Answer:
Acetate ion
Explanation:
CH₃COOH ↔ CH₃COO⁻ + H⁺
HCl -> H⁺ + Cl⁻
The H+ ions from added HCl reacts with acetate ions and more amount of acetic acid will be formed due to common ion effect.
The pH of the buffer solution does not decrease drastically because the added HCl reacts with the sodium acetate present in the buffer, maintaining the buffer's capacity to neutralize added protons.
Sodium acetate acts as a conjugate base in the buffer system, reacting with added HCl to form acetic acid and water. This reaction prevents the pH from decreasing drastically because the added protons from HCl are consumed in the formation of acetic acid, thereby maintaining the buffer capacity.
A buffer system works to maintain a relatively constant pH upon the addition of acids or bases, demonstrating its capacity to minimize changes in the hydrogen ion concentration of a solution.
A 1.000 kg sample of nitroglycerine, C3H5N3O9, explodes and releases gases with a temperature of 1985°C at 1.100 atm. What is the volume of gas produced? 4 C3H5N3O9(s) → 12 CO2(g) + 10 H2O(g) + 6 N2(g) + O2(g) A 1.000 kg sample of nitroglycerine, C3H5N3O9, explodes and releases gases with a temperature of 1985°C at 1.100 atm. What is the volume of gas produced? 4 C3H5N3O9(s) → 12 CO2(g) + 10 H2O(g) + 6 N2(g) + O2(g) 4730 L 5378 L 3525 L 742.2 L
Answer:
742.2 L
Explanation:
First we must find the number of moles of nitroglycerine reacted.
Molar mass of nitroglycerine= 227.0865 g/mol
Mass of nitroglycerine involved = 1×10^3 g
Number of moles of nitroglycerine= 1×10^3g/227.0865 g/mol
n= 4.40361 moles
T= 1985°C + 273= 2258K
P= 1.100atm
R= 0.082atmLmol-1K-1
Using the ideal gas equation:
PV= nRT
V= nRT/P
V= 4.40361× 0.082× 2258/1.1
V= 742 L
Considering the reaction stoichiometry and ideal gas law, the volume of gas produced is 5375.626 L.
The balanced reaction is:
4 C₃H₅N₃O₉(s) → 12 CO₂(g) + 10 H₂O(g) + 6 N₂(g) + O₂(g)
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
C₃H₅N₃O₉(s): 4 moles O₂(g): 1 moles CO₂(g): 12 moles H₂O(g): 10 molesN₂(g): 6 molesThen 4 moles of nitroglycerine C₃H₅N₃O₉(s) produce in total 29 moles of gas [12 moles of CO₂(g) + 10 moles of H₂O(g) + 6 moles of N₂(g) + 1 mole O₂(g)]
Being the molar mass of nitroglycerine C₃H₅N₃O₉(s) 227 g/mole, then the amount of moles that 1 kg (1000 g) of the compound contains can be calculated as:
[tex]1000 gramsx\frac{1 mole}{227 grams}= 4.405 moles[/tex]
Then you can apply the following rule of three: if by stoichiometry 4 moles of nitroglycerine C₃H₅N₃O₉(s) produce in total 29 moles of gas, 4.405 moles of C₃H₅N₃O₉(s) will produce how many moles of gas?
[tex]amount of moles of gas=\frac{4.405 moles of nitroglycerinex29 moles of gas}{4 moles of nitroglycerine}[/tex]
amount of moles of gas= 31.93625 moles
On the other hand, an ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P× V = n× R× T
In this case, you know:
P= 1.1 atmV= ?n= 31.93625 molesR= 0.082[tex]\frac{atmL}{molK}[/tex]T= 1985 C= 2258 K (being 0 C=273 K)Replacing:
1.1 atm× V= 31.93625 moles× 0.082 [tex]\frac{atmL}{molK}[/tex]× 2258 K
Solving:
V= (31.93625 moles× 0.082 [tex]\frac{atmL}{molK}[/tex]× 2258 K) ÷ 1.1 atm
V= 5375.625 L
Finally, the volume of gas produced is 5375.626 L.
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Answer:
(a) Distillate (D) = 0.597 kg
Bottom product (B) = 0.403 kg
(b) Compositions of distillate product:
Benzene = 0.995
Toluene = 0.05
Compositions of bottom product:
Benzene = 0.0149
Naphthalene = 0.1241
Toluene = 0.8610
(c) Fraction of toluene fed that is recovered in the bottom product = 99.14%
Explanation:
See the attached file for the calculation
g Identify which of the following statements about human glycogen debranching enzyme are true based on the HPLC results. The (α‑1→6) glucosidase catalytic center is in the C‑terminal half. The transferase catalytic center can hydrolyze α‑1,6 glycosidic bonds. The oligo‑(α1→4)‑(α1→4) glucanotransferase catalytic center is in the C‑terminal half. Oligo‑(α1→4)‑(α1→4) glucanotransferase activity creates a substrate for (α‑1→6) glucosidase. Based on the peaks in the HPLC charts, what do you think is the most likely substrate for the oligo‑(α1→4)‑(α1→4) glucanotransferase catalytic center of glycogen debranching enzyme? maltotetraosyl and glucosyl residues 6‑O‑α‑glucosyl cyclomaltoheptaose maltosyl and maltotriosyl residues maltoheptaosyl and maltooctaosyl residues cyclomaltoheptaose (β‑cyclodextrin)
The most likely substrates for the oligo-(α1→4)-(α1→4) glucanotransferase catalytic center of glycogen debranching enzyme are maltosyl and maltotriosyl residues.
Explanation:The glycogen debranching enzyme has two activities; it acts as a glucosidase and a glucanotransferase. The (α‑6) glucosidase activity hydrolyzes α-1,6 glycosidic bonds, whereas the oligo-(α1→4)-(α1→4) glucanotransferase activity shifts α-1,4-linked glucose chains from one branch to another, often creating a substrate that the glucosidase can act upon. Based on the mechanism of glycogenolysis, the enzyme glycogen phosphorylase releases glucose units from the linear chain until a few are left near the branching point, and it is here that the glucan transferase action is relevant. The glucan transferase shifts the remaining α-1,4 linked glucose units, leaving a single α-1,6 linked glucose that the glucosidase can then release. Hence, the most likely substrates for the oligo-(α1→4)-(α1→4) glucanotransferase catalytic center are maltosyl and maltotriosyl residues, as these are the short α-1,4 linked glucose chains that are left after phosphorylase action.
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Construct a simulated 1H NMR spectrum for 1-chloropropane by dragging and dropping the appropriate splitting patterns into the boxes on the chemical shift baseline, and by dragging integration values into the small box above each signal. Items may be used more than once. Peak heights do not represent integration.
To construct a simulated 1H NMR spectrum for 1-chloropropane, you need to consider the chemical shifts and splitting patterns of the hydrogen atoms. Follow these steps: identify the different types of hydrogen atoms, determine the chemical shifts, assign integration values, and drag and drop splitting patterns.
Explanation:To construct a simulated 1H NMR spectrum for 1-chloropropane, you need to consider the chemical shifts and splitting patterns of the hydrogen atoms. Here's a step-by-step guide:
Identify the different types of hydrogen atoms in 1-chloropropane. In this molecule, there are three types: H on the methyl group, H on the second carbon, and H on the third carbon.Determine the chemical shifts of the hydrogen atoms. The chemical shift values for H on the methyl group, H on the second carbon, and H on the third carbon are typically around 0.9-1.2 ppm, 1.2-1.4 ppm, and 2.5-3.0 ppm, respectively.Assign integration values to the signals. The integration values represent the relative number of hydrogen atoms in each type. For example, the H on the methyl group typically has an integration value of 3, indicating that there are three hydrogen atoms in the methyl group.Drag and drop the appropriate splitting patterns into the boxes on the chemical shift baseline. The splitting pattern depends on the neighboring hydrogen atoms and follows the n+1 rule. For example, if a hydrogen atom has two neighboring hydrogen atoms, it will show a triplet splitting pattern.Learn more about simulated 1H NMR spectrum for 1-chloropropane here:https://brainly.com/question/9812005
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COLLEGE CHEMISTRY 35 POINTS
Determine the percent yield when 12 g of CO2 are formed experimentally from the reaction of 0.5 moles of C8H18 reacting with excess oxygen according to the following balanced equation:
2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(l)
Please show your work.
6.81% is the percent yield when 12 g of CO2 are formed experimentally from the reaction of 0.5 moles of C8H18 reacting with excess oxygen.
Explanation:
Data given:
mass of carbon dioxide formed = 12 grams (actual yield)
atomic mass of CO2 = 44.01 grams/mole
moles of [tex]C_{8} H_{18}[/tex] = 0.5
Balanced chemical reaction:
2 [tex]C_{8} H_{18}[/tex] + 25 [tex]O_{2}[/tex] ⇒ 16 C[tex]O_{2}[/tex] + 18 [tex]H_{2} _O{}[/tex]
number of moles of carbon dioxide given is
number of moles = [tex]\frac{mass}{atomic mass of 1 mole}[/tex]
number of moles= [tex]\frac{12}{44}[/tex]
number of moles of carbon dioxide gas = 0.27 moles
from the reaction 2 moles of [tex]C_{8} H_{18}[/tex] reacts to produce 16 moles of C[tex]O_{2}[/tex]
So, when 0.5 moles reacted it produces x moles
[tex]\frac{16}{2}[/tex] = [tex]\frac{x}{0.5}[/tex]
x = 4
4 moles of carbon dioxide formed, mass from it will give theoretical yield.
mass = number of moles x molar mass
mass = 4 x 44.01
= 176.04 grams
percent yield =[tex]\frac{actual yield}{theoretical yield}[/tex] x 100
percent yield = [tex]\frac{12}{176.04}[/tex] x 100
percent yield = 6.81 %
Determine the ground-state electron configuration and bond order for each of the Period 2 diatomic molecules. Specify which MO is the HOMO and which is the LUMO. List the molecules in order of increasing dissociation and list them in order of bond length. You may need to THINK a bit to get this correct
Answer:
Explanation:
check the attachment below for correct explanations.
After treating cyclohexanone with LDA, he then added tert-butyl bromide to the same reaction vessel, which one would normally do for a one-pot two-step reaction sequence. However, none of the desired compound formed. Instead, he isolated three organic compounds in a 1:1:1 ratio. What is the structure of each of these organic compounds?
Answer:
2-Tert-butyl-cyclohexanone, 2,6-Di-Tert-Butylcyclohexanone and Di-isopropyl-amine.
Explanation:
Suppose the formation of nitryl fluoride proceeds by the following mechanism:
Step. Elementary reaction. Rate constant
1. NO2(g)+F2(g)=NO2F(g)+F(g). K1
2. F(g)+NO2(g)=NO2F(g). K2
Suppose also k1<
Write the balanced chemical equation for the overall chemical reaction:
Write the experimentally observable rate law for the overall chemical reaction:
Note your answer should not contain concentrations or any intermediates.
Express the rate constant k for the overall chemical reaction in terms of k1, k2, and If necessary the rate constants k-1 and k-2 for the reverse of two elementary reactions in the mechanism.
The balanced chemical equation for the overall chemical reaction is 2NO2(g) + F2(g) → 2NO2F(g). The experimentally observable rate law for the overall chemical reaction is Rate = k [NO2]^2 [F2]. The rate constant k for the overall chemical reaction can be expressed in terms of k1, k2, k-1, and k-2 as k = k1 (k2/k-1).
Explanation:The balanced chemical equation for the overall chemical reaction is:
2NO2(g) + F2(g) → 2NO2F(g)
The experimentally observable rate law for the overall chemical reaction is:
Rate = k [NO2]2 [F2]
The rate constant k for the overall chemical reaction can be expressed in terms of k1, k2, k-1, and k-2 as:
k = k1 (k2/k-1)
Why does a higher concentration make a reaction faster?
There are more collisions per second and the collisions are of greater energy.
Collisions occur with greater energy.
There are more collisions per second.
Higher concentration make a reaction faster due to faster collisions. The correct answer is "There are more collisions per second."
When the concentration of reactants is increased, it leads to a higher number of particles or molecules in the reaction mixture. As a result, there are more frequent collisions between the reactant particles per unit time.
In a chemical reaction, the reactant particles must collide with sufficient energy and proper orientation for a successful reaction to occur. By increasing the concentration, the chances of successful collisions increase because there are more particles available to collide with each other.
However, it's important to note that while a higher concentration increases the frequency of collisions, it does not necessarily guarantee that all collisions will result in a reaction. The energy of the collisions and proper orientation are also essential factors for a successful reaction.
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For many purposes we can treat ammonia (NH) as an ideal gas at temperatures above its boiling point of -33.C. Suppose the temperature of a sample of ammonia gas is raised from -21.0°C to 23.0°C, and at the same time the pressure is changed. If the initial pressure was 4.6 atm and the volume decreased by 50.0%, what is the final pressure? Round your answer to 2 significant digits. atm OP x 5 ?
The final pressure of the ammonia gas, given that the the volume decreased by 50.0%, is 11 atm
How to calculate the final pressure f the ammonia gas?
To solve this question in the most simplified way, we shall begin by listing out the given data from the question. This is shown below:
Initial temperature of ammonia gas (T₁) = -21 °C = -21 + 273 = 252 KFinal temperature of ammonia gas (T₂) = 23 °C = 23 + 273 = 296 KInitial volume of ammonia gas (V₁) = VInitial pressure of ammonia gas (P₁) = 4.6 atmFinal volume of ammonia gas (V₂) = 50% of V₁ = 0.5VFinal pressure of ammonia gas (P₂) = ?The final pressure of the ammonia gas can be calculated as shown below:
[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\\\\\frac{4.6\ \times\ V}{252} = \frac{P_2\ \times\ 0.5V}{296}\\\\252\ \times\ P_2\ \times\ 0.5V = 4.6\ \times\ V\ \times\ 296\\\\P_2 = \frac{4.6\ \times\ V\ \times\ 296}{252\ \times\ 0.5V} \\\\P_2 = 11\ atm[/tex]
Thus, the final pressure is 11 atm
The phosphorylation of glucose to glucose 6-phosphate Group of answer choices is so strongly exergonic that it does not require a catalyst. is an exergonic reaction not coupled to any other reaction. is an endergonic reaction that takes place because it is coupled to the exergonic hydrolysis of ATP. is an exergonic reaction that is coupled to the endergonic hydrolysis of ATP.
Answer:
The phosphorylation of glucose to glucose-6-phosphate is endergonic reaction that is coupled to the exergonic hydrolysis of ATP.
Explanation:
In glycosis, the first reaction that takes place is the phosphorylation of glucose to glucose-6-phosphate by the enzyme hexokinase. This is an exergenic reaction. This is a coupled reaction in which phosphorylation of glucose is coupled to ATP hydrolysis. The free energy of ATP hydrolysis fuels glucose phosphorylation.
For most solids at room temperature, the specific heat is determined by oscillations of the atom cores in the lattice (each oscillating lattice site contributes 3kT of energy, by equipartition), as well as a contribution from the mobile electrons (if it's a metal). At room temperature the latter contribution is typically much smaller than the former, so we will ignore it here. In other words, you can reasonably estimate the specific heat simply by counting the number of atoms! Use this fact to estimate the specific heat of copper (atomic mass = 63.6), given that the specific heat of aluminum (atomic mass = 27.0) is 900 J/kg-K.
Answer:
The specific heat of copper is [tex]C= 392 J/kg\cdot ^o K[/tex]
Explanation:
From the question we are told that
The amount of energy contributed by each oscillating lattice site is [tex]E =3 kT[/tex]
The atomic mass of copper is [tex]M = 63.6 g/mol[/tex]
The atomic mass of aluminum is [tex]m_a = 27.0g/mol[/tex]
The specific heat of aluminum is [tex]c_a = 900 J/kg-K[/tex]
The objective of this solution is to obtain the specific heat of copper
Now specific heat can be defined as the heat required to raise the temperature of 1 kg of a substance by [tex]1 ^o K[/tex]
The general equation for specific heat is
[tex]C = \frac{dU}{dT}[/tex]
Where [tex]dT[/tex] is the change in temperature
[tex]dU[/tex] is the change in internal energy
The internal energy is mathematically evaluated as
[tex]U = 3nk_BT[/tex]
Where [tex]k_B[/tex] is the Boltzmann constant with a value of [tex]1.38*10^{-23} kg \cdot m^2 /s^2 \cdot ^o K[/tex]
T is the room temperature
n is the number of atoms in a substance
Generally number of atoms in mass of an element can be obtained using the mathematical operation
[tex]n = \frac{m}{M} * N_A[/tex]
Where [tex]N_A[/tex] is the Avogadro's number with a constant value of [tex]6.022*10^{23} / mol[/tex]
M is the atomic mass of the element
m actual mass of the element
So the number of atoms in 1 kg of copper is evaluated as
[tex]m = 1 kg = 1 kg * \frac{10000 g}{1kg } = 1000g[/tex]
The number of atom is
[tex]n = \frac{1000}{63.6} * (6.0*0^{23})[/tex]
[tex]= 9.46*10^{24} \ atoms[/tex]
Now substituting the equation for internal energy into the equation for specific heat
[tex]C = \frac{d}{dT} (3 n k_B T)[/tex]
[tex]=3nk_B[/tex]
Substituting values
[tex]C = 3 (9.46*10^{24} )(1.38 *10^{-23})[/tex]
[tex]C= 392 J/kg\cdot ^o K[/tex]
How many moles of gas are present in 1.13 L of gas at 2.09 atm and 291 K?
n = 0.0989 moles
Explanation:n = PV / RT
P = 2.09atm
V = 1.13L
R = 0.08206
T = 291K
Plug the numbers in the equation.
n = (2.09atm)(1.13L) / (0.08206)(291K)
n = 0.0989 moles
Final answer:
Using the ideal gas law PV = nRT, and rearranging for n (moles), we plug in the given values of pressure, volume, and temperature alongside the ideal gas constant to calculate the number of moles of gas present. There are approximately 0.0988 moles of gas present in 1.13 L of gas at [tex]\(2.09 \, \text{atm}\) and \(291 \, \text{K}\).[/tex]
Explanation:
To calculate the number of moles of gas present, we can use the ideal gas law equation:
[tex]\[ PV = nRT \][/tex]
First, we need to convert the given pressure to atm and the volume to liters if they are not already in those units.
Given:
- [tex]\( P = 2.09 \, \text{atm} \)[/tex]
-[tex]\( V = 1.13 \, \text{L} \[/tex]
- [tex]\( T = 291 \, \text{K} \)[/tex]
Now, we can rearrange the ideal gas law equation to solve for \( n \):
[tex]\[ n = \frac{PV}{RT} \][/tex]
[tex]\[ n = \frac{(2.09 \, \text{atm})(1.13 \, \text{L})}{(0.0821 \, \text{atm} \cdot \text{L} / \text{mol} \cdot \text{K})(291 \, \text{K})} \][/tex]
[tex]\[ n = \frac{2.3617}{23.9211} \][/tex]
[tex]\[ n \approx 0.0988 \, \text{mol} \][/tex]
Therefore, There are approximately 0.0988 moles of gas present in 1.13 L of gas at [tex]\(2.09 \, \text{atm}\) and \(291 \, \text{K}\).[/tex]
What is the daily cost of incandescent lamp
Mole ratios for a solving a stoichiometry problem are obtained from the
1. Molar Mass
2. Periodic Table
3. Balanced Equation
4. Total Mass of the Products
a sample of a radioactive isotope initially contains 20 x 10^10 atoms. After 16 days, 5 x 10^10 atoms remain. what is the half-life of the isotope?
A)12 days
B)20 days
C)10 days
D)16 days
E)8 days
The half-life of the radioactive isotope is 8 days because after 16 days (which is two half-lives), the sample is reduced from 20 x 10^10 atoms to 5 x 10^10 atoms.
Explanation:The question deals with the concept of the half-life of a radioactive isotope, which is the time required for half of the radioactive atoms in a sample to decay. According to the problem statement, an initial sample containing 20 x 1010 atoms decays to 5 x 1010 atoms in 16 days. To find the half-life (t1/2), we can use the fact that after one half-life, the number of atoms would be halved. Starting with 20 x 1010 atoms, after one half-life, there would be 10 x 1010 atoms, and after two half-lives, there would be 5 x 1010. This indicates that two half-lives have passed in 16 days, making the half-life 8 days.
The correct answer is 8 days, option E.
Buffer capacity is a measure of a buffer solution's resistance to changes in pH as strong acid or base is added. Suppose that you have 125 mL of a buffer that is 0.360 M in both acetic acid ( CH 3 COOH ) and its conjugate base ( CH 3 COO − ) . Calculate the maximum volume of 0.300 M HCl that can be added to the buffer before its buffering capacity is lost.
Answer:
The maximum volume is 122.73 mL
Explanation:
125 mL of butter that is 0.360 M in CH₃COOH and CH₃COO⁻.
pKa of acetic acid = 4.76
Using Henderson Hasselbalch equation.
pH =pKa + log [tex]\frac{CH_3COO^-}{CH_3COOH}[/tex]
= 4.76 + [tex]log \frac{0.3600}{0.3600}[/tex]
= 4.76
Assuming the pH of the buffer changes by a unit of 1 , then it will lose its' buffering capacity.
When HCl is added , it reacts with CH₃COO⁻ to give CH₃COOH.
However, [CH₃COOH] increases and the log term results in a negative value.
Let assume, the new pH is less than 4.76
Let say 3.76; calculating the concentration when the pH is 3.76; we have
3.76 = 4.76 + log [tex]\frac{CH_3COO^-}{CH_3COOH}[/tex]
log [tex]\frac{CH_3COO^-}{CH_3COOH}[/tex] = 4.76 - 3.76
log [tex]\frac{CH_3COO^-}{CH_3COOH}[/tex] = 1.00
[tex]\frac{CH_3COO^-}{CH_3COOH}[/tex] = 0.1
Let number of moles of acid be x (i.e change in moles be x); &
moles of acetic acid and conjugate base present be = Molarity × Volume
= 0.360 M × 125 mL
= 45 mmol
replacing initial concentrations and change in the above expression; we have;
[tex]\frac{[CH_3COO^-]}{[CH_3COOH]} = 0.1[/tex]
[tex]\frac{45-x}{45+x} =0.1[/tex]
0.1(45 + x) = 45 - x
4.5 + 0.1 x = 45 -x
0.1 x + x = 45 -4.5
1.1 x = 40.5
x = 40.5/1.1
x = 36.82
So, moles of acid added = 36.82 mmol
Molarity = 0.300 M
So, volume of acid = [tex]\frac{moles}{molarity }[/tex] = [tex]\frac{36.82 \ mmol}{0.300}[/tex]
= 122.73 mL
2. Calculate the mass of 3.47x1023 gold atoms.
3.47 x [tex]10^{23}[/tex] atoms of gold have mass of 113.44 grams.
Explanation:
Data given:
number of atoms of gold = 3.47 x [tex]10^{23}[/tex]
mass of the gold in given number of atoms = ?
atomic mass of gold =196.96 grams/mole
Avagadro's number = 6.022 X [tex]10^{23}[/tex]
from the relation,
1 mole of element contains 6.022 x [tex]10^{23}[/tex] atoms.
so no of moles of gold given = [tex]\frac{3.47 X 10^{23} }{6.022 X 10^{23} }[/tex]
0.57 moles of gold.
from the relation:
number of moles = [tex]\frac{mass}{atomic mass of 1 mole}[/tex]
rearranging the equation,
mass = number of moles x atomic mass
mass = 0.57 x 196.96
mass = 113.44 grams
thus, 3.47 x [tex]10^{23}[/tex] atoms of gold have mass of 113.44 grams
The mass of 3.47 x 10^23 gold atoms calculated using Avogadro's number and the molar mass of gold equals approximately 113.52 grams. To calculate this, the given number of atoms was first converted into moles and the number of moles were then multiplied by the molar mass.
Explanation:In chemistry, to calculate the mass of a certain number of atoms, we use the concept of a mole and Avogadro's number (6.022 x 1023). For gold atoms, the molar mass is 197 g/mol. Given that the number of gold atoms is 3.47 x 1023, we can use the relationship that 1 mole of gold atoms = 6.022 x 1023 gold atoms. Therefore, the mass of 3.47 x 1023 gold atoms can be calculated as follows:
First, we convert the number of atoms to moles using Avogadro's number: 3.47 x 1023 atoms / 6.022 x 1023 atoms/mol = 0.576 moles of gold.
Then, we multiply this moles by the molar mass of gold to get the mass in grams: 0.576 moles * 197 g/mol = 113.52 g.
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How many moles are in 12.0 grams of O2
Answer:
Moles = 0.375
Explanation:
Moles= m/M
= 12/32 = 0.375mol
There are approximately 0.375 moles of O₂ in 12.0 grams of O₂.
To find the number of moles in a given mass of a substance, one can use the formula:
[tex]\[ \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \][/tex]
The molar mass of oxygen (O₂) is approximately 32.0 g/mol, since one oxygen atom has an atomic mass of approximately 16.0 g/mol and O₂ has two oxygen atoms.
Given the mass of O₂ is 12.0 g, we can calculate the moles as follows:
[tex]\[ \text{moles of \ } {O_2} = \frac{12.0 \text{ g}}{32.0 \text{ g/mol}} \][/tex]
[tex]\[ \text{moles of} \ {O_2} = 0.375 \text{ mol} \][/tex]
Therefore, there are 0.375 moles of O₂ in 12.0 grams of O₂.
Calculate the amount of energy in kilojoules needed to change 369 gg of water ice at −−10 ∘C∘C to steam at 125 ∘C∘C. The following constants may be useful: Cm (ice)=36.57 J/(mol⋅∘C)Cm (ice)=36.57 J/(mol⋅∘C) Cm (water)=75.40 J/(mol⋅∘C)Cm (water)=75.40 J/(mol⋅∘C) Cm (steam)=36.04 J/(mol⋅∘C)Cm (steam)=36.04 J/(mol⋅∘C) ΔHfus=+6.01 kJ/molΔHfus=+6.01 kJ/mol ΔHvap=+40.67 kJ/molΔHvap=+40.67 kJ/mol Express your answer with the appropriate units.
Answer:
We need 1136.4 kJ energy
Explanation:
Step 1: Data given
The mass of water = 369 grams
the initial temperature = -10°C
The finaltemperature = 125 °C
Cm (ice)=36.57 J/(mol⋅∘C)
Cm (water)=75.40 J/(mol⋅∘C)
Cm (steam)=36.04 J/(mol⋅∘C)
ΔHfus=+6010 J/mol
ΔHvap=+40670 J/mol
Step 2: :Calculate the energy needed to heat ice from -10 °C to 0°C
Q = n*C*ΔT
⇒Q = the energy needed to heat ice to 0°C
⇒with n = the moles of ice = 369 grams / 18.02 g/mol = 20.48 moles
⇒with C = the specific heat of ice = 36.57 / J/mol°C
⇒ ΔT = the change of temperature = 10°C
Q = 20.48 moles * 36.57 J/mol°C * 10°C
Q = 7489.5 J = 7.490 kJ
Step 3: calculate the energy needed to melt ice to water at 0°C
Q = n* ΔHfus
Q = 20.48 moles * 6010 J/mol
Q = 123084.8 J = 123.08 kJ
Step 4: Calculate energy needed to heat water from 0°C to 100 °C
Q = n*C*ΔT
⇒Q = the energy needed to heat water from 0°C to 100 °C
⇒with n = the moles of water = 369 grams / 18.02 g/mol = 20.48 moles
⇒with C = the specific heat of water =75.40 J/(mol⋅∘C)
⇒ ΔT = the change of temperature = 100°C
Q = 20.48 moles * 75.40 J/mol°C* 100°C
Q = 154419.2 J = 154.419 kJ
Step 5: Calculate energy needed to vapourize water to steam at 100°C
Q = 20.48 moles * 40670 J/mol
Q = 832921.6 J = 832.922 kJ
Step 6: Calculate the energy to heat steam from 100 °C to 125 °C
⇒Q = the energy needed to heat steam from 100 °C to 125 °C
⇒with n = the moles of steam = 369 grams / 18.02 g/mol = 20.48 moles
⇒with C = the specific heat of steam = 36.04 J/(mol⋅∘C)
⇒ ΔT = the change of temperature = 25 °C
Q = 20.48 moles * 36.04 J/mol*°C * 25°C
Q = 18452.5 J = 18.453 kJ
Step 7: Calculate total energy needed
Q = 7489.5 J + 123084.8 J + 154419.2 J + 832921.6 J + 18452.5 J
Q = 1136367.6 J
Q = 1136.4 kJ
We need 1136.4 kJ energy
The amount of energy needed is [tex]\( \\1120 \text{ kJ}} \).[/tex]
To calculate the amount of energy required to change 369 g of water from ice at[tex]\(-10^\circ \text{C}\)[/tex] to steam at [tex]\(125^\circ \text{C}\),[/tex] we need to consider the different phases of water and the energy required for each phase change and temperature change.
Let's break down the calculation step by step.
1. Energy to heat ice from [tex]\(-10^\circ \text{C}\) to \(0^\circ \text{C}\).[/tex]
First, calculate the energy required to heat the ice initially at[tex]\(-10^\circ \text{C}\)[/tex] to its melting point [tex]( \(0^\circ \text{C}\) ).[/tex]
Specific heat capacity of ice [tex](\(C_m\)): \(36.57 \text{ J/(mol} \cdot \text{°C)}\).[/tex]
Molar mass of water (H₂O). [tex]\(18.015 \text{ g/mol}\).[/tex]
Amount of ice in moles:
[tex]\[ \text{moles of ice} = \frac{369 \text{ g}}{18.015 \text{ g/mol}} \approx 20.49 \text{ mol} \][/tex]
Energy required to heat ice.
[tex]\[ Q_1 = n \times C_m \times \Delta T = 20.49 \text{ mol} \times 36.57 \text{ J/(mol} \cdot \text{°C)} \times (0^\circ \text{C} - (-10^\circ \text{C})) = 7485.39 \text{ J} \][/tex]
[tex]\( Q_1 = 7.48539 \text{ kJ} \)[/tex]
2. Energy to melt the ice at [tex]\(0^\circ \text{C}\).[/tex]
Latent heat of fusion [tex](\(\Delta H_{\text{fus}}\)) for ice: \(6.01 \text{ kJ/mol}\)[/tex].
Energy required to melt ice:
[tex]\[ Q_2 = \text{moles of ice} \times \Delta H_{\text{fus}} = 20.49 \text{ mol} \times 6.01 \text{ kJ/mol} = 123.05 \text{ kJ} \][/tex]
[tex]\( Q_2 = 123.05 \text{ kJ} \)[/tex]
3.Energy to heat water from[tex]\(0^\circ \text{C}\) to \(100^\circ \text{C}\)[/tex].
Specific heat capacity of water[tex](\(C_m\))[/tex]: [tex]\(75.40 \text{ J/(mol} \cdot \text{°C)}\)[/tex].
Energy required to heat water:
[tex]\[ Q_3 = \text{moles of water} \times C_m \times \Delta T = 20.49 \text{ mol} \times 75.40 \text{ J/(mol} \cdot \text{°C)} \times (100^\circ \text{C} - 0^\circ \text{C}) = 155,297.80 \text{ J} \][/tex] [tex]\( Q_3 = 155.2978 \text{ kJ} \)[/tex]
4. Energy to vaporize water at [tex]\(100^\circ \text{C}\)[/tex]:
Latent heat of vaporization [tex](\(\Delta H_{\text{vap}}\)) for water: \(40.67 \text{ kJ/mol}\).[/tex]
Energy required to vaporize water:
[tex]\[ Q_4 = \text{moles of water} \times \Delta H_{\text{vap}} = 20.49 \text{ mol} \times 40.67 \text{ kJ/mol} = 833.34 \text{ kJ} \][/tex]
[tex]\( Q_4 = 833.34 \text{ kJ} \)[/tex]
5. Energy to heat steam from [tex]\(100^\circ \text{C}\) to \(125^\circ \text{C}\):[/tex]
Specific heat capacity of steam [tex](\(C_m\)): \(36.04 \text{ J/(mol} \cdot \text{°C)}\).[/tex]
Energy required to heat steam:
[tex]\[ Q_5 = \text{moles of steam} \times C_m \times \Delta T = 20.49 \text{ mol} \times 36.04 \text{ J/(mol} \cdot \text{°C)} \times (125^\circ \text{C} - 100^\circ \text{C}) = 1831.79 \text{ J} \][/tex]
[tex]\( Q_5 = 1.83179 \text{ kJ} \)[/tex]
6. Total energy required.
Summing up all the energy contributions.[tex]\[ Q_{\text{total}} = Q_1 + Q_2 + Q_3 + Q_4 + Q_5 = 7.48539 \text{ kJ} + 123.05 \text{ kJ} + 155.2978 \text{ kJ} + 833.34 \text{ kJ} + 1.83179 \text{ kJ} = 1120.00598 \text{ kJ} \][/tex]
A sealed copper container with a mass of 0.3 kg is filled with 1.5 mole of helium gas. Initially, the helium gas is at a temperature of 124 oC and the copper container is at 20 oC. The helium-copper system is thermally isolated. Note that the specific heat of copper is 386 J/(kgK) and the molar specific heat of helium is 12.5 J/(molK). Find the equilibrium temperature of the system.
Answer:
[tex]T = 34.493\,^{\textdegree}C[/tex]
Explanation:
The equilibrium temperature of the gas-container system is:
[tex]Q_{Cu} = -Q_{g}[/tex]
[tex](0.3\,kg)\cdot\left(386\,\frac{J}{kg\cdot ^{\textdegree}C} \right)\cdot (T-20\,^{\textdegree}C) = (1.5\,mol)\cdot \left(12.5\,\frac{J}{mol\cdot ^{\textdegree}C} \right)\cdot (124\,^{\textdegree}C-T)[/tex]
[tex]\left(115.8\,\frac{J}{^{\textdegree}C} \right)\cdot (T-20\,^{\textdegree}C) = \left(18.75\,\frac{J}{^{\textdegree}C} \right)\cdot (124\,^{\textdegree}C-T)[/tex]
[tex]134.55\cdot T = 4641[/tex]
[tex]T = 34.493\,^{\textdegree}C[/tex]
Consider the following initial rate data (at 273 K) for the decomposition of a substrate (substrate 1) which decomposes to product 1 and product 2: [Substrate 1] (M) Initial Rate (M/s) 0.4 0.183 0.8 0.183 2 0.183 Determine the half-life for the decomposition of substrate 1 when the initial concentration of the substrate is 2.77 M.
Answer:
15.1 seconds is the half life of the reaction when concentration of the substrate is 2.77 M.
Explanation:
A → B + C
The rate law of the reaction will be :
[tex]R=k[A]^x[/tex]
Initial rate of the reaction when concentration of the substrate was 0.4 M:
[tex]0.183 M/s=k[0.4 M]^x[/tex]..[1]
Initial rate of the reaction when concentration of the substrate was 0.8 M:
[tex]0.183 M/s=k[0.8 M]^x[/tex]...[2]
[1] ÷ [2] :
[tex]\frac{0.183 M/s}{0.183 M/s}=\frac{k[0.4 M]^x}{k[0.8 M]^x}[/tex]
x = 0
The order of the reaction is zero.
For the value of rate constant ,k:
[tex]0.183 M/s=k[0.4 M]^x[/tex]..[1]
x = 0
[tex]0.183 M/s=k[0.4 M]^0[/tex]
k= 0.183 M/s
The half life of the zero order kinetics is given by :
[tex]t_{1/2}=\frac{[A_o]}{2k}[/tex]
Where:
[tex][A_o][/tex] = Initial concentration of A
k = Rate constant of the reaction
So, the half-life for the decomposition of substrate 1 when the initial concentration of the substrate is 2.77 M:
[tex]t_{1/2}=\frac{2.77 M}{0.183 M/s}=15.1 s[/tex]
15.1 seconds is the half life of the reaction when concentration of the substrate is 2.77 M.
The half-life for the decomposition of substrate 1 when the initial concentration is 2.77 M is approximately 18 minutes.
Explanation:The half-life of a reaction is the time required for one-half of a given amount of reactant to be consumed. In this case, we have initial rate data for the decomposition of substrate 1 with varying concentrations. To determine the half-life of substrate 1, we need to find the time it takes for the concentration to decrease to half its initial value.
Based on the given data, we can see that the initial rate of the reaction is constant at 0.183 M/s for different initial concentrations of substrate 1 (0.4 M, 0.8 M, and 2 M). Since the initial concentration of substrate 1 is given as 2.77 M, it will take the same amount of time as the other initial concentrations to reach half its initial concentration.
Therefore, the half-life for the decomposition of substrate 1 when the initial concentration of the substrate is 2.77 M is the same as the half-life observed when the initial concentration is 0.4 M, 0.8 M, or 2 M, which is approximately 18 minutes.
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Treatment of ethyl acetoacetate with NaOEt (2 equiv) and BrCH2CH2Br forms compound X. This reaction is the first step in the synthesis of illudin-S, an antitumor substance isolated from the jack-o'-lantern, a saffron-colored mushroom. What is the structure of X
Answer:
See explanation below
Explanation:
In this reaction we have the ethyl acetoacetate which is reacting with 2 eq of sodium etoxide. The sodium etoxide is a base and it usually behaves as a nucleophyle of many reactions. Therefore, it will atract all the acidics protons in a molecule.
In the case of the ethyl acetoacetate, the protons that are in the methylene group (CH3 - CO - CH2 - COOCH2CH3) are the more acidic protons, therefore the etoxide will substract these protons instead of the protons of the methyl groups. This is because those hydrogens (in the methylene group) are between two carbonile groups, which make them more available and acidic for any reaction. As we have 2 equivalents of etoxide, means that it will substract both of the hydrogen atoms there, and then, reacts with the Br - CH2CH2 - Br and form a product of an aldolic condensation.
The mechanism of this reaction to reach X is shown in the attached picture.
At a certain temperature the rate of this reaction is first order in HI with a rate constant of :7.21/s2HI(g) = H2(g) + I2(g)Suppose a vessel contains HI at a concentration of 0.440M. Calculate the concentration of HI in the vessel 0.210 seconds later. You may assume no other reaction is important.
Answer:
C HI = 0.0968 M
Explanation:
2HI(g) → H2(g) + I2(g)- ra = K(Ca)∧α = - δCa/δt∴ a: HI(g)
∴ order (α) = 1
∴ rate constant (K) = 7.21/s
∴ Initial concenttration HI(g) (Cao) = 0.440 M
∴ t = 0.210 s ⇒ Ca = ?
⇒ - δCa/δt = KCa
⇒ - ∫δCa/Ca = K*∫δt
⇒ Ln(Cao/Ca) = K*t
⇒ Cao/Ca = e∧(K*t)
⇒ Cao/e∧(K*t) = Ca
⇒ Ca = (0.440)/e∧((7.21*0,21))
⇒ Ca = 0.440/4.54533
⇒ Ca = 0.0968 M
6. Air at 300°C and 130kPa ows through a horizontal 7-cm ID pipe at a velocity of 42.0m/s. (a) Calculate _ Ek W, assuming ideal-gas behavior. (b) If the air is heated to 400°C at constant pressure, what is Δ _ Ek _ Ek 400°C _ Ek 300°C? (c) Why would it be incorrect to say that the rate of transfer of heat to the gas in Part (b) must equal the rate of change of kinetic energy?
Answer:
Check the explanation
Explanation:
kindly check the attached image below to see the answer to question A and B
(c) The energy balance equation if the pressure is constant (ΔE=0) is,
ΔE= ΔU + Δ[tex]E_{k}[/tex]
From the above relation, it is clear that some heat energy used to raises the temperature of the air.
Hence, the internal energy is not equal to zero. Therefore, the rate of transfer of heat to air is not equal to the rate of thane in kinetic energy of the air.
That is ΔE ≠ Δ[tex]E_{k}[/tex]
Answer:
a) [tex]\delta E_k[/tex] = 112.164 W
b) [tex]\delta E_k[/tex] = 42.567 W
c) From what we've explained in part b; increase in temperature of the system is caused by rate of heat transfer . Therefore, not all heat is used to increase kinetic energy. Hence, since not all the heat is used to increase the kinetic energy . It is not valid and it is incorrect to say that rate of heat transfer is equal to the change in kinetic energy.
Explanation:
The given data include:
Inlet diameter [tex](d_1)[/tex] = 7 cm = 0.7 m
Inlet velocity [tex](v_1)[/tex] = 42 m/s
Inlet pressure [tex](P_1)[/tex] = 130 KPa
Inlet temoerature [tex](T_1)[/tex] = 300°C = (300 + 273.15) = 573.15 K
a) Assuming Ideal gas behaviour
Inlet Volumetric flowrate [tex](V_1)[/tex] = Inlet velocity [tex](v_1)[/tex] × area of the tube
[tex]= v_1*(\frac{\pi}{4})0.07^2\\\\= 0.161635 \ m^3/s[/tex]
Using Ideal gas law at Inlet 1
[tex]P_1V_1 =nRT_1[/tex]
where ; n = molar flow rate of steam
making n the subject of the formula; we have:
[tex]n = \frac{P_1V_1}{RT_1}[/tex]
[tex]n = \frac{130*42}{8.314*573.15}[/tex]
[tex]n= 4.4096*10^{-3} \ Kmol/s[/tex]
Moleular weight of air = 28.84 g/mol
The mass flow rate = molar flowrate × molecular weight of air
[tex]= 4.4096*10^{-3} \ * 28.84\\= 0.12717 \ kg/s[/tex]
Finally: the kinetic energy at Inlet [tex](\delta E_k) = \frac{1}{2}*mass \ flowrate *v_1^2[/tex]
= [tex]0.5*0.12717*42^2[/tex]
= 112.164 W
b) If the air is heated to 400°C;
Then temperature at 400°C = (400 + 273.15)K = 673.15 K
Thee pressure is also said to be constant ;
i.e [tex]P_1= P_2[/tex] = 130 KPa
Therefore; the mass flow rate is also the same ; so as the molar flow rate:
Thus; [tex]n= 4.4096*10^{-3} \ Kmol/s[/tex]
Using Ideal gas law at Inlet 2
[tex]P_2V_2 = nRT_2[/tex]
making [tex]V_2[/tex] the subject of the formula; we have:
[tex]V_2 = \frac{nRT_2}{P_2}[/tex]
[tex]V_2 = \frac{4.4906*10^{-3}*8.314*673.15}{130}[/tex]
[tex]V_2 = 0.189836 \ m^3/s[/tex]
Assuming that the diameter is constant
[tex]d_1 = d_2 = 0.07 \ cm[/tex]
Now; the velocity at outlet = [tex]\frac{V_2}{\frac{\pi}{4}d_2^2}[/tex]
= [tex]\frac{0.0189836}{\frac{\pi}{4}(0.07)^2}[/tex]
= 49.33 m/s
Change in kinetic energy [tex]\delta E_k[/tex] = [tex]\frac{1}{2}*mass \ flowrate * \delta V[/tex]
= [tex]0.5*0.12717 *(49.33^2 -42^2)[/tex]
= 42.567 W
c).
From what we've explained in part b; increase in temperature of the system is caused by rate of heat transfer . Therefore, not all heat is used to increase kinetic energy. Hence, since not all the heat is used to increase the kinetic energy . It is not valid and it is incorrect to say that rate of heat transfer is equal to the change in kinetic energy.
Find the pHpH of a solution prepared from 1.0 LL of a 0.15 MM solution of Ba(OH)2Ba(OH)2 and excess Zn(OH)2(s)Zn(OH)2(s). The KspKsp of Zn(OH)2Zn(OH)2 is 3×10−153×10−15 and the KfKf of Zn(OH)2−4Zn(OH)42− is 2×10152×1015.
Answer:
pH = 13.09
Explanation:
Zn(OH)2 --> Zn+2 + 2OH- Ksp = 3X10^-15
Zn+2 + 4OH- --> Zn(OH)4-2 Kf = 2X10^15
K = Ksp X Kf
= 3*2*10^-15 * 10^15
= 6
Concentration of OH⁻ = 2[Ba(OH)₂] = 2 * 0.15 = 3 M
Zn(OH)₂ + 2OH⁻(aq) --> Zn(OH)₄²⁻(aq)
Initial: 0 0.3 0
Change: -2x +x
Equilibrium: 0.3 - 2x x
K = Zn(OH)₄²⁻/[OH⁻]²
6 = x/(0.3 - 2x)²
6 = x/(0.3 -2x)(0.3 -2x)
6(0.09 -1.2x + 4x²) = x
0.54 - 7.2x + 24x² = x
24x² - 8.2x + 0.54 = 0
Upon solving as quadratic equation, we obtain;
x = 0.089
Therefore,
Concentration of (OH⁻) = 0.3 - 2x
= 0.3 -(2*0.089)
= 0.122
pOH = -log[OH⁻]
= -log 0.122
= 0.91
pH = 14-0.91
= 13.09
In this chemical Formula for Ammonia, the Subscripts indicate what?
A. When added together, we know that Nitrogen and Hydrogen combined create four molecules of Ammonia
B. There are two atoms of Nitrogen and two atoms of Hydrogen combined to make Ammonia.
C. There are two molecules of Nitrogen and two molecules of Hydrogen combined to make Ammonia
D. The compound N2 and the compound H2 make Ammonia
Answer:
B
Explanation:
B. There are two atoms of Nitrogen and two atoms of Hydrogen combined to make Ammonia.
Provide a preliminary design o fa n air stripping column t o remove toluene from ground water. Levels of toluene range from 0.1 to 2.1 mglL and this must be reduced to 50 f.LglL. A hydrogeologic study of the area indicates that a flow rate of 1 1 0 gal/min is required to ensure that contamination not spread. Laboratory investigations have determined the overall transfer constant, KLa = 0.020 s-' . Use a column diameter of 2.0 feet and an air-to-water ratio of 15. Specifically determine: liquid loading rate, stripping factor, and height of the tower. Provide a sketch of the unit indicating all required appurtenances.
Answer:
Explanation:
Find attach the solution
Nitrate salts (NO3), when heated, can produce nitrites (NO2) plus oxygen (O2). A sample of potassium nitrate is heated, and the 02 gas produced is collected in a 700 mL flask. The pressure of the gas in the flask is 2.7 atm, and the temperature is recorded to be 329 K. The value of R= 0.0821 atm L/(mol K) How many moles of O2 gas were produced? moles After a few hours, the 700 mL flask cools to a temperature of 293K. What is the new pressure due to the O2 gas?
Answer: a) 0.070 moles of oxygen were produced.
b) New pressure due to the oxygen gas is 2.4 atm
Explanation:
According to ideal gas equation:
[tex]PV=nRT[/tex]
P = pressure of gas = 2.7 atm
V = Volume of gas = 700 ml = 0.7 L
n = number of moles = ?
R = gas constant =[tex]0.0821Latm/Kmol[/tex]
T =temperature = 329 K
[tex]n=\frac{PV}{RT}[/tex]
[tex]n=\frac{2.7atm\times 0.7L}{0.0821 L atm/K mol\times 329K}=0.070moles[/tex]
Thus 0.070 moles of oxygen were produced.
When the 700 mL flask cools to a temperature of 293K.
[tex]PV=nRT[/tex]
[tex]P=\frac{nRT}{V}[/tex]
[tex]P=\frac{0.070\times 0.0821\times 293}{0.7}[/tex]
[tex]P=2.4atm[/tex]
The new pressure due to the oxygen gas is 2.4 atm
The pH at 25 °C of an aqueous solution of the sodium salt of p-monochlorophenol (NaC6H4ClO) is 11.05. Calculate the concentration of C6H4ClO- in this solution, in moles per liter. Ka for HC6H4ClO is equal to 6.6×10-10.
Answer:
Approximately [tex]8.3 \times 10^{-2}\; \rm mol \cdot L^{-1}[/tex].
Explanation:
The [tex]K_a[/tex] in this question refers the dissociation equilibrium of [tex]\rm HC_6H_4ClO[/tex] as an acid:
[tex]\rm HC_6H_4ClO\, (aq) \rightleftharpoons H^{+} \, (aq) + C_6H_4ClO^{-}\, (aq)[/tex].
[tex]\displaystyle K_a\left(\mathrm{HC_6H_4ClO}\right) = \frac{\left[\mathrm{H^{+}}\right] \cdot \left[\mathrm{C_6H_4ClO^{-}}\right]}{\left[\mathrm{HC_6H_4ClO}\right]}[/tex].
However, the question also states that the solution here has a [tex]\rm pH[/tex] of [tex]11.05[/tex], which means that this solution is basic. In basic solutions at [tex]\rm 25\;^\circ C[/tex], the concentration of [tex]\rm H^{+}[/tex] ions is considerably small (typically less than [tex]10^{-7}\;\rm mol \cdot L^{-1}[/tex].) Therefore, it is likely not very appropriate to use an equilibrium involving the concentration of [tex]\rm H^{+}[/tex] ions.
Here's the workaround: note that [tex]\rm C_6H_4ClO^{-}\, (aq)[/tex] is the conjugate base of the weak acid [tex]\rm HC_6H_4ClO\, (aq)[/tex]. Therefore, when [tex]\rm C_6H_4ClO^{-}\, (aq)[/tex] dissociates in water as a base, its [tex]K_b[/tex] would be equal to [tex]\displaystyle \frac{K_w}{K_a} \approx \frac{10^{-14}}{K_a}[/tex]. ([tex]K_w[/tex] is the self-ionization constant of water. [tex]K_w \approx 10^{-14}[/tex] at [tex]\rm 25\;^\circ C[/tex].)
In other words,
[tex]\begin{aligned} & K_b\left(\mathrm{C_6H_4ClO^{-}}\right) \\ &= \frac{K_w}{K_a\left(\mathrm{HC_6H_4ClO}\right)} \\ &\approx \frac{10^{-14}}{6.6 \times 10^{-10}} \\ & \approx 1.51515 \times 10^{-5}\end{aligned}[/tex].
And that [tex]K_b[/tex] value corresponds to the equilibrium:
[tex]\rm C_6H_4ClO^{-}\, (aq) + H_2O\, (l) \rightleftharpoons HC_6H_4ClO\, (aq) + OH^{-}\, (aq)[/tex].
[tex]\displaystyle K_b\left(\mathrm{C_6H_4ClO^{-}}\right) = \frac{\left[\mathrm{HC_6H_4ClO}\right]\cdot \left[\mathrm{OH^{-}}\right]}{\left[\mathrm{C_6H_4ClO^{-}}\right]}[/tex].
The value of [tex]K_b[/tex] has already been found.
The [tex]\rm OH^{-}[/tex] concentration of this solution can be found from its [tex]\rm pH[/tex] value:
[tex]\begin{aligned}& \left[\mathrm{OH^{-}}\right] \\ &= \frac{K_w}{\left[\mathrm{H}^{+}\right]} \\ & = \frac{K_w}{10^{-\mathrm{pH}}} \\ &\approx \frac{10^{-14}}{10^{-11.05}} \\ &\approx 1.1220 \times 10^{-3}\; \rm mol\cdot L^{-1} \end{aligned}[/tex].
To determine the concentration of [tex]\left[\mathrm{HC_6H_4ClO}\right][/tex], consider the following table:
[tex]\begin{array}{cccccc}\textbf{R} &\rm C_6H_4ClO^{-}\, (aq) & \rm + H_2O\, (l) \rightleftharpoons & \rm HC_6H_4ClO\, (aq) & + & \rm OH^{-}\, (aq) \\ \textbf{I} & (?) & \\ \textbf{C} & -x & & + x& & +x \\ \textbf{E} & (?) - x & & x & & x\end{array}[/tex]
Before hydrolysis, the concentration of both [tex]\mathrm{HC_6H_4ClO}[/tex] and [tex]\rm OH^{-}[/tex] are approximately zero. Refer to the chemical equation. The coefficient of [tex]\mathrm{HC_6H_4ClO}[/tex] and [tex]\mathrm{HC_6H_4ClO}[/tex] are the same. As a result, this equilibrium will produce [tex]\rm OH^{-}[/tex] and [tex]\mathrm{HC_6H_4ClO}[/tex] at the exact same rate. Therefore, at equilibrium, [tex]\left[\mathrm{HC_6H_4ClO}\right] \approx \left[\mathrm{OH^{-}}\right] \approx 1.1220 \times 10^{-3}\; \rm mol\cdot L^{-1}[/tex].
Calculate the equilibrium concentration of [tex]\left[\mathrm{C_6H_4ClO^{-}}\right][/tex] from [tex]K_b\left(\mathrm{C_6H_4ClO^{-}}\right)[/tex]:
[tex]\begin{aligned} & \left[\mathrm{C_6H_4ClO^{-}}\right] \\ &= \frac{\left[\mathrm{HC_6H_4ClO}\right]\cdot \left[\mathrm{OH^{-}}\right]}{K_b}\\&\approx \frac{\left(1.1220 \times 10^{-3}\right) \times \left(1.1220 \times 10^{-3}\right)}{1.51515\times 10^{-5}}\; \rm mol \cdot L^{-1} \\ &\approx 8.3 \times 10^{-2}\; \rm mol \cdot L^{-1}\end{aligned}[/tex].