Explanation:
It is known that [tex]Fe(OH)_{3}[/tex] is insoluble in water. As a result, plants are not able to absorb [tex]Fe^{3+}[/tex] readily through osmosis.
Therefore, the [tex]Fe^{3+}[/tex] in [tex]Fe(OH)_{3}[/tex] would be released in acidic environments, using neutralization, Iron(III) ions can be released.
Hence, the easiest way is to add low concentrations of [tex]H_{2}SO_{4}[/tex] to the soil is as follows.
[tex]Fe(OH)_{3} + H_{2}SO_{4} \rightarrow H_{2}O + Fe_{2}(SO_{4})_{3}[/tex]
Thus, we can conclude that [tex]Fe_{2}(SO_{4})_{3}[/tex] is soluble and is good for plants too.
Plants can be iron deficient even when grown in soils with high iron content because the form of iron present, Fe(OH)3, is insoluble and not easily accessible by plants. Therefore, they may not be able to absorb enough iron to meet their requirements.
Explanation:The fact that plants can still be iron deficient even when grown in soils with high iron content can be explained by the form of iron present. In this case, the iron is in the form of Fe(OH)3. While plants require iron for proper growth and development, they can only absorb it in a specific form, called Fe²+. Fe(OH)3 is insoluble and cannot be easily accessed by plants. Therefore, even with high iron content, the plants may still not be able to absorb enough iron to meet their requirements, resulting in iron deficiency.
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A sucrose solution is prepared to a final concentration of 0.210 MM . Convert this value into terms of g/Lg/L, molality, and mass %%. (Use the following values: molecular weight MWsucroseMWsucrose = 342.296 g/molg/mol ; density rhosol′nrhosol′n = 1.02 g/mLg/mL ; and mass of water, mwatmwat = 948.1 gg ). Note that the mass of solute is included in the density of the solution.
Answer:
1) 71.9 g/L
2) 0.221 m olal
3) 7.05% by mass
Explanation:
Step 1: Data given
Concentration of sucrose = 0.210 M
Molar weight of sucrose = 342.3 g/mol
Density of solution = 1.02 g/mL
Mass of water = 948.1 grams
Step 2: Convert this value into terms of g/L
(0.210 mol/L) * (342.3 g/mol) = 71.9 g/L
Calculate the molality
Step 1: Calculate mass water
Suppose we have a volume of 1.00L
Mass of the solution = 1000 mL * 1.02 g/mL = 1020 g solution
We know that there are 71.9 g of solute in a liter of solution from the first calculation. This means
(1020 grams solution) - (71.9 g solute) = 948.1 g = 0.9481 kg water
Step 2: Calculate molality
Molality = moles sucrose / mass water
(0.210 mol) / (0.9481 kg) = 0.221 mol/kg = 0.221 m olal
Mass %
% MAss = (mass solute / mass solution)*100%
(71.9 g) / (1020 g) *100% = 7.05% by mass
A solution contains 10. mmol of H3PO4 and 5.0 mmol of NaH2PO4. How many milliliters of 0.10 M NaOH must be added to reach the second equivalence point of the titration of the H3PO4 with NaOH
Explanation:
Reaction equations for the given species is as follows.
[tex]H_{3}PO_{4} + 2NaOH \righleftharpoons Na_{2}HPO_{4} + 2H_{2}O[/tex]
[tex]NaH_{2}PO_{4} + NaOH \rightleftharpoons Na_{2}HPO_{4} + H_{2}O[/tex]
At the first equivalence point we need 2 × 10 mmol NaOH.
At the second equivalence point we need 5 mmol of NaOH.
Hence, total moles of NaOH required is as follows.
(20 + 5) mmol = 25 mmol
We assume that volume of NaOH required is V.
[tex]25 mmol NaOH \times \frac{10^{-3} mol NaOH}{1 mmol NaOH} \times \frac{1000 ml V}{0.10 \text{mol NaOH}}[/tex]
= 250 ml V
Thus, we can conclude that 250 ml of 0.10 M NaOH must be added to reach second equivalence point of the titration of the [tex]H_{3}PO_{4}[/tex] and NaOH.
In the given case, 250 ml of 0.10 M NaOH must be added to attain the second equivalence point.
Calculation based on equivalence point:The reactions taking place in the given case are:
H₃PO₄ + 2NaOH ⇔ Na₂HPO₄ + 2H₂ONaH₂PO₄ + NaOH ⇔ Na₂HPO₄ + H₂OBased on the given information, a solution comprises 10 mmol of H₃PO₄ and 5 mmol of NaH₂PO₄.
For the first equivalence point, there is a need of 2 × 10 mmol NaOH.
For the second equivalence point, there is a need of 5 mmol NaOH.
Now the total moles of NaOH needed is,
= 20 + 5 mmol
= 25 mmol
Now let the volume of NaOH required be V. Now putting the values we get,
[tex]= 25 mmol NaOH * \frac{10^{-3} mole NaOH}{1 mmol NaOH} * \frac{1000 mLV}{0.10 mol NaOH} \\= 250 ml[/tex]
Thus, 250 ml of 0.10 M NaOH is need to be added to attain the second equivalence point of the titration.
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Write the balanced equation for the combustion of isooctane (C8H18) to produce carbon dioxide and water. Use the smallest possible integers to balance the equation. Also, separate the sign with 1 space and enter the reaction arrow as hyphen greater than sign: ->
The balanced equation for the combustion of isooctane (C8H18) is C8H18 + 12.5O2 -> 8CO2 + 9H2O.
Explanation:The balanced equation for the combustion of isooctane (C8H18) to produce carbon dioxide (CO2) and water (H2O) is:
C8H18 + 12.5O2 -> 8CO2 + 9H2O
In this equation, the coefficients have been adjusted to balance the number of atoms on both sides of the equation. There are 8 carbon atoms, 18 hydrogen atoms, and 25 oxygen atoms on both sides of the equation.
Final answer:
The balanced chemical equation for the complete combustion of isooctane (C₈H₁₈) is 2 C₈H₁₈(l) + 25 O₂(g) → 16 CO₂(g) + 18 H₂O(g), comprising a step-by-step process of first balancing the carbon, then hydrogen, and finally the oxygen atoms.
Explanation:
The balanced chemical equation for the combustion of isooctane (C₈H₁₈) to produce carbon dioxide (CO₂) and water (H₂O) can be represented as follows:
2 C₈H₁₈(l) + 25 O₂(g) → 16 CO₂(g) + 18 H₂O(g)
Here is a step-by-step explanation:
Start by writing the unbalanced equation: C₈H₁₈ + O₂ → CO₂ + H₂O.Balance the carbon (C) atoms first: As there are 8 C atoms in isooctane, you need to have 8 CO₂ molecules to balance the carbons.Next, balance the hydrogen (H) atoms: With 18 H atoms in isooctane, you will need 9 H₂O molecules.Finally, balance the oxygen (O) atoms: There are 16 O atoms in 8 CO₂ molecules and 18 O atoms in 9 H₂O molecules, giving a total of 25 O₂ molecules needed.After balancing, you end up with the smallest possible integers that balance the equation.
at body temperature (37∘c) the rate constant of an enzyme-catalyzed reaction is 2.3 x 1014 times greater than the rate constant of the uncatalyzed reaction. assuming the frequency factor A is the same for both reaction, y how much does the enzyme lower the activation energ
Answer:
The activation energy is lowered by +85.35 KJ/mol
Explanation:
Rate constant is related to the activation energy and the temperature of the reaction through
K = Ae⁻ᴱᵃ/ᴿᵀ
In K = (In A) - (Ea/RT)
In K = (-Ea/RT) + In A
where k = the activity constant
Ea = activation energy
R = molar gas constant
T = absolute temperature in Kelvin
₁₂
At point 1, without the catalyst
In K₁ = (-Ea₁/RT) + In A (eqn 1)
At point 2 with the catalyst
In K₂ = (-Ea₂/RT) + In A (eqn 2)
Note that the molar gas constant, the absolute temperature in Kelvin and the activity constant are all the same for both points.
Subtract (eqn 1) from (eqn 2)
In K₂ - In K₁ = (-Ea₂/RT) - (-Ea₁/RT) + In A - In A
In (K₂/K₁) = (1/RT) (Ea₁ - Ea₂)
We were told that K₂/K₁ = 2.3 × 10¹⁴, then find the difference in Ea
R = 8.314 J/mol.K, T = 37°C = 310.15 K
In (2.3 × 10¹⁴) = (1/(8.314×310.15) (Ea₁ - Ea₂)
33.1 = (1/2578.5871) (Ea₁ - Ea₂)
(Ea₁ - Ea₂) = 33.1 × 2578.5871 = 85351.23 J/mol
Therefore, the activation energy is lowered by +85.35 KJ/mol
The activation energy is lowered by +85.35 KJ/mol
Relation between Rate constant and Activation Energy:[tex]K = Ae^{\frac{-E_a}{RT} }\\\\In K = (In A) -( \frac{-E_a}{RT})\\\\In K = (\frac{-E_a}{RT}) + In A[/tex]
where
k = the activity constantEa = activation energyR = molar gas constantT = absolute temperature in Kelvin1. At point 1, without the catalyst
[tex]In K_1 = (-Ea_1/RT) + In A[/tex]............(i)
2. At point 2 with the catalyst
[tex]In K_ = (-Ea_2/RT) + In A[/tex]...........(ii)
On subtracting equation (i) from (ii)
[tex]In K_2 - In K_1 = (-Ea_2/RT) - (-Ea_1/RT) + In A - In A\\\\In (K_2/K_2) = (1/RT) (Ea_1 - Ea_2)[/tex]
Given:
K₂/K₁ = 2.3 × 10¹⁴R = 8.314 J/mol.K, T = 37°C = 310.15 KOn solving:
[tex]In (2.3 * 10^{14}) = (1/(8.314*310.15) (Ea_1 - Ea_2)\\\\33.1 = (1/2578.5871) (Ea_1 - Ea_2)\\\\(Ea_1 - Ea_2) = 33.1 * 2578.5871\\\\ (Ea_1 - Ea_2)= 85351.23 J/mol[/tex]
Thus, the activation energy is lowered by +85.35 KJ/mol.
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A mixture of 0.438 M H2, 0.444 M I2 , and 0.895 M HI is enclosed in a vessel and heated to 430 °C. H2 (g) + I2 (g) <-----> 2 HI (g) Kc = 54.3 at 430∘C Calculate the equilibrium concentrations of each gas at 430∘C.
Answer:
[H₂] = 0.178 M
[I₂] = 0.184 M
[HI] = 1.415 M
Explanation:
For the equilibrium:
H₂(g) + I₂(g) ⇄ 2 HI(g)
the equilibrium constant is given by the equation:
Kc = [ HI]² / [H₂][I₂]
Lets use first the reaction quotient which has the same expression as the equilibrium constant to predict the direction the reaction will take, i.e towards reactants or product side.
Q =( 0.895)²/(0.438)(0.444) = 4.12
Q is less than Kc so the reaction will favor the product side.
We can set up the following table to account for all the species at equilibrium:
H₂ I₂ HI
initial 0.438 0.444 0.895
change -x -x +2x
equilibrium 0.438 - x 0.444 - x 0.895 + 2x
Now we are in position to express these concentrations in terms of the equilibrium conctant, Kc
54.3 = (0.895 + 2x)² / (0.438 -x)(0.444 - x)
performing the calculatiopns will result in a quadratic equation:
0.801 + 3.580x +4x² = (0.194 - 0.882x + x²)x 54.3
Upon rearrangement and some algebra, we have
0.801 + 3.580 x + 4x² = 10.534 - 47.893x + 54.3 x²
0 = 9.733 - 51.473 x + 54.3 x²
This equation has two roots X₁ = 0.687 and X₂ = 0.26
The first is physically impossible since it will imply that more 0.687 will make the quantity at equilibrium for both H₂ and I₂ negative.
Therefore the concentrations at equilibrium of each gas are:
[H₂] = (0.438 - 0.260) = 0.178 M
[I₂] = (0.444 - 0.260) M = 0.184 M
[HI] = [0.895 + 2x(0.260)] M = 1.415 M
Note if we plug these values into the equilibrium expression we get 61 which is due to the rounding errors propagating in the quadratic equation.
The equilibrium concentrations of H2, I2, and HI at 430 °C are calculated using an expression derived from the reaction quotient equation, plugged into the Kc equation, which is then solved for 'x'. The solutions found are the changes in molarities which applied to the initial molarities give the equilibrium concentrations
Explanation:Let's denote the change in molarity of H2, I2, and HI as 'x'. At equilibrium, the molarities of H2, I2, and HI will be 0.438+x, 0.444+x, and 0.895-2x respectively. We know the equilibrium constant, Kc = 54.3. Thus, (0.895-2x)2/(0.438+x)(0.444+x) = 54.3. This is a quadratic equation in 'x' and needs to be solved to get the value of 'x'.
After finding 'x', put this value back into the equilibrium concentrations of the gases, i.e., 0.438+x for H2, 0.444+x for I2, and 0.895-2x for HI. These will give you the equilibrium concentrations of the gases at 430 °C.
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At a given temperature, 4.92 atm of Cl2 and 4.65 atm of Br2 are mixed and allowed to come to equilibrium. The equilibrium pressure of BrCl is found to be 1.597 atm. Calculate Kp for the reaction at this temperature. Cl2(g) + Br2(g) <=> 2 BrCl(g). Give answer to 2 decimal places.
Answer:
Kp = 0.16
Explanation:
Step 1: Data given
Initial pressure of Cl2 = 4.92 atm
Initial pressure of Br2 = 4.65 atm
The equilibrium pressure of BrCl is found to be 1.597 atm
Step 2: The balanced equation
Cl2(g) + Br2(g) ⇔ 2 BrCl(g
Step 3: The initial pressures
pCl2 = 4.92 atm
pBr2 = 4.65 atm
pBrCl = 0 atm
Step 4: The pressure at the equilibrium
For 1 mol Cl2 we need 1 mol Br2 to produce 2 moles BrCl
pCl2 = 4.92 - X atm
pBr2 = 4.65 - Xatm
pBrCl = 2X atm = 1.597 atm
X = 1.597/2 = 0.7985 atm
pCl2 = 4.92 - X atm = 4.92 - 0.7985 = 4.1215 atm
pBr2 = 4.65 - Xatm = 3.8515 atm
Step 5: Calculate Kp
Kp = (BrCl)² / (Cl2)*(Br2)
Kp = 1.597² / (4.1215*3.8515)
Kp = 0.16
Final answer:
To calculate Kp for the given reaction, start with the initial pressures of Cl2 and Br2, calculate the change in pressure due to the formation of BrCl, and apply these values in the Kp equation. The result is Kp = 0.084.
Explanation:
The student asked how to calculate Kp for the reaction Cl2(g) + Br2(g) ⇌ 2 BrCl(g) at a given temperature, given initial pressures for Cl2 and Br2, and the equilibrium pressure of BrCl. To solve for Kp, we utilize the change in concentration according to the reaction's stoichiometry and apply it to the equation for the equilibrium constant in terms of pressure (Kp).
Initial pressures: Cl2 = 4.92 atm, Br2 = 4.65 atm. At equilibrium, BrCl = 1.597 atm. The change in pressure for Cl2 and Br2 to form 2 BrCl is equal to the pressure of BrCl divided by 2, since the stoichiometry of the reaction dictates twice the amount of BrCl for each reactant consumed. Thus, the change (δ) is 1.597 atm / 2 = 0.7985 atm. Therefore, the pressures of Cl2 and Br2 at equilibrium are 4.92 - 0.7985 atm and 4.65 - 0.7985 atm, respectively.
To find Kp, the equation is Kp = (PBrCl)2 / (PCl2 × PBr2). Substituting the equilibrium pressures into this equation gives Kp = (1.5972) / ((4.92 - 0.7985) × (4.65 - 0.7985)). Solving this yields Kp = 0.084 (to two decimal places).
If a 32.4 gram sample of sodium sulfate (Na2SO4) reacts with a 65.3 gram sample of barium chloride (BaCl2) according to the reaction below: Na2SO4 (aq) + BaCl2 (aq) → BaSO4 (s) + 2NaCl (aq) What is the theoretical yield of barium sulfate (BaSO4) in grams?
Answer: The theoretical yield of barium sulfate is 50.9 grams
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
For sodium sulfate:Given mass of sodium sulfate = 32.4 g
Molar mass of sodium sulfate = 142 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of sodium sulfate}=\frac{32.4g}{142g/mol}=0.228mol[/tex]
For barium chloride:Given mass of barium chloride = 65.3 g
Molar mass of barium chloride = 208.23 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of barium chloride}=\frac{65.3g}{208.23g/mol}=0.314mol[/tex]
The chemical equation for the reaction of barium chloride and sodium sulfate follows:
[tex]Na_2SO_4+BaCl_2\rightarrow BaSO_4+2NaCl[/tex]
By Stoichiometry of the reaction:
1 mole of sodium sulfate reacts with 1 mole of barium chloride
So, 0.228 moles of sodium sulfate will react with = [tex]\frac{1}{1}\times 0.228=0.228mol[/tex] of barium chloride
As, given amount of barium chloride is more than the required amount. So, it is considered as an excess reagent.
Thus, sodium sulfate is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 mole of sodium sulfate produces 1 mole of barium sulfate.
So, 0.228 moles of sodium sulfate will produce = [tex]\frac{1}{1}\times 0.228=0.228moles[/tex] of barium sulfate
Now, calculating the mass of barium sulfate from equation 1, we get:
Molar mass of barium sulfate = 233.4 g/mol
Moles of barium sulfate = 0.228 moles
Putting values in equation 1, we get:
[tex]0.228mol=\frac{\text{Mass of barium sulfate}}{223.4g/mol}\\\\\text{Mass of barium sulfate}=(0.228mol\times 223.4g/mol)=50.9g[/tex]
Hence, the theoretical yield of barium sulfate is 50.9 grams
Methylamine, CH3NH2, is a weak base and one of several substances that give herring brine its pungent odor. In .100 M CH3NH2, only 6.4 percent of the base has undergone ionization. What are Kb and pKb of methylamine
Answer:
[tex]K_{b}[/tex] is 0.000438 and [tex]pK_{b}[/tex] is 3.36
Explanation:
Methylamine is a monoprotic base.
For a monoprotic base, [tex]K_{b}=\frac{ca^{2}}{(1-a)}[/tex]
where, c is concentration of base in molarity and a is it's degree ionization
Here [tex]a=\frac{6.4}{100}=0.064[/tex] and c = 0.100 M
So, [tex]K_{b}=\frac{(0.100)\times (0.064)^{2}}{(1-0.064)}=0.000438[/tex]
We know, [tex]pK_{b}=-logK_{b}[/tex]
Hence, [tex]pK_{b}=-log(0.000438)=3.36[/tex]
A student dissolves 14.g of benzoic acid C7H6O2 in 425.mL of a solvent with a density of 0.92 g/mL. The student notices that the volume of the solvent does not change when the benzoic acid dissolves in it.Calculate the molarity and molality of the student's solution. Round both of your answers to 2 significant digits.
The molarity and molality of a solution of 14. g of benzoic acid dissolved in 425.mL of a solvent with a density of 0.92 g/mL are 0.26 M and 0.28 m respectively.
Explanation:To calculate the molarity and molality of the solution, we'll first need to know how many moles of benzoic acid (C7H6O2) are present. Simply converting grams to moles using the molar mass of benzoic acid (122.12 g/mol), we get 14.g ÷ 122.12 g/mol = 0.11 mol.
Molarity is defined as moles of solute divided by liters of solution. Therefore, the molarity of the solution would be 0.11 mol ÷ 0.425 L = 0.26 M.
On the other hand, molality is calculated as moles of solute divided by kilograms of solvent. To obtain the mass of the solvent in kg, we need to multiply the volume by the density, 425.mL x 0.92 g/mL = 391 g = 0.391 kg. Consequently, the molality of the solution would be 0.11 mol ÷ 0.391 kg = 0.28 m.
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Mirex(MW = 540) is a fully chlorinated organic pesticide that was manufactured to control fire ants. Due to its structure, mirex is very unreactive; thus, it persists in the environment. Lake Erie water samples have had mirex measured as high as 0.002 μg/L and lake trout samples with 0.002 μg/g. (a) (10 points) In the water samples, what is the aqueous concentration of mirex in units of (i) ppb, (ii) ppt, (iii) μM? (b) (10 points) In the fish samples, what is the concentration of mirex in (i) ppm, (ii) pp
Answer:
A) i) 0.002 ppb (ii) 2ppt (iii) 3.7 x 10^(-6) μM
B) i)0.002 ppm. (ii) 2ppb
Explanation:
A) We know that 1 ppb = 1 μgram per liter, and so the concentration of Mirex in ppb would be 0.002 ppb.
1 ppt = 1 nanogram per liter of water, so the concentration of Mirex in ppt would be 2 ppt;
(0.002 μg/L) (100ng/μg) (1ppt/ng/L) = 2ppt.
Now, MW of Mirex = 540 g/mol ≡ μg/μmol
Thus, 1 μmole = 540 μgram,
Hence, the concentration of Mirex in μmoles would be;
(0.002 μg/L)/(540 μg/μmol) = 3.7 x 10^(-6) μM
B) i) 1 ppm = 1 μgram per gram.
Thus, the concentration of Mirex in ppm would be = 0.002 ppm.
ii) Now, 1 ppb = 1 nanogram per gram.
Thus, concentration of Mirex in ppb would be = (0.002 μg/g) (100ng/μg) (1ppb ng/g) = 2ppb
Final answer:
The concentration of mirex in water samples is 0.002 ppb, 2 ppt, and approximately 3.70 x 10⁻³ μM. In fish samples, the concentration of mirex is 0.002 ppm and 2 ppb.
Explanation:
The student is asking about converting the concentration of a pesticide, mirex, in water and fish samples to various units. The concentration in water samples is given as 0.002 μg/L. The conversions are as follows:
Parts per billion (ppb) is equivalent to micrograms per liter (μg/L), so the concentration is 0.002 ppb.To convert to parts per trillion (ppt), we multiply the ppb value by 1,000. Thus, the concentration is 2 ppt.The concentration in micromolar (μM) can be found by dividing the microgram concentration by the molecular weight of mirex and then dividing by the volume of the solution in liters. Since the molecular weight (MW) of mirex is 540 g/mol, the calculation is 0.002 μg/L × (1 mg/1,000 μg) × (1 mol/540,000 mg) × (1,000,000 μM/1 M), resulting in approximately 3.70 x 10⁻³ μM.For the fish samples:
The concentration as parts per million (ppm) is the same as micrograms per gram (ug/g), so it is 0.002 ppm.The parts per billion (ppb) value is 0.002 ppb, as 1 ppm is equal to 1,000 ppb, and the value for mirex in fish is already less than 1 ppm.The heat of fusion ΔHf of ethyl acetate C4H8O2 is 10.5 /kJmol. Calculate the change in entropy ΔS when 95.g of ethyl acetate freezes at −84.0°C. Be sure your answer contains a unit symbol. Round your answer to 2 significant digits.
The change in entropy when 95 g of ethyl acetate freezes at -84.0°C is -59.8 J/K, rounded to two significant digits.
The question involves calculating the change in entropy when 95 g of ethyl acetate freezes at -84.0℃. Given the heat of fusion ( Hf) for ethyl acetate is 10.5 kJ/mol, we first need to convert grams to moles using the molar mass of ethyl acetate (C4H8O2) which is 88.11 g/mol.
Next, we use the formula:
moles = mass (g) / molar mass (g/mol)
For 95 g ethyl acetate, moles = 95 g / 88.11 g/mol = 1.078 moles.
The change in entropy ( S) can be found using the formula:
S = H / T
where T is the temperature in Kelvin. To convert -84.0℃ to Kelvin, we add 273.15:
T = -84.0 + 273.15 = 189.15 K
The entropy change then is:
S = 10.5 kJ/mol / 189.15 K
S per mole = 0.0555 kJ/K∙mol
Finally, for 1.078 moles, the total entropy change is:
S total = 0.0555 kJ/K∙mol × 1.078 mol
S total = 0.0598 kJ/K
Expressed in J/K (as 1 kJ = 1000 J),
S total = 59.8 J/K.
Since we are calculating the entropy change for freezing, the sign should be negative because entropy decreases during freezing.
Therefore, the change in entropy when 95 g of ethyl acetate freezes at -84.0℃ is -59.8 J/K, rounded to two significant digits.
In the laboratory, a general chemistry student measured the pH of a 0.329 M aqueous solution of benzoic acid, C6H5COOH to be 2.327. Use the information she obtained to determine the Ka for this acid. Ka(experiment) =
Answer:
The dissociation constant for the acid ( experimental ) is 1.45 lit/mol
Explanation:
The value of dissociation constant can be calculated as,
[tex]K_{a}[/tex] = C × ∝²
Where, C = concentration of the solution = 0.329M
∝ = Degree of dissociation
again , Degree of dissociation can be obtained form :
[tex]p_{H}[/tex] = C × ∝
∝ = [tex]\frac{p_{H} }{C}[/tex]
∝ = [tex]\frac{2.327}{0.329}[/tex] = 7.072
So, now [tex]K_{a}[/tex] = C × ∝²
= 0.329 ×( 7.072)²
= 1.45 lit/ mol
Answer:
The Ka = 6.74 * 10^-5
Explanation:
Step 1: Data given
Concentration of benzoic acid = 0.329 M
pH = 2.327
Step 2: Calculate the Ka
pH = -log (√([HA]*Ka))
2.327 = -log (√(0.329*Ka))
10 ^ - 2.327 = √(0.329*Ka))
0.0047098 = √(0.329*Ka))
2.218 * 10^-5 = 0.329 * Ka
Ka = 6.74 * 10^-5
The Ka = 6.74 * 10^-5
You want to determine the protein content in milk with the Kjeldahl method. You take 100 g whole milk and use 100 mL of 0.5 M hydrochloric acid to collect ammonia. You needed 34.50 mL of 0.3512 M NaOH for the back-titration. Calculate the percentage of protein in the sample?
Answer:
82.53 % protein in the milk.
Explanation:
Titration formula:
Conc (acid) × Vol (acid) = Conc (base) × Vol (base)
Beginning from back-titration:
0.5 M HCl × Vol (HCl) = 0.3512 (conc of NaOH) × 34.50 mL (vol of NaOH)
= [tex]\frac{12.1164}{0.5}[/tex] = 24.2328 mL
⇒ Vol of HCl used in initial titration is 24.2328 mL
So from Initial Titration:
Using same titration formula,
0.5 × 24.2328 = Conc. (base) × 100 ml
Conc (base) = 12.1164 ÷ 100
= 0.121164 M.
But concentration = mass ÷ molar mass
0.121164 = [tex]\frac{100 g}{ molar mass}[/tex]
= 825.3277 g/mol of protein
⇒ [tex]\frac{825.32765}{1000}[/tex] × 100
= 82.53 % protein in the milk.
The rate of SN2 reaction of the type Nu- RX -> R-Nu X- can be increased Select one: a. using a solvent of high polarity b. using tertiary substrate c. using weak nucleophile d. using higher concentration of the nucleophile
Answer:
using higher concentration of the nucleophile
Explanation:
In SN2 reaction, the attack of the nucleophile on the substrate occurs simultaneously as the leaving group departs. The entering group normally attacks through the back side of the molecule. The reaction is concerted and bimolecular. This implies that the concentration of the nucleophile is important in the rate equation for the reaction. Hence increasing the concentration of the nucleophile will increase the rate of SN2 reaction.
Molten solder flows between the two base metals being soldered because ____. A. the atmospheric pressure pushes it in B. of capillary action C. there are no surface pores in the base metal D. the tubing surface melts and the pressure differential forces it in
Answer:
Molten solder flows between the two base metals being soldered because of capillary action.
Explanation:
Capillary molten solder takes place when a metal and a melt is brought into contact with the tube and an accessory after heating. Due to the phenomenon of capillary action, the molten metal rises and extends in any direction, due to the small space that remains between the wall of the tube and that of the fitting; With this, when cooling, a completely hermetic union is achieved.
Molten solder flows between two base metals due to capillary action, which allows the solder to spread evenly and adhere to the bases, metal ensuring a strong bond.
Explanation:Molten solder flows between the two base metals being soldered mainly because of a process known as capillary action. Capillary action is a natural occurrence where a liquid, in this case the molten solder, moves along a narrow space, such as between the two pieces of metal, against the force of gravity. The solder is drawn into the space and spread evenly between the two base metals because of the adhesive forces between the liquid and the surrounding materials are stronger than the cohesive forces within the liquid.
This phenomenon plays a fundamental role in the soldering process, as it allows the solder to spread evenly and adhere to the base metals, ensuring a strong and stable bond between them. This principle is used not only in soldering but also in various other fields such as plant biology, painting, and inkjet printing.
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For the oxidation–reduction reaction equation 2 Rb + Br 2 ⟶ 2 RbBr 2Rb+Br2⟶2RbBr indicate how many electrons are transferred in the formation of one formula unit of product.
For the given reaction between rubidium and Boron, only one electron will be transferred in the formation [tex]2 RbBr[/tex].
The potential difference between the two half-cells of an electrochemical cell is known as cell potential. The transfer of electrons from one-half cell to another produces the potential difference.
The given chemical reaction is:
[tex]2Rb_{(s)} + Br_{(l)}[/tex] ⇒ [tex]2RbBr_{(s)}[/tex]
The following electrochemical reaction can be divided into the reaction occurring at the cathode and the reaction occurring at the anode:
Oxidation reaction at the anode: [tex]2Rb_{(s)}[/tex] ⇒ [tex]2Rb^+_{(aq)} + 2e^-[/tex]
Reduction reaction at the cathode: [tex]Br_{2(l)} + 2e^-[/tex] ⇒ [tex]2Br^-_{(aq)}[/tex]
Since there are two units of RbBr in the aforementioned equation and two electrons were transported, only one electron is needed to generate one unit of RbBr.
Thus, only one electron is involved in the formation of RbBr.
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Final answer:
In the formation of one formula unit of RbBr, one electron is transferred from rubidium (Rb) to bromine (Br), as Rb loses one electron to form Rb+ and Br gains one electron to form Br-.
Explanation:
The question regards the number of electrons transferred in the formation of one formula unit of RbBr from the reaction 2 Rb + Br2 → 2 RbBr. In order to determine the electron transfer, we need to consider the oxidation states of the elements involved. Rubidium (Rb) is oxidized, meaning it loses electrons, and bromine (Br2) is reduced, meaning it gains electrons.
Rubidium starts with an oxidation state of 0 and becomes Rb+ in RbBr. This indicates that each Rb atom loses one electron. Bromine starts with an oxidation state of 0 as diatomic Br2 and each Br atom gains one electron to become Br- in RbBr. Therefore, for each RbBr formula unit produced, one electron is transferred from Rb to Br.
Be sure to answer all parts. A buffer consists of 0.38 M KHCO3 and 0.71 M K2CO3. Carbonic acid is a diprotic acid with Ka1 = 4.5 × 10−7 and Ka2 = 4.7 × 10−11. (a) Which Ka value is more important to this buffer?
Explanation:
The dissociation equations for the carbonic acid are as follows.
[tex]H_{2}CO_{3}(aq) + H_{2}O(l) \rightarrow HCO^{-}_{3}(aq) + H_{3}O^{+}(aq)[/tex]
[tex]HCO^{-}_{3}(aq) + H_{2}O(l) \rightarrow (CO^{2-}_{3})(aq) + H_{3}O^{+}(aq)[/tex]
The given buffer contains [tex]KHCO_{3}[/tex] and [tex]K_{2}CO_{3}[/tex]. This means that there will be [tex]HCO^{-}_{3}[/tex] ions and [tex]CO^{2-}_{3}[/tex] ions and since, both of them are present in the second step.
Therefore, [tex]K_{a2}[/tex] is more significant with respect to this reaction.
Thus, we can conclude that [tex]K_{a2}[/tex] is more important to this buffer.
A buffer resists changes in pH. When considering carbonic acid, which is a diprotic acid i.e., it can donate two protons, the first ionization is represented by Ka1, and the second by Ka2. In a buffer solution with physiological pH close to 7, Ka1 is more important because the first ionization of carbonic acid happens more readily at this pH.
Explanation:Your question asks which Ka value is more important for this buffer, which contains 0.38 M KHCO3 and 0.71 M K2CO3. Carbonic acid is a diprotic acid with Ka1 = 4.5 × 10^−7 and Ka2 = 4.7 × 10^−11.
A buffer solution resists changes in pH by neutralizing added acids and bases. The effectiveness of a buffer is determined by the pKa of the acid component of the buffer, which is the negative logarithm of its Ka. Generally, the most effective buffers are those wherein the pKa is close to the desired pH.
In our case, since carbonic acid is diprotic it has two associated Ka values. The first dissociation (giving HCO3^-) has Ka1 and the second dissociation (giving CO3^2-) has Ka2. Typically in physiological circumstances such as in our blood, where pH is close to 7, the first ionization of carbonic acid matters more because it is the one that happens readily at physiological pH. Therefore, Ka1 would be more important in this buffer solution as it represents the first ionization of the acid.
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the osmotic pressure ofa solution containing 5.87 mg of an unknown protein per 10ml of solution is 2.45 torr at 25 degree celsius. Find the molar mass of the unknown protein.
Final answer:
The molar mass of the unknown protein is 5.79 g/mol.
Explanation:
To find the molar mass of the unknown protein, we can use the formula for osmotic pressure:
π = MRT
Where π is the osmotic pressure, M is the molarity of the solution, R is the ideal gas constant, and T is the temperature in Kelvin. Rearranging the formula, we have:
M = π / (RT)
Substituting the given values:
M = 2.45 torr / (0.0821 L.atm/(mol.K) * 298 K)
M = 0.0996 mol/L
To convert the molarity to grams per liter, we need to multiply it by the molar mass
Molar mass = (0.0996 mol/L) * (10 mL/1 L) * (5.87 mg/1 mL)
Molar mass = 5.79 g/mol
A volume of 90.0 mL mL of a 0.590 M M HN O 3 HNO3 solution is titrated with 0.350 M M KOH KOH . Calculate the volume of KOH KOH required to reach the equivalence point. Express your answer to three significant figures and include the appropriate units.
Answer:
152 mL is the volume of KOH required to reach the equivalence point.
Explanation:
[tex]HNO_3(aq)+KOH(aq)\rightarrow KNO_3(aq)+H_2O(l)[/tex]
To calculate the concentration of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HNO_3[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is KOH.
We are given:
[tex]n_1=1\\M_1=0.590 M\\V_1=90.0 mL\\n_2=1\\M_2=0.350 M\\V_2=?[/tex]
Putting values in above equation, we get:
[tex]1\times 0.590 M\times 90.00=1\times 0.350 M\times V_2[/tex]
[tex]V_2=\frac{1\times 0.590 M\times 90.0 mL}{1\times 0.350 M}=151 .7 mL\approx 152 mL[/tex]
152 mL is the volume of KOH required to reach the equivalence point.
Answer:
152 ml.
Explanation:
Given:
Volume of HNO3 = 90 ml
Molar concentration of HNO3 = 0.59 M
Molar concentration of KOH = 0.35 M
Equation of the reaction
KOH + HNO3 --> KNO3 + H2O
Number of moles of HNO3 = molar concentration × volume
= 0.59 × 0.09
= 0.0531 moles.
By stoichiometry, 1 mole of HNO3 reacts with 1 mole of KOH. Therefore,
Number of moles of KOH = 0.0531 moles.
Volume = 0.0531 ÷ 0.350
= 0.152 l
= 152 ml.
A solution contains 4.08 g of chloroform (CHCl3) and 9.29 g of acetone (CH3COCH3). The vapor pressures at 35 ∘C of pure chloroform and pure acetone are 295 torr and 332 torr, respectively. Assuming ideal behavior, calculate the vapor pressures of each of the components and the total vapor pressure above the solution.
The vapor pressures of chloroform and acetone in the solution are 52.02 torr and 273.50 torr, respectively, and the total vapor pressure above the solution is 325.52 torr.
Explanation:To calculate the vapor pressures of chloroform and acetone above the solution, we should make use of Raoult's law. According to Raoult's law, the partial pressure of each component of a mixture is the product of the vapor pressure of the pure component and its mole fraction in the mixture.
First, calculate the mole fractions of each substance. The molar mass of chloroform (CHCl3) is about 119.38 g/mol, and the molar mass of acetone (CH3COCH3) is about 58.08 g/mol. Therefore, the mole fractions of chloroform and acetone are 4.08 g CHCl3 * (1 mol / 119.38 g) = 0.0342 mol and 9.29 g CH3COCH3 * (1 mol / 58.08 g) = 0.1599 mol, respectively. The total moles are 0.0342 mol + 0.1599 mol = 0.1941 mol, so the mole fraction of chloroform is 0.0342/0.1941 = 0.176 and that of acetone is 0.1599/0.1941 = 0.824.
Next, apply Raoult's Law to find the partial pressure of each component in the mixture. The vapor pressure of chloroform is 0.176 * 295 torr = 52.02 torr, and the vapor pressure of acetone is 0.824 * 332 torr = 273.50 torr.
The total vapor pressure above the solution is the sum of the vapor pressures of chloroform and acetone, that is, 52.02 torr + 273.50 torr = 325.52 torr.
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Assume the solubility of benzoic acid in ice-cold water is 1.70 g/L and the solubility of benzoic acid in hot water is 68.0 g/L. Calculate the minimum volume of water (in mL) needed to recrystallize 0.700 g of benzoic acid.
Answer:
The volume of water needed to recrystallize 0.700 g of benzoic acid is 10.29 mL.
Explanation:
For the complete recrystalization,
the amount of hot water should be such that, the benzoic acid is completely soluble in it.
As we are given that the solubility of benzoic acid in hot water is 68.0 g/L. Now we have to determine the volume of water is needed to recrystallize 0.700 g of benzoic acid.
we conclude that,
As, 68.0 grams of benzoic acid soluble in 1 L of water.
So, 0.700 grams of benzoic acid soluble in of water.
The volume of water needed = 0.01029 L = 10.29 mL
conversion used : (1 L = 1000 mL)
Therefore, the volume of water needed to recrystallize 0.700 g of benzoic acid is 10.29 mL.
for the hypothetical reaction 2A B -> 2C D, the initial rate of disappearance of A is 1.6x10^-1 mol/(L s) . what is the initial rate of disappearance of B
Answer:
0.8 x10^-1 mol/(L s)
Explanation:
2A + B -> 2C + D
For the reaction above, the differential rate is usually expressed as;
Rate = - (1 / 2) Δ[A] / Δt = - Δ[B] / Δt
The negative sign denotes disappearance.
Upon comparing ; (1 / 2) Δ[A] / Δt = Δ[B] / Δt
If initial rate of disappearance of A is 1.6x10^-1 mol/(L s);
That means
Δ[B] / Δt = 1.6x10^-1 mol/(L s) / 2
Δ[B] / Δt = 0.8 x10^-1 mol/(L s)
The initial rate of disappearance of B = 1.6x10^-1 mol/(L s)
In preparation for a demonstration, your professor brings a 1.50−L bottle of sulfur dioxide into the lecture hall before class to allow the gas to reach room temperature. If the pressure gauge reads 173 psi and the lecture hall is 20°C, how many moles of sulfur dioxide are in the bottle? In order to solve this problem, you will first need to calculate the pressure of the gas. Hint: The gauge reads zero when 14.7 psi of gas remains.
Answer:
0.66 mol
Explanation:
Zero Gauge pressure = 14.7 psi
Pressure read = 173 psi
Actual pressure = 173 psi - 14.7 psi = 158.3 psi
P (psi) = 1/14.696 P(atm)
So, Pressure = 10.77 atm
Given that:
Temperature = 20 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (20+ 273.15) K = 298.15 K
V = 1.50 L
Using ideal gas equation as:
[tex]PV=nRT[/tex]
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L atm/ K mol
Applying the equation as:
10.77 atm × 1.50 L = n ×0.0821 L atm/ K mol × 298.15 K
⇒n = 0.66 mol
Select the correct definition of a glycolipid. A glycolipid is: a lipid molecule that contains a glycerol backbone and three fatty acids a lipid molecule that contains at least one carbohydrate unit a lipid molecule produced during glycolysis a molecule produced during the breakdown of glycogen
Answer:
a lipid molecule that contains at least one carbohydrate unit
Explanation:
A glycolipid -
It refers to the lipid which consists of the carbohyde group attached via a glycosidic bond , which are basically covalent bonds , is referred to as a glycolipid .
They are present on the surface of the eukaryotic cell membranes .
Glycolipids are important to connect from one tissue to another and facilitate cellular recognition .
Hence , from the given information of the question ,
The correct answer is a lipid molecule that contains at least one carbohydrate unit .
A particular reaction, A- products, has a rate that slows down as the reaction proceeds. The half-life of the reaction is found to depend on the initial concentration of A. Determine whether each statement is likely to be true or false for this reaction.
a. A doubling of the concentration of A doubles the rate of the reaction.
b. A plot of 1/[A] versus time is linear.
c. The half-life of the reaction gets longer as the initial concen- tration of A increases.
d. A plot of the concentration of A versus time has a constant slope.
Explanation:
Half life of zero order and second order depends on the initial concentration. But as the given reaction slows down as the reaction proceeds, therefore, it must be second order reaction. This is because rate of reaction does not depend upon the initial concentration of the reactant.
a. As it is a second order reaction, therefore, doubling reactant concentration, will increase the rate of reaction 4 times. Therefore, the statement a is wrong.
b. Expression for second order reaction is as follows:
[tex]\frac{1}{[A]} =\frac{1}{[A]_0} +kt[/tex]
the above equation can be written in the form of Y = mx + C
so, the plot between 1/[A] and t is linear. So the statement b is true.
c.
Expression for half life is as follows:
[tex]t_{1/2}=\frac{1}{k[A]_0}[/tex]
As half-life is inversely proportional to initial concentration, therefore, increase in concentration will decrease the half life. Therefore statement c is wrong.
d.
Plot between A and t is exponential, therefore there is no constant slope. Therefore, the statement d is wrong
The beta-ketoester Claisen product is generated from the product of this final step by addition of dilute HCl. Write the final step of the Claisen condensation using curved arrows to show electron reorganization.
Answer:
This shown on the second uploaded image
Explanation:
What is occurring in this reaction is the further deprotonation of the base and this would now react with HCl to give the final product
If 50.0 g of N2O4 is introduced into an empty 2.12 L container, what are the partial pressures of NO2 and N2O4 after equilibrium has been achieved at 45∘C?
Answer:
p(N2O4) = 0.318 atm
p(NO2) = 7.17 atm
Explanation:
Kc for the reaction N2O4 <=> 2NO2 is 0.619 at 45 degrees C If 50.0g of N2O4 is introduced into an empty 2.10L container, what are the partial pressures of NO2 and N2O4 after equilibrium has been achieved at 45 degrees C?
Step 1: Data given
Kc = 0.619
Temperature = 45.0 °C
Mass of N2O4 = 50.0 grams
Volume = 2.10 L
Molar mass N2O4 = 92.01 g/mol
Step 2: The balanced equation
N2O4 ⇔ 2NO2
Step 3: Calculate moles N2O4
Moles N2O4 = 50.0 grams / 92.01 g/mol
Moles N2O4 = 0.543 moles
Step 4: The initial concentration
[N2O4] = 0.543 moles/2.10 L = 0.259 M
[NO2]= 0 M
Step 5: Calculate concentration at the equilibrium
For 1 mol N2O4 we'll have 2 moles NO2
[N2O4] = (0.259 -x)M
[NO2]= 2x
Step 6: Calculate Kc
Kc = 0.619= [NO2]² / [N2O4]
0.619 = (2x)² / (0.259-x)
0.619 = 4x² / (0.259 -x)
x = 0.1373
Step 7: Calculate concentrations
[N2O4] = (0.259 -x)M = 0.1217 M
[NO2]= 2x = 0.2746 M
Step 8: The moles
Moles = molarity * volume
Moles N2O4 = 0.1217 M * 2.10 = 0.0256 moles
Moles NO2 = 0.2746 M * 2.10 = 0.577 moles
Step 9: Calculate partial pressure
p*V = n*R*T
⇒ with p = the partial pressure
⇒ with V = the volume = 2.10 L
⇒ with n = the number of moles
⇒ with R = the gas constant = 0.08206 L*atm/mol*K
⇒ with T = the temperature = 45 °C = 318 K
p = (nRT)/V
p(N2O4) = (0.0256 *0.08206 * 318)/ 2.10
p(N2O4) = 0.318 atm
p(NO2) = (0.577 *0.08206 * 318)/ 2.10
p(NO2) = 7.17 atm
Write the expression for the equilibrium constant for the reaction represented by the equation CaCO3(s)⇌Ca2+(aq)+CO32−(aq). Is Kc > 1, < 1, or ≈ 1? Explain your answer.
Answer: For the given reaction, the value of [tex]K_c[/tex] is greater than 1
Explanation:
For the given chemical equation:
[tex]CaCO3(s)\rightleftharpoons Ca^{2+}(aq.)+CO_3^{2-}(aq.)[/tex]
The expression of [tex]K_c[/tex] for above equation follows:
[tex]K_c=\frac{[Ca^{2+}]\times [CO_3^{2-}]}{[CaCO_3}]\\\\K_c=[Ca^{2+}]\times [CO_3^{2-}][/tex]
The concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression
As, the denominator is missing and the numerator is the only part left in the expression. So, the value of [tex]K_c[/tex] will be greater than 1.
Hence, for the given reaction, the value of [tex]K_c[/tex] is greater than 1
Solvents at disposal:
Hexanes, Toluene, dichloromethane, diethyl, ethyl acetate, methanol, water, acetic acid
From the solvents, choose a single solvent that would give the best chance of separating benzophenone, diphenyl methanol and biphenyl using TLC on silica gel. Explain reasoning behind the choice. (phenyl groups are fairly non-polar)
Explanation:
In the given solvents -
Hexanes, Toulene, and dichloromethane have lesser eluting power. Diethyl ether and Ethyl acetate have medium eluting power. Methanol, water, and acetic acid have very little eluting power. Greater is the eluting power of the solvent lesser is the amount to be added. For solvents like Diethyl ether and Ethyl acetate have to be added in the percentage of 1% - 50% in addition to the solvent of higher eluting power. Hexanes and Toulene are often used in TLC (Thin Layer Chromatography ) as a Base. The reason behind this is they are non-polar compounds. The best-used solvent is Hexane.Final answer:
Toluene is the best solvent choice for separating benzophenone, diphenyl methanol, and biphenyl using TLC on silica gel due to its balanced polarity, which will allow for better differentiation between the non-polar compounds.
Explanation:
In choosing a solvent for separating benzophenone, diphenyl methanol, and biphenyl using TLC on silica gel, toluene appears to be the most appropriate choice. Since phenyl groups are fairly non-polar, a non-polar or moderately polar solvent is needed to achieve good separation. Toluene is a good choice because it has a higher polarity than hexanes but is less polar than the other solvents listed, providing a balanced environment to distinguish the three compounds on the basis of their differing polarities. A solvent like methanol or water would be too polar, causing the compounds to stay near the baseline, while hexanes might be too non-polar, failing to separate the compounds efficiently. Therefore, toluene is the recommended solvent for TLC in this scenario.
Calculate the value of K p for the equation C ( s ) + CO 2 ( g ) − ⇀ ↽ − 2 CO ( g ) K p = ? given that at a certain temperature C ( s ) + 2 H 2 O ( g ) − ⇀ ↽ − CO 2 ( g ) + 2 H 2 ( g ) K p 1 = 3.23 H 2 ( g ) + CO 2 ( g ) − ⇀ ↽ − H 2 O ( g ) + CO ( g ) K p 2 = 0.693
Answer:
kp = 1.55
Explanation:
To get to this reaction, we just need to sum the other two reactions, and multiply coefficients if it's needed so:
C(s) + 2H₂O(g) ⇄CO₂(g) + 2H₂(g) Kp1 = 3.23 (1)
H₂(g) + CO₂(g) ⇄ H₂O(g) + CO(g) Kp2 = 0.601 (2)
we will multiply eqn 2 by 2 because we need to eliminate H₂
2H2(g) + 2CO2(g) ⇄ 2H2O(g) + 2CO(g) Kp = (0.601)²
Now, add both equation
C(s) + 2H₂O(g) ⇄CO₂(g) + 2H₂(g) Kp1 = 3.23
2H2(g) + 2CO2(g) ⇄ 2H2O(g) + 2CO(g) Kp = (0.601)²
we have,
C(s) + CO₂(g) ⇄ 2CO(g)
Now the value of Kp will be:
kp = 3.23 × (0.693)²
kp = 1.55