E-2 M/s

Rate3 = k[P]3 = (5.27 E-2 s⁻¹)(0.90 M) = 4.74 E-2 M/s

These expected rates are close to the experimental rates, which suggests that the rate law is a good approximation for the reaction under these conditions.

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If in a reaction between vinegar and antacid tablets, the antacid is the limiting reagent. cm of gas. At constant pressure and temperature, three tablets produce 600 cm³ . The  volume that four tablets will produce is: C. 800 cm³.

Since the antacid is the limiting reagent, the amount of gas produced will be directly proportional to the number of tablets used.

We know that three tablets produced 600 cm³ of gas. Therefore, we can set up a proportion:

3 tablets produce 600 cm³ of gas

4 tablets produce x cm³ of gas

To solve for x, we can use cross-multiplication:

3 tablets × x cm³ of gas = 4 tablets × 600 cm³ of gas

3x = 2400

x = 800 cm³

Therefore the answer is C. 800 cm³.

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The molar equilibrium concentration of Cu² ⁺ is [tex]3.3 x 10^-26[/tex]M when 0.100 mol NH₃ is added to 1.00 L of 0.800 M with Kf = [tex]2.1 x 10^13[/tex].

The issue includes computing the molar balance convergence of Cu² ⁺ in an answer that has 0.100 mol NH3 added to 1.00 L of 0.800 M [Cu(NH₃)₄]²⁺ . To tackle the issue, we can involve the balance consistent articulation for the arrangement of[Cu(NH₃)₄]²⁺:

Kf =[Cu(NH₃)₄]²⁺/(Cu² ⁺)(NH₃)₄

Since the volume of the arrangement doesn't change when NH₃ is added, the underlying grouping of [Cu(NH₃)₄]²⁺ is as yet 0.800 M, while the underlying convergences of Cu² ⁺ and NH₃ are zero. Let x be the molar centralization of Cu² ⁺ at balance, and (0.100 - 4x) be the molar grouping of NH₃ at harmony. Then, we can compose the harmony consistent articulation as:

[tex]2.1 x 10^13 = (x)(0.800 - x)^4/(0.100 - 4x)^4[/tex]

Settling for x gives the molar balance convergence of Cu² ⁺ as 3.3 x [tex]10^-26[/tex] M. This tiny focus demonstrates that the majority of the copper in the arrangement is as [Cu(NH₃)₄]²⁺ and that the expansion of NH₃ has moved the harmony towards the development of this complex.

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If a skimmer and a gull eat the same fish and the skimmer is more successful at catching the fish over time, it is likely that the skimmer population would increase, while the gull population may decrease.

The skimmer's success in catching the fish would give it an advantage in obtaining the necessary nutrients for survival and reproduction. As a result, the skimmer population would likely grow over time as more individuals are able to survive and reproduce due to the abundance of food.

On the other hand, the gull population may decrease due to the competition with the skimmer for the same food source. If the skimmer population grows significantly, it may lead to a reduction in the availability of fish for the gulls to feed on. Over time, this could result in a decline in the gull population due to reduced food availability.

However, it is important to note that the impact on the bird populations may depend on various factors such as the size of the populations, availability of other food sources, and environmental factors. Therefore, the outcome of this scenario cannot be predicted with certainty and would require further analysis and investigation.

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The volume of O2 that can be released at STP from the given sample of silver oxide is 23.45 mL.

Creating the balanced chemical equation for the breakdown of silver oxide is the first step in tackling this issue:

2Ag2O(s) → 4Ag(s) + O2(g)

We can deduce from the equation that 2 moles of AgO will result in 1 mole of O2. Since Ag2O has a molar mass of 231.735 g/mol, 0.4856 g of Ag2O is equivalent to:

0.4856 g Ag2O x (1 mol Ag2O/231.735 g Ag2O) = 0.002095 mol Ag2O

Therefore, the number of moles of O2 that can be produced from 0.4856 g of Ag2O is:

0.002095 mol Ag2O x (1 mol O2/2 mol Ag2O) = 0.0010475 mol O2

1 mole of any gas takes up 22.4 L of space at STP As a result, 0.4856 g of Ag2O can generate the following amount of O2 at STP:

0.0010475 mol O2 x 22.4 L/mol = 0.02345 L or 23.45 mL

Therefore, the volume of O2 that can be released at STP from the given sample of silver oxide is 23.45 mL.

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Chemical reactions nearly always include a change in energy between the products and reactants.

Thus, Energy is released when chemical bonds are created and is released when chemical bonds are destroyed.

The overall energy of a system, however, must remain constant according to the Law of Conservation of Energy, and chemical reactions frequently absorb or release energy in the form of heat, light, or both.

The difference in the amounts of chemical energy that are stored in the products and reactants accounts for the energy change in a chemical reaction. Enthalpy refers to the system's heat content or stored chemical energy.

Thus, Chemical reactions nearly always include a change in energy between the products and reactants.

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The mole fractions of each species at equilibrium are 0.25 for A and D, 0.5 for B, and 0.225 for C.

we can use the principles of chemical equilibrium and the mole fraction formula.

First, we need to write the balanced chemical equation for the reaction involving A, B, C, and D. Let's assume that the reaction is:

A + 2B <=> C + D

where A, B, C, and D are the chemical species, and the coefficients indicate their stoichiometric ratios.

Next, we need to write the expression for the equilibrium constant, Kc, for this reaction:

Kc = [C][D] / [A][B]²

where [X] denotes the molar concentration of species X at equilibrium.

Since we know the initial moles of A, B, and D, we can calculate their total moles in the mixture:

Total moles = 1 mol A + 2 mol B + 1 mol D = 4 mol

We also know that the final mixture contains 0.9 mol of C. Therefore, the molar concentration of C at equilibrium is:

[C] = 0.9 mol / 4 L = 0.225 M

Since we have only one unknown, we can use the equilibrium constant expression to calculate the molar concentration of D:

Kc = [C][D] / [A][B]²

0.9 = (0.225)(D) / (1)(2²)

D = 1.8

Therefore, the molar concentration of D at equilibrium is 1.8 M.

Using the law of conservation of mass, we can also calculate the molar concentration of A and B at equilibrium:

[A] = 1 mol / 4 L = 0.25 M

[B] = 2 mol / 4 L = 0.5 M

Mole fraction of X = moles of X / total moles

Mole fraction of A = 1 mol / 4 mol = 0.25

Mole fraction of B = 2 mol / 4 mol = 0.5

Mole fraction of C = 0.9 mol / 4 mol = 0.225

Mole fraction of D = 1 mol / 4 mol = 0.25

Therefore, the mole fractions of each species at equilibrium are 0.25 for A and D, 0.5 for B, and 0.225 for C.

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Answer:

3.566 mol

Explanation:

Since 60.05 is grams divided by mol to cancel out the grams to get only mols it must be divided by 16.84 g

[tex]\frac{60.05 g}{mol} *\frac{1 }{16.84g} =3.566[/tex] mols acetic acid

The product that is formed first is product B and the product that will be the major product if given enough time would also be product B.

In an energy diagram, the vertical axis represents the overall energy of the reactants, while the horizontal axis is the ‘reaction coordinate’, tracing from left to right the progress of the reaction from starting compounds to final products.

The activation energy of the reaction can be shown on a diagram as the energy between the reactants that the transition state.

The product with lesser activation energy is the product that is formed first and the major product is decided by the stability of the product which depends on the energy. Lesser is the energy of the product, greater is its stability.

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The Gibbs free energy change for the given reaction at 25°C and the given concentrations is -25.5 kJ/mol

The Gibbs free energy change (∆G) of a reaction can be calculated using the equation:

∆G = -RT ln(K)

Where R is the gas constant (8.314 J/molK), T is the temperature in Kelvin, and K is the equilibrium constant.

The equilibrium constant (K) can be calculated from the acid dissociation constant (Ka) as:

K = [C] ÷ ([A] × [B])

Substituting the given values, we get:

K = (5.00 x 10⁻⁵) ÷ (1.50 x 1.00) = 3.33 x 10⁻⁵

Therefore,

∆G = - (8.314 J/molK) × (298 K) × ln(3.33 x 10⁻⁵)

= 25.5 kJ/mol

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Answer:

The answer is 966.67 mm of Hg.

Explanation:

To solve this problem, we can use Boyle's Law which states that the pressure of a gas is inversely proportional to its volume when the temperature and the amount of gas are kept constant. The formula for Boyle's Law is:

P1V1 = P2V2

where P1 is the initial pressure, V1 is the initial volume, P2 is the final pressure, and V2 is the final volume.

Using the given values:

P1 = 580 mmHg

V1 = 500 cm³

V2 = 300 cm³

We can solve for P2:

P1V1 = P2V2

580 mmHg x 500 cm³ = P2 x 300 cm³

290,000 mmHg·cm³ = P2 x 300 cm³

P2 = 290,000 mmHg·cm³ / 300 cm³

P2 = 966.67 mmHg (rounded to the nearest hundredth)

Therefore, the pressure when the volume is reduced to 300 cm³ is approximately 966.67 mmHg.

Answer:

Atomic number of sodium is 11

Electronic configuration of a sodium atom :

Since sodium has one electron in its outermost shell, Therefore, sodium can easily donate it's one electron. As the result it becomes sodium cation with + 1 charge.

Electronic configuration of a sodium cation,[tex] \: \sf ({Na}^{+1}) [/tex]

In case of sodium cation, it has fully filled electronic configuration.

Cations - Atoms that carry postive charge are called cations. Cations are formed when an atom loses its electron.

For example : [tex]\sf {Na}^{+} [/tex]

Anions - Atoms that carry negative charge are called anions. Anions are formed when an atom gains a electron.

For example : [tex]\sf {Cl}^{-} [/tex]

The volume of C₂H₂ that is collected over water at 23 ∘C by the reaction of 1.52 g of CaC₂ is 0.61 L

First, we shall determine the mole of CaC₂ that reacted. Details below:

Mole = mass / molar mass

Mole of CaC₂ = 1.52 / 64

Mole of CaC₂ = 0.024 mole

Next, we shall determine the mole of C₂H₂. obtained. Details below:

CaC₂ + 2H₂O -> C₂H₂ + Ca(OH)₂

From the balanced equation above,

1 mole of CaC₂ reacted to produced 1 mole of C₂H₂

Therefore,

0.024 mole of CaC₂ will also react to produce 0.024 mole of C₂H₂

Finally, we shall determine the volume of C₂H₂ collected. This is shown below:

PV = nRT

729.93 × V = 0.024 × 62.36 × 296

Divide both sides by 729.93

V = (0.024 × 62.36 × 296) / 729.93

V = 0.61 L

Thus, the volume of the C₂H₂ gas collected is 0.61 L

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A sunscreen made with Benzophenone-5 ([tex]C_1_4H_1_1NaO_6S[/tex]) would be expected to be more waterproof than benzophenone-3 ([tex]C_1_4H_1_2O_3[/tex]).

This is due to the presence of a sodium salt group (Na) and a sulfonic acid group ([tex]SO_3H[/tex] ) in benzophenone-5, which makes it more polar than benzophenone-3. Polar molecules interact more strongly with water molecules and are less likely to dissolve in nonpolar solvents such as oils.

Because sunscreen is designed to be water-resistant, the more polar benzophenone-5 should have stronger interactions with water and give more water resistance than benzophenone-3.

Moreover, the sulfonic acid group in benzophenone-5 may allow it to make stronger hydrogen bonds with water, increasing its water resistance even further.

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The equation that represents a balanced reaction involving 14 atoms of oxygen is:

2C2H6 + 7O2 --> 4CO2 + 6H2O

a. The relationship between variables in this equation is that 2 moles of C2H6 (ethane) react with 7 moles of O2 (oxygen) to produce 4 moles of CO2 (carbon dioxide) and 6 moles of H2O (water). This equation follows the law of conservation of mass, where the total number of atoms on both sides of the equation is the same, indicating a balanced reaction.

b. The graph is linear, as the coefficients of the reactants and products in the equation are whole numbers and form a consistent ratio. The coefficients of 2, 7, 4, and 6 represent the stoichiometry of the reaction, indicating a fixed relationship between the reactants and products.

c. An example of a situation where this balanced equation could be applicable is the combustion of ethane (C2H6) in the presence of excess oxygen (O2) to produce carbon dioxide (CO2) and water (H2O), which is a common reaction in the combustion of hydrocarbon fuels. The equation represents the balanced stoichiometry of this reaction, where 2 moles of ethane react with 7 moles of oxygen to produce 4 moles of carbon dioxide and 6 moles of water, involving a total of 14 atoms of oxygen in the reaction.

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Explanation:

Table H lists vapor pressure data for pure water at various temperatures. We can use this data to estimate the vapor pressure of the water in the given system at its final temperature of 80.0°C.

First, we need to calculate the heat absorbed by the water sample during the heating process. We can use the specific heat capacity of water to do this:

q = m * c * ΔT

where q is the heat absorbed, m is the mass of water (100 g), c is the specific heat capacity of water, and ΔT is the temperature change (80°C - 30°C = 50°C).

Plugging in the values, we get:

q = 100 g * 4.18 J/(g*C) * 50 C

q = 20900 J

This tells us that 20,900 joules of energy were absorbed by the water sample during heating.

Next, we need to consider the saturated solution of KCIO3 in the water sample. At 80.0°C, the water is already close to boiling, so it is likely that the vapor pressure of the water in the system is close to the vapor pressure of pure water at this temperature. From Table H, we can see that the vapor pressure of pure water is approximately 356 mmHg at 80.0°C.

Therefore, the vapor pressure of the water in the given system at its final temperature of 80.0°C is approximately 356 mmHg.

Answer:

Explanation:

[tex]\frac{918.7 g}{1} *\frac{1}{216.59m } = 4.241 mol[/tex] To start off the mol of HgO must be found.

After that the molar ratio between HgO and O must be found but in this case its 1:1

[tex]4.241 mol HgO*\frac{1 molO}{1molHgO} = 4.241 mol O[/tex] the mols of HgO is put on the bottom to cancel out  with the other one leaving just mols of oxygen. Finally to find g of oxygen it must be multiplied by its molar mass.

[tex]\frac{4.241 molO}{1} * \frac{15.999 g}{mol} = 67.85 g[/tex] Oxygen

The reaction rates for trial 1 is 8.22 x 10⁻² M⁻² s⁻¹ and 1.10 M⁻² s⁻¹ for trail 2 and 3

Keep the concentration of W constant while varying the concentrations of U and S while measuring the reaction rate in order to determine the reaction rate with regard to U and S.

Select trial 1 as the reference trial and calculate the reaction's rate constant (k) with respect to U and S, assuming that the concentration of W is constant throughout all three trials.

For trial 1:

[W] = 0.13 M

Rate = 4.72 x 10⁻⁴ M/s

For trial 2:

[W] = 0.13 M

Rate = 1.18 x 10⁻² M/s

From the equation rate = k[U][S], set up the following ratio of rates:

Rate2/Rate1 = (k[U]2[S]2)/(k[U]1[S]1)

Simplifying:

k = (Rate2/Rate1) x (1/[U]2) x (1/[S]2) x ([U]1) x ([S]1)

Substituting the values from trials 1 and 2:

k = (1.18 x 10⁻² M/s) / (4.72 x 10⁻⁴ M/s) x (1/0.65 M) x (1/1 M) x (0.13 M) x (1 M)

k = 8.22 x 10⁻²M⁻² s⁻¹

Similarly, for trial 3:

[W] = 0.13 M

Rate = 2.95 x 10⁻¹ M/s

Again, using trial 1 as the reference trial, figure out the reaction's rate constant (k) in relation to U and S:

k = (Rate3/Rate1) x (1/[U]3) x (1/[S]3) x ([U]1) x ([S]1)

k = (2.95 x 10⁻¹ M/s) / (4.72 x 10⁻⁴ M/s) x (1/3.25 M) x (1/1 M) x (0.13 M) x (1 M)

k = 1.10 M⁻² s⁻¹

Therefore, the equation states the reaction rate in relation to U and S is k = 8.22 x 10⁻² M⁻² s⁻¹ and 1.10 M⁻² s⁻¹ for trials 2 and 3, respectively.

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