The question involves calculation of equilibrium concentrations in a complex formation reaction in a solution, where the reaction is between a metal ion M²⁺ and cyanide ions CN⁻ to form a complex [M(CN)4]²⁻. The stated formation constant for the reaction allows formulating an equation for calculating the equilibrium concentration of M²⁺.
Explanation:To determine the concentration of M²⁺ ions at equilibrium, we need to consider the process of complex formation here, which is M²⁺ + 4CN⁻ → [M(CN)4]²⁻. The formation constant for this reaction is given as 7.7 x 10¹⁶. Given a 0.150 mole quantity of M(NO3)2 is added to a liter of 0.820 M NaCN solution, initially we will have [M²⁺] = 0.150 M and [CN⁻] = 0.820 M.
As the reaction proceeds to equilibrium, [M²⁺] drops by x amount reacting with 4x amount of [CN⁻]. At equilibrium, [M²⁺] = 0.150 - x, [CN-] = 0.820 - 4x and [M(CN)4]²⁻ = x. Applying the formation constant: 7.7 x 10¹⁶ = [M(CN)4]²⁻ / ([M²⁺][CN-]⁴) = x / (0.150 - x)(0.820 - 4x)⁴
This is a difficult equation to solve directly, but if we make the assumption that x, which relates to the equilibrium concentration of [M²⁺], is much less than initial concentrations of M²⁺ and CN⁻, this simplifies the equation and allows determination of the equilibrium concentration of M²⁺.
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To find the concentration of M2+ ions at equilibrium, use the formation constant and initial concentrations of M(NO3)2 and NaCN.
Explanation:The concentration of M2+ ions at equilibrium can be determined using the formation constant and the initial concentrations of M(NO3)2 and NaCN. First, calculate the initial concentration of M2+ ions by multiplying the concentration of M(NO3)2 (0.150 M) by its stoichiometric coefficient (2). Then, use the formation constant (7.70 × 10^16) to calculate the concentration of [M(CN)4]2- ions at equilibrium. Since the stoichiometric coefficient of M2+ ions is also 1, the concentration of M2+ ions at equilibrium will be equal to the concentration of [M(CN)4]2- ions.
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Even though we have known of many nuclear energy technologies for about one century some of these technologies have not yet been fully developed. Obtaining energy from thorium remains a challenge for many reasons. In a reactor, thorium is bombarded with neutrons and becomes U-233 and then U-234, which can undergo fission and produce energy. Complete the following U-234 fission reaction. Hint: Do not ignore the amount of energy given off by the reaction! 234.0 tlu . i37 154Xe + 3?Sr + 3 1.0087n + 180 MeV
Answer: The nuclear fission reaction of U-234.043 is written below.
Explanation:
In a nuclear reaction, the total mass and total atomic number remains the same.
For the given fission reaction:
[tex]^{234.043}_{92}\textrm{U}\rightarrow ^{137.159}_{54}\textrm{Xe}+^{A}_{38}\textrm{Sr}+3^{1.0087}_0\textrm{n}+180MeV[/tex]
To calculate A:
Total mass on reactant side = total mass on product side
234.043 = 137.159 + A + 3(1.0087)
A = 93.858
Now, the chemical equation becomes:
[tex]^{234.043}_{92}\textrm{U}\rightarrow ^{137.159}_{54}\textrm{Xe}+^{93.858}_{38}\textrm{Sr}+3^{1.0087}_0\textrm{n}+180MeV[/tex]
Hence, the nuclear fission reaction of U-234.043 is written above.
A system at equilibrium has two aqueous products and two aqueous reactants. An additional amount of one of the products is added to the system. After the addition, which of the following will change?Select the correct answer below:a. the amount of the added productb. the amount of the other productc. the amounts of the reactantsd. all of the above
If an extra amount of a product is added to a system at equilibrium, the system will shift towards the reactants to counteract this change. This shift can cause changes in the amounts of all substances involved, including the added product, the other product, and the reactants.
Explanation:When an extra amount of a product is added to a system at equilibrium, the system shifts to counteract this change and re-establish equilibrium, as described by Le Châtelier's principle. Specifically, the system will shift in the direction that reduces the excess amount of the product, i.e., towards the reactants. With this shift, the quantities of all substances in the system can change.
Therefore, the appropriate answer is d. all of the above. Adding more of one product will indeed increase the amount of that product initially (a), but the system will shift to remove the added product by moving towards the reactants. In this process, the amount of the other product may also change as it is consumed or generated (b), and the amounts of reactants will also adjust as they are generated or consumed (c).
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Calculate the relative atomic mass of this sample of Germanium, giving your answer to two decimal places. ......................................................................................................................... ......................................................................................................................... ......................................................................................................................... .........................................................................................................................
Answer:
Germanium is an element in the same group with Carbon and Silicon. The atomic number is 32. The relative atomic mass is usually measured with the Sample of an isotope. In this case Germanium has a relative atomic mass of 72.63
Answer:
The relative atomic mass of the sample of Germanium is 72.89 u
Explanation:
For a sample of germanium we have the following compositions:
m70Ge = 70 u
%70Ge = 22.6%
m72Ge = 72 u
%72Ge = 25.45%
m74Ge = 74 u
%74Ge = 36.73%
m76Ge = 76 u
%76Ge = 15.22%
To calculate the atomic mass of Germanium we must add all the atomic masses of its isotopes multiplied by its percentage of abundance and all this divided by 100, in this way:
mGe = [(m70Gex%70Ge)+(m72Gex%72Ge)+(m74Gex%74Ge)+(m76Gex%76Ge)]/100
replacing values:
mGe = [(70x22.6)+(72x25.45)+(74x36.73)+(76x15.22)]/100 = 72.89 u
Air is about 78.0% nitrogen molecules and 21.0% oxygen molecules. Several other gases make up the remaining 1% of air molecules. What is the partial pressure of the other gases in air (excluding oxygen and nitrogen) at an atmospheric pressure of 0.90 atm
Answer:
The partial pressure of the other gases is 0.009 atm
Explanation:
Step 1: Data given
Air is about 78.0% nitrogen molecules and 21.0% oxygen molecules and 1% of other gases.
The atmospheric pressure = 0.90 atm
Step 2: Calculate mol fraction
If wehave 100 moles of air, 78 moles will be nitrogen,
21 moles will be oxygen, and 1 mol will be other gases.
Mol fraction = 1/100 = 0.01
Step 3: Calculate the partial pressure of the other gases
Pgas = Xgas * Ptotal
⇒ Pgas = the partial pressure = ?
⇒ Xgas = the mol fraction of the gas = 0.01
⇒Ptotal = the total pressure of the pressure = 0.90 atm
Pgas = 0.01 * 0.90 atm
Pgas = 0.009 atm
The partial pressure of the other gases is 0.009 atm
Which of the following is a false statement about the isotopes carbon-12 (12C) and carbon-13 (13C)? Please choose the correct answer from the following choices, and then select the submit answer button. 12C and 13C have the same atomic weight. 12C and 13C have different numbers of neutrons. 13C has one more electron than 12C. 13C has more subatomic particles in its nucleus than 12C. 13C has the same atomic number as 12C.
The given isotopes of carbon, i.e., 12C and 13C, have the same number of protons but the number of neutrons differs. So, the statement 12C and 13C have different numbers of neutrons, is correct.
Isotopes are members of the same element's family but have variable numbers of neutrons despite having the same number of protons.
The element carbon can have different isotopes, based on the number of neutrons.
The isotopes of carbon include, C10, C11, C12, C13, C14, C15, and C16.
The given isotope, 12C and 13C, possess similar chemical properties and the same atomic number.
But the mass of the 13C isotope of carbon is more, because it possesses, one extra neutron, as compared to the 12C isotope of carbon.
Thus, the statement about the isotopes, “12C and 13C have different numbers of neutrons”, is correct.
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The statement that 13C has one more electron than 12C is incorrect. Both carbon-12 (12C) and carbon-13 (13C) isotopes have the same number of electrons due to both having the same atomic number. They only differ in the number of neutrons they possess.
Explanation:The statement '13C has one more electron than 12C' is false. Both isotopes, carbon-12 (12C) and carbon-13 (13C) have the same atomic number '6', which represents the number of protons and, under stable conditions, also views it as the same number of electrons.
So, while 12C and 13C indeed differ in the number of neutrons (12C has 6 neutrons and 13C has 7), they possess the same number of protons and electrons. Furthermore, their atomic weights are different as a result of the difference in the number of neutrons. The atomic weight of 12C is closer to 12 and that of 13C is closer to 13.
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PCL was mixed with gelatin to make a blend for elctrospun fibrous scaffold encapsulating growth factor that was admixed in the polymer solution and then filled in the syringe for electrospinning. Three scaffolds were made with (A: only PCL), (B: PCL:gelatin=3:1), (C: PCL:gelatin = 1:1). The scaffolds were immersed in PBS at 37 °C! and its degradation rate was noted: Scaffold A, B and C degraded 1%, 25% and 50% respectively, in 21 days. Assuming the growth factor is trapped and cannot freely diffuse out of the scaffold. Assuming, growth factor release is dependent on scaffold degradation A. Which scaffold can be used for delivering growth factor (encapsulated in the scaffold) needed to be delivered in first 7 days of incubation in an in-vivo experiment and why? (5pts) B. The scaffold A and C were prepared and freeze dried. Their weights were Dry weight of A: 10 mg Dry weight of C: 10 mg Both (A and C) the scaffolds were kept in PBS for 7 days at 37 °C. The scaffolds were freeze dried and their weight was recorded. Weight of A after degradation: 9.967 mg Weight of C after degradation: 8.33 mg Calculate the percentage weight remaining for A and C scaffold and comment on why one scaffold degraded faster than the other.
Answer:
Explanation:
First PCL to form a scaffold, combined with gelatin.
They are made by first three forms A) made by just PCL.B) made by gelatin ratio PCL is 3:1 and last is C) made by gelatin PCL.
The decoration rate is 1%, 25% and 50% respectively.
A) Growth factor is stuck in 21 days, and can not spread. In this case in vivo experiment for 7 days used highly degradable scaffold use the ability to break down due to decomposition in vivo degradation rate depends on the scaffold's acid byproduct impact.
B) the amount of scaffolded degradation.
First, with scaffold A.
10 mg scaffold weight A= 1 per cent degradation.
Following degradation wt is 9.967=? Degradation per cent.
So, degradation (9.967* 1)/10= 0.9967 per cent.
Likewise for C) scaffold c 10 mg wt. Loss of 50 per cent.
After wt 8.33=? Degradation per cent.
Degradation (8.33* 50)/10=41.65 per cent.
Scaffold c degrades significantly, since the loss of wt is even greater.
a metal sample has a mass of 7.56 g. the sample is placed into a graduated cylinder previously filled with 20.00 mL of water. If the final volume in the cylinder is 21.68 mL, what is the density of the substance?
Answer:
4.5g/ml
Explanation:
metal sample has a mass = 7.56 g
cylinder previously filled with water = 20.00 mL
final volume in the cylinder = 21.68 mL
[tex]density = \frac{mass}{V_f - V_i}[/tex]
[tex]density = \frac{7.56}{(21.68 - 20.00)}[/tex]
density = 4.5g/ml
Write the balanced chemical equation for the combustion of ethane, , and answer these questions. (Use the lowest possible coefficients. Omit states of matter.) How many molecules of oxygen would combine with 16 molecules of ethane in this reaction
Answer:
1. 2C2H6 + 7O2 —> 4CO2 + 6H2O
2. 56moles of O2
Explanation:
Ethane undergo Combustion to produce CO2 and H20 according the equation below:
C2H6 + O2 —> CO2 + H2O
Let us balance the equation. There are 6 atoms of H on the left side and 2 atoms on the right side. It can be balanced by putting 3 in front of H2O as shown below:
C2H6 + O2 —> CO2 + 3H2O
There are 2 atoms of C on left and 1atom on the right. It can be balanced by putting 2 in front of CO2 as shown below:
C2H6 + O2 —> 2CO2 + 3H2O
There are a total of 7 atoms of O on the right and 2 atoms on the left. It can be balanced by putting 7/2 in front of O2 as shown below:
C2H6 + 7/2O2 —> 2CO2 + 3H2O
Now we multiply through by 2 to remove the fraction as shown below
2C2H6 + 7O2 —> 4CO2 + 6H2O
Now the equation is balanced
2. 2C2H6 + 7O2 —> 4CO2 + 6H2O
From the equation above,
2 moles of ethane(C2H6) combined with 7moles of O2.
Therefore, 16moles of ethane(C2H6) will combine with = (16x7)/2 = 56moles of O2
With respect to NAD+ and NADP+, which electron carrier is generally preferred for anabolic reactions? Group of answer choices Just NAD+ Just NADP+ Either of these Neither of these
Answer:
Just NADP+.
Explanation:
Two main type of biochemical reaction are anabolic reaction and catabolic reaction. Anabolic reactions joins the small molecules to form the large products.
The electron carrier may be defined as the molecule that has the ability to transport the electron from one molecule to the other molecule. NADP+ acts as electron carrier in the synthesis of molecule especially in the cholesterol metabolism. NAD+ acts as electron carrier in the catabolic reactions.
Thus, the correct answer is option (2).
How many mL of 0.05 M sodium acetate should be added to 100 mL of 0.05M acetic acid to make a buffer of pH 5.1? What is the molarity of the resulting buffer with respect to acetate (Acetate + Acetic Acid)? pKa acetic = 4.76
Explanation:
Let us assume that volume of acetic acid added is V ml.
So, [tex][CH_{3}COOH] = \frac{0.05 \times 100}{100 + V}[/tex]
and, [tex][CH_{3}COONa] = \frac{0.05 \times V}{100 + V}[/tex]
Expression for the buffer solution is as follows.
pH = [tex]pK_{a} + log \frac{[CH_{3}COONa]}{[CH_{3}COOH]}[/tex]
5.1 = [tex]4.76 + log \frac{0.05 \times V}{0.05 \times 100}[/tex]
0.34 = log V - 2
log V = 2.34
or, V = 218.77 ml
Now, we will calculate the molarity of the buffer with respect to acetate as follows.
= [tex][CH_{3}COO^{-}] + [CH_{3}COOH][/tex]
= [tex]\frac{0.05 \times 218.77}{318.77} + \frac{0.05 \times 100}{318.77}[/tex]
= 0.0499 M
or, = 0.05 M (approx)
Thus, we can conclude that molarity of the resulting buffer with respect to acetate is 0.05 M.
Final answer:
To make a buffer of pH 5.1 using 0.05 M acetic acid, the Henderson-Hasselbalch equation is used to calculate the amount of 0.05 M sodium acetate needed. The result is based on achieving the correct ratio of acetate ion to acetic acid. The molarity of the resulting buffer depends on the total moles of acetate and acid in the final solution volume.
Explanation:
To determine how many mL of 0.05 M sodium acetate should be added to 100 mL of 0.05M acetic acid to make a buffer of pH 5.1, we can use the Henderson-Hasselbalch equation, which is pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the acetate ion and [HA] is the concentration of acetic acid. Given that the pKa of acetic acid is 4.76, plugging in the pH 5.1 gives us the equation 5.1 = 4.76 + log([A-]/[0.05]).
Solving for [A-], we find that the ratio of [A-] to [HA] needed is approximately 2.2. Since the volume of the acetic acid solution is 100 mL and its concentration is 0.05M, to achieve the desired ratio, the amount of sodium acetate needed can be calculated based on the molarity and the final volume of the solution. The resulting molarity of the buffer with respect to acetate (acetate + acetic acid) will be determined by the total moles of acetate ions and acetic acid divided by the total volume of the solution after the addition of sodium acetate.
Place the following substances in order of increasing boiling point. Ne Cl2 O2
a. O2 < Cl2 < Ne
b. Cl2 < Ne < O2
c. Cl2 < O2 < Ne
d. Ne < O2 < Cl2
e. Ne < Cl2 < O2
Answer: = D
Explanation:
The atomic mass increases from Ne to O2 to Cl2 hence the boiling point also increases, therefore
Ne < O2 < Cl2
A student has 500.0 mL of a 0.1133 M aqueous solution of BaI2 to use in an experiment. She accidentally leaves the container uncovered and comes back the next week to find only a solid residue. The mass of the residue is 28.28 g. Determine the chemical formula of this residue.
Answer:
[tex]BaI_2\ ^.6H_2O[/tex]
Explanation:
Hello,
In this case, by knowing the volume and the molarity of the barium iodide, is it possible to compute the residue's mass as shown below:
[tex]m_{BaI_2}=0.500L*0.1133\frac{molBaI_2}{L}*\frac{391.136gBaI_2}{1molBaI_2} =22.16gBaI_2[/tex]
Nevertheless, the obtained value is lower than the obtained by 6.133 g which means that mass corresponds to water forming a hydrate. In such a way, one could know how many waters in form of hydrate remain with the residue by a trial-error procedure as shown below:
[tex]m=28.28g=0.500L*0.1133\frac{molBaI_2}{L}*\frac{(391.136+18)gBaI_2\ ^.H_2O}{1molBaI_2} =23.17g\rightarrow No\\m=28.28g=0.500L*0.1133\frac{molBaI_2}{L}*\frac{(391.136+2*18)gBaI_2\ ^.2H_2O}{1molBaI_2} =24.20g\rightarrow No\\m=28.28g=0.500L*0.1133\frac{molBaI_2}{L}*\frac{(391.136+3*18)gBaI_2\ ^.3H_2O}{1molBaI_2} =25.22g\rightarrow No\\m=28.28g=0.500L*0.1133\frac{molBaI_2}{L}*\frac{(391.136+4*18)gBaI_2\ ^.4H_2O}{1molBaI_2} =26.24g\rightarrow No\\[/tex]
[tex]m=28.28g=0.500L*0.1133\frac{molBaI_2}{L}*\frac{(391.136+5*18)gBaI_2\ ^.5H_2O}{1molBaI_2} =27.26g\rightarrow No\\m=28.28g=0.500L*0.1133\frac{molBaI_2}{L}*\frac{(391.136+6*18)gBaI_2\ ^.6H_2O}{1molBaI_2} =28.28g\rightarrow Yes\\[/tex]
Therefore, the formula is barium iodide hexahydrate:
[tex]BaI_2\ ^.6H_2O[/tex]
Best regards.
A silver rod and a SHE are dipped into a saturated aqueous solution of silver oxalate, Ag2C2O4, at 25°C. The measured potential difference between the rod and the SHE is 0.5812 V, the rod being positive. Calculate the solubility product constant for silver oxalate.
The solubility product constant (Ksp) is calculated as 2.392.
To solve this problem, we can use the Nernst equation:
Ecell = E°cell - (RT/nF) * ln(Q)
where:
Ecell is the measured potential difference between the rod and the SHE (0.5812 V)
E°cell is the standard cell potential
R is the ideal gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin (25°C = 298 K)
n is the number of electrons transferred in the balanced redox reaction
F is Faraday's constant (96485 C/mol)
Q is the reaction quotient
In this case, the balanced redox reaction is:
Ag2C2O4(s) ⇌ 2Ag+(aq) + C2O4^2-(aq)
The solubility product constant (Ksp) for silver oxalate can be expressed as:
Ksp = [Ag+]^2 * [C2O4^2-]
Since the rod is positive, it means that Ag+ ions are being reduced to Ag(s) on the rod, so we can write:
E°cell = E°red,cathode - E°red,anode
The standard reduction potential for Ag+ to Ag(s) is 0.7996 V, and the standard reduction potential for C2O4^2- to CO2(g) is 0.1936 V.
Therefore, E°cell = 0.7996 V - 0.1936 V = 0.606 V
Now we can substitute all the values into the Nernst equation:
0.5812 V = 0.606 V - (8.314 J/(mol·K) * 298 K / (2 * 96485 C/mol)) * ln(Q)
Simplifying:
0.5812 V = 0.606 V - (0.0257 V) * ln(Q)
0.0257 V * ln(Q) = 0.606 V - 0.5812 V
0.0257 V * ln(Q) = 0.0248 V
ln(Q) = 0.0248 V / 0.0257 V
ln(Q) = 0.964
Q = e^(0.964)
Q ≈ 2.622
Since Q = [Ag+]^2 * [C2O4^2-], we can write:
[Ag+]^2 * [C2O4^2-] ≈ 2.622
The solubility product constant (Ksp) is the product of the concentrations of the ions at equilibrium, so we can assume that the concentrations are equal to each other:
[Ag+] ≈ [C2O4^2-]
Therefore, we can substitute [Ag+] for [C2O4^2-]:
[Ag+]^2 * [Ag+] ≈ 2.622
[Ag+]^3 ≈ 2.622
Taking the cube root of both sides:
[Ag+] ≈ (2.622)^(1/3)
[Ag+] ≈ 1.378
Therefore, the approximate solubility of silver oxalate is 1.378 M.
The solubility product constant (Ksp) can now be calculated as:
Ksp = [Ag+]^2 * [C2O4^2-] ≈ (1.378)^2 * (1.378) ≈ 2.392
The solubility product constant for silver oxalate at [tex]25\°C[/tex] is approximately [tex]\( 10^{-22.125} \)[/tex].
The solubility product constant [tex](\( K_{sp} \))[/tex] for silver oxalate can be calculated using the Nernst equation and the measured potential difference between the silver rod and the standard hydrogen electrode (SHE). The reaction occurring at the silver rod can be represented as:
[tex]\[ \text{Ag}_2\text{C}_2\text{O}_4(s) \rightleftharpoons 2\text{Ag}^+(aq) + \text{C}_2\text{O}_4^{2-}(aq) \][/tex]
The Nernst equation for this reaction at [tex]25\°C[/tex] is given by:
[tex]\[ E = E^{\circ} - \frac{0.0592}{n} \log \frac{1}{[\text{Ag}^+]^2[\text{C}_2\text{O}_4^{2-}]} \][/tex]
where [tex]\( E \)[/tex] is the measured potential (0.5812 V), [tex]\( E^{\circ} \)[/tex] is the standard reduction potential for the [tex]\( \text{Ag}^+ \)[/tex] / Ag couple, [tex]\( n \)[/tex] is the number of electrons transferred in the half-reaction (which is 2 for this reaction), and the concentrations of [tex]\( \text{Ag}^+ \)[/tex] and [tex]\( \text{C}_2\text{O}_4^{2-} \)[/tex] are equal to the solubility of silver oxalate since they come from its dissolution.
First, we need to find the standard reduction potential [tex]\( E^{\circ} \)[/tex] for the [tex]\( \text{Ag}^+ \)[/tex] / Ag couple, which is 0.7996 V.
Rearranging the Nernst equation to solve for the solubility \( S \), we get:
[tex]\[ S = [\text{Ag}^+] = [\text{C}_2\text{O}_4^{2-}] = 10^{\frac{(E^{\circ} - E)n}{0.0592}} \][/tex]
Plugging in the values:
[tex]\[ S = 10^{\frac{(0.7996 - 0.5812) \times 2}{0.0592}} \][/tex]
[tex]\[ S = 10^{\frac{0.2184 \times 2}{0.0592}} \][/tex]
[tex]\[ S = 10^{\frac{0.4368}{0.0592}} \][/tex]
[tex]\[ S = 10^{7.375} \][/tex]
[tex]\[ S \approx 10^{-7.375} \][/tex]
The solubility product constant [tex]\( K_{sp} \)[/tex] is given by:
[tex]\[ K_{sp} = [\text{Ag}^+]^2[\text{C}_2\text{O}_4^{2-}] \][/tex]
[tex]\[ K_{sp} = (S)^2(S) \][/tex]
[tex]\[ K_{sp} = (10^{-7.375})^2(10^{-7.375}) \][/tex]
[tex]\[ K_{sp} = 10^{-7.375 \times 3} \][/tex]
[tex]\[ K_{sp} = 10^{-22.125} \][/tex]
Suppose a current of 60. A flows through a copper wire for 22.0 minutes. Calculate how many moles of electrons travel through the wire. Be sure your answer has the correct unit symbol and the correct number of significant digits.
Answer: 0.821 moles
Explanation:
Moles of electron = 1 mole
According to mole concept:
1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of particles.
We know that:
Charge on 1 electron = [tex]1.6\times 10^{-19}C[/tex]
Charge on 1 mole of electrons = [tex]1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C[/tex]
To calculate the charge passed, we use the equation:
[tex]I=\frac{q}{t}[/tex]
where,
I = current passed = 60 A
q = total charge = ?
t = time required = 22.0 minutes =[tex]22.0\times 60=1320sec[/tex]
Putting values in above equation, we get:
[tex]60A=\frac{q}{1320}\\\\q={60A}\times {1320s}=79200C[/tex]
As 96500 C contains = 1 mole of electrons
79200 C contains = [tex]\frac{1}{96500}\times 79200 =0.821[/tex] mole of electrons
Thus 0.821 moles of electrons travel through the wire.
Grey Goose ® vodka has an alcohol content of 40.0 % (v/v). Assuming that vodka is composed of only ethanol and water answer the following questions. Note: The molar masses of water and ethanol are 18.0 g and 46.0 g, respectively. The densities of water, ethanol, and this vodka mixture are 1.00 g/mL, 0.789 g/mL, and 0.935 g/mL, respectively
a. Calculate the molarity of ethanol in this vodka, assuming that water is the solvent.
b. Calculate the percent by mass of ethanol % (m/m) in this vodka.
c. Calculate the molality of ethanol in this vodka assuming that water is the solvent.
d. Calculate the mole fractions of ethanol and water in this vodka.
e. Calculate the vapor pressure, in torr, of this vodka at 25.0 oC if the vapor pressures of pure water and ethanol are 23.8 torr and 45.0 torr, respectively?
Explanation:
Grey Goose vodka has an alcohol content of 40.0 % (v/v).
Volume of vodka = V = 100 mL
This means that 40.0 mL of alcohol is present 100 mL of vodka.
Volume of ethanol=V' = 40.0 mL
Mass of ethanol = m
Density of the ethanol = d = 0.789 g/mL
[tex]m=d\times V' = 0.789 g/ml\times 40.0 mL=31.56 g[/tex]
Volume of water = V''= 100 ml - 40.0 mL = 60.0 mL
Mass of water = m'
Density of the water = d' = 1.00 g/mL
[tex]m'=d'\times V'' = 1.00 g/ml\times 60.0 mL=60.0 g[/tex]
a.)
Moles of ethanol = n= [tex]\frac{31.56 g}{46g/mol}=0.6861 mol[/tex]
Volume of vodka = V = 100 mL = 0.100 L ( 1mL=0.001 L)
Molarity of the ethanol:
[tex]=\frac{0.6861 mol}{0.100 L}=6.861 M[/tex]
6.861 M the molarity of ethanol in this vodka.
b) Mass of ethanol = 31.56 g
Moles of ethanol = n= [tex]\frac{31.56 g}{46g/mol}=0.6861 mol[/tex]
Volume of vodka = V = 100 mL
Mass of vodka = m
Density of the water = D = 0.935 g/mL
[tex]M=D\times V=0.935 g/ml\times 100 ml=93.5 g[/tex]
The percent by mass of ethanol % (m/m):
[tex]\frac{31.56 g}{93.5 g}\times 100=33.75\%[/tex]
33.75% is the percent by mass of ethanol % (m/m) in this vodka.
c)
Moles of ethanol = n= [tex]\frac{31.56 g}{46g/mol}=0.6861 mol[/tex]
Mass of solvent that is water = 60.0 g = 0.060 kg ( 1g = 0.001 kg)
Molality of ethanol in vodka :
[tex]m=\frac{0.6861 mol}{0.060 kg}=11.435 m[/tex]
11.435 m is the molality of ethanol in this vodka.
d)
Moles of ethanol = [tex]n_1=\frac{31.56 g}{46g/mol}=0.6861 mol[/tex]
Moles of water = [tex]n_2=\frac{60.0 g}{18 g/mol}=3.333 mol[/tex]
Mole fraction of ethanol = [tex]\chi_1[/tex]
[tex]\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.6861 mol}{0.6861 mol+3.333 mol}[/tex]
= 0.1707
Mole fraction of water = [tex]\chi_2[/tex]
[tex]\chi_2=\frac{n_2}{n_1+n_2}=\frac{3.3333 mol}{0.6861 mol+3.333 mol}[/tex]
= 0.8290
e)
The vapor pressure of vodka = P
Mole fraction of ethanol = [tex]\chi_1=0.1707[/tex]
Mole fraction of water = [tex]\chi_2=0.8290[/tex]
The vapor pressures of ethanol = [tex]p_1=45.0 Torr[/tex]
The vapor pressures of pure water = [tex]p_2=23.8Torr[/tex]
[tex]P=\chi_1\times p_1+\chi_2\times p_2[/tex]
[tex]P=0.1707\times 45.0torr+0.8290\times 23.8 Torr=27.41 torr[/tex]
The vapor pressure of vodka is 27.41 Torr.
In a coffee-cup calorimeter experiment, 10.00 g of a soluble ionic compound was added to the calorimeter containing 75.0 g H2O initially at 23.2°C. The temperature of the water increased to 31.8°C. What was the change in enthalpy for the dissolution of the compound? Give your answer in units of joules per gram of compound. Assume that the specific heat of the solution is the same as that of pure water, 4.18 J ⁄ (g ⋅ °C).
The change in enthalpy for the dissolution of the compound is 344.11 J/g.
Explanation:In a coffee-cup calorimeter experiment, the change in enthalpy for the dissolution of a compound can be determined using the equation q = mCΔT, where q is the amount of heat absorbed or released, m is the mass of the compound, C is the specific heat of the solution, and ΔT is the change in temperature. Given that the initial temperature of the water is 23.2°C and the final temperature is 31.8°C, the change in temperature is ΔT = 31.8°C - 23.2°C = 8.6°C.
Since 10.00 g of the compound was added to 75.0 g of water, the total mass of the solution is 75.0 g + 10.00 g = 85.0 g.
Using the equation q = mCΔT, where C is the specific heat of water (4.18 J/(g·°C)), the amount of heat involved in the dissolution can be calculated as q = (85.0 g)(4.18 J/(g·°C))(8.6°C) = 3,441.14 J.
The change in enthalpy for the dissolution of the compound is thus 3,441.14 J/10.00 g = 344.11 J/g of the compound.
Acetylenic compounds may be used as dienophiles in the Diels-Alder reaction. Write the structure for an adduct that you expect from the reaction of 1,3-butadiene with hexafluoro-2-butyne:
Answer:
Explanation:
An electron-donating heteroatom substituent at position-2 of a furan promotes regiospecific opening of the 7-oxa bridge of the Diels-Alder cycloadduct with hexafluoro-2-butyne, producing a 4-heterosubstituted 2,3-di(trifluoromethyl)phenol building block in a single step. The phenol and heteroatom substituent are easily transformed to the corresponding iodide or triflate that readily undergoes Heck, Suzuki, and Stille reactions to install a variety of substituents in high yields. This methodology provides a facile and general synthesis of 1,4-disubsituted 2, 3-di(trifluoromethyl)benzenes.
Final answer:
The Diels-Alder reaction of 1,3-butadiene with hexafluoro-2-butyne would form a six-membered ring with a hexafluoroalkyl side group, yielding a bicyclic compound.
Explanation:
The Diels-Alder reaction between 1,3-butadiene and hexafluoro-2-butyne would yield a six-membered ring product, specifically a bicyclic compound due to the formation of a cyclohexene derivative where the double bond is part of the ring and the hexafluoroalkyl group is attached to two adjacent carbons in the ring. It involves the 1,3-butadiene acting as the diene and the hexafluoro-2-butyne acting as the dienophile in an intermolecular reaction. The Diels-Alder reaction creates two new sigma bonds and a pi bond across the alkene and the alkyne to form the new cyclohexene ring.
Consider the following mechanism for the decomposition of NO2Cl to NO2 and Cl2: (1) NO2Cl ⇌ NO2 + Cl (2) NO2Cl + Cl → NO2 + Cl2 (a) Use steady-state approximation to express the rate of Cl2 production. Select the single best answer
Final answer:
The rate of Cl2 production can be determined using the steady-state approximation, which states that the rate of the forward reaction in the first step is equal to the rate of the reverse reaction. Therefore, the rate of Cl2 production is given by the rate of the forward reaction of step 1. This can be expressed as k1 [NO2Cl] [Cl].
Explanation:
The rate of Cl2 production can be determined using the steady-state approximation. According to the given mechanism, the first step is in equilibrium, so the forward and reverse reaction rates are equal. This allows us to express the rate of the forward reaction of step 1 as:
rate of forward reaction of step 1 = k1 [NO2Cl] [Cl]
Since the reverse reaction of step 1 is negligible, the overall rate of Cl2 production is equal to the rate of the forward reaction of step 1. Therefore, the rate of Cl2 production is given by:
rate of Cl2 production = k1 [NO2Cl] [Cl]
Final answer:
To express the rate of Cl2 production using the steady-state approximation, one must balance the rate of chlorine atom creation with its consumption and solve for its concentration in terms of the reactants and rate constants. This value is then used in the rate equation for Cl2 production.
Explanation:
The question pertains to the mechanism of the decomposition of nitryl chloride (NO2Cl) to nitrogen dioxide (NO2) and chlorine gas (Cl2). Since the steady-state approximation is used, we consider the intermediate species such as chlorine atoms (Cl) to be in a steady state, meaning their concentrations do not change over time. The provided steps do not correspond directly to the decomposition of NO2Cl, but they outline a similar mechanism which can be analyzed in the same manner using steady-state approximation.
If we follow a similar approach for NO2Cl decomposition, we would set the rate of creation of Cl equal to the rate of its consumption. This would lead to a system of equations that can then be solved to express the rate of Cl2 production in terms of the concentrations of the reactants and the rate constants of the elementary steps.
The rate of reaction (2) can be determined using the rate law, which is determined experimentally. However, since we are not given any experimental data or rate law, we cannot provide a specific answer.
To summarize, to find the rate of Cl2 production in the decomposition of NO2Cl, we need the rate law for reaction (2). Without the rate law, we cannot determine the specific rate of Cl2 production.
Consider the reaction: N2(g) + 2O2(g)2NO2(g) Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.90 moles of N2(g) react at standard conditions. S°system = J/K Submit Answer
Explanation:
It is known that standard entropies for [tex]N_{2}(g)[/tex] is 191.6 J/mol K, [tex]O_{2}(g)[/tex] = 205 J/mol K, and [tex]NO_{2}(g)[/tex] is 239.7 J/mol K at 298 K.
Therefore, we will calculate the value of [tex]\Delta S^{o}_{reaction}[/tex] from standard absolute entropies as follows.
[tex]\Delta S^{o}_{reaction} = \sum \Delta S^{o}_{products} - \sum \Delta S^{o}_{reactants}[/tex]
= 2 mole of [tex]NO_{2}(g)[/tex] - 1 mole of [tex]N_{2}(g)[/tex] + 2 mole of [tex]O_{2}(g)[/tex]
= [tex]2 \times 239.7 J/mol K - 1 \times 191.6 J/mol K + 2 \times 205 J/mol K[/tex]
= -122.2 J/K
The entropy change for 1.90 moles of [tex]N_{2}(g)[/tex] reacting is as follows.
[tex]\Delta S^{o}_{system}[/tex] = 1.90 moles of [tex]N_{2}(g) \times 122.2 J/K/ 1 \text{mol of} N_{2}(g)[/tex]
= 232.18 J/K
Thus, we can conclude that the entropy change for the given system is 232.18 J/K.
The aggregation of nonpolar molecules or groups in water is thermodynamically due to the ________.
A. increased entropy of the water molecules.
B. decreased enthalpy of the system.
C. very strong van der Waals forces among the nonpolar molecules or groups.
D. increased entropy of the nonpolar molecules when they associate.
Answer:
A. Increased entropy of the water molecules.
Explanation:
Entropy is the quantitative measure of disorder or randomness in a system or element. In thermodynamics, entropy or hydrophobic effect is the free energy change of water enclosing a solute. Hence the existing negative free charges enhances the effect of hydrophilicty hence the aggregation of non-polar molecules.
The aggregation of nonpolar molecules in water is primarily due to the increased entropy of the water molecules, as they rearrange themselves to accommodate nonpolar molecules.
Explanation:The aggregation of nonpolar molecules or groups in water is thermodynamically due to the increased entropy of the water molecules. When nonpolar molecules or groups are added to water, water molecules arrange themselves in order to minimize the disruption of hydrogen bonds, which increases the entropy of water. In other words, the introduction of nonpolar substances leads to a more disordered and random arrangement of water molecules, increasing the system's overall entropy.
Options B, C, and D although relevant, are not the primary reason. While enthalpy may decrease due to aggregation and Van der Waals forces might influence nonpolar association, the primary factor contributing to the phenomenon is the change in water's entropy.
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Determine the rate constant for each of the following fi rst-order reactions, in each case expressed for the rate of loss of A: (a) A S B, given that the concentration of A decreases to one-half its initial value in 1000. s; (b) A S B, given that the concentration of A decreases from 0.67 molL1 to 0.53 molL1 in 25 s; (c) 2 A S B C, given that [A]0 0.153 molL1 and that after 115 s the concentration of B rises to 0.034 molL1
Explanation:
The integrated first law is given by :
[tex][A]=[A]_o\times e^{-k\times t}[/tex]
Where:
[tex][A]_o[/tex] = initial concentration of reactant
[A] = concentration of reactant after t time
k = rate constant
a)
[tex][A_o]=x[/tex]
[tex][A]=\frac{x}{2}[/tex]
t = 1000 s
[tex]\frac{x}{2}=x\times e^{-k\times 1000 s}[/tex]
Solving for k:
[tex]k=0.06934 s^{-1}[/tex]
The rate constant for this reaction is [tex]0.06934 s^{-1}[/tex].
b)
[tex][A_o]=0.67 mol/L[/tex]
[tex][A]=0.53 mol/L[/tex]
t = 25 s
[tex]0.53 mol/L=0.67 mol/L\times e^{-k\times 25s}[/tex]
Solving for k:
[tex]k=0.009376 s^{-1}[/tex]
The rate constant for this reaction is [tex]0.009376 s^{-1}[/tex].
c) 2 A → B +C
[tex][A_o]=0.153 mol/L[/tex]
[tex][A]=?[/tex]
[tex][B]=0.034 mol/L[/tex]
According to reaction, 1 mole of B is obtained from 2 moles of A.
Then 0.034 mole of B will be obtained from:
[tex]\frac{2}{1}\times 0.034 mol= 0.068 mol[/tex] of A
So, the concentration left after 115 seconds:
[tex][A]=0.153 mol/L-0.068 mol/L=0.085 mol/L[/tex]
t = 115 s
[tex]0.085mol/L=0.53 mol/L\times e^{-k\times 115 s}[/tex]
Solving for k:
[tex]k=0.01592 s^{-1}[/tex]
The rate constant for this reaction is [tex]0.01592 s^{-1}[/tex].
An aqueous solution of sodium sulfate is allowed to react with an aqueous solution of barium nitrate. What is the coefficient of the solid in the balanced equation
Answer: The coefficient of the solid in the balanced equation is 1.
Explanation:
A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.
The balanced chemical equation is:
[tex]Na_2SO_4(aq)+Ba(NO_3)_2(aq)\rightarrow BaSO_4(s)+2NaNO_3(aq)[/tex]
Thus a coefficient of 1 is placed in front of the solid.
The coefficient of the solid in the balanced equation is 1.
The double substitution reaction is a reaction in which ion exchange takes place. Salts that are soluble in water are indicated by the symbol (aq), and salts that are insoluble in water and remain solid are represented by (s) according to the chemical formula.The balanced chemical equation is:
Ba(NO3)2 + Na2SO4 → BaSO4 + NaNO3
Thus, the coefficient of the solid in the balanced equation is 1.
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Lab group B determines that 0.003 moles of bicarbonate was present in their portion of an antacid tablet. How many grams of bicarbonate are in their tablet? Report your answer to 3 decimal places.
Answer:
There are 0.183 grams of bicarbonate in their tablet
Explanation:
Step 1: Data given
0.003 moles of bicarbonate was present in their portion of an antacid tablet
Bicarbonate = HCO3-
Molar mass of bicarbonate = 61.02 g/mol
Step 2: Calculate mass bicarbonate
Mass bicarbonate = moles bicarbonate * molar mass bicarbonate
Mass bicarbonate = 0.003 moles * 61.02 g/mol
Mass bicarbonate = 0.183 grams
There are 0.183 grams of bicarbonate in their tablet
Determine whether the following reaction is a Lewis acid/base reaction and if so, what is the Lewis acid: Fe3+(aq) + 6 CN−(aq) → Fe(CN)63−(aq) Group of answer choices This is not a Lewis acid/base reaction It is a Lewis acid/base reaction, and Fe3+ is the Lewis acid It is a Lewis acid/base reaction, and CN- is the Lewis acid
Answer: It is a Lewis acid/base reaction, and [tex]Fe^{3+}[/tex] is the Lewis acid.
Explanation:
According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.
[tex]Fe^{3+}[/tex] can readily accept electrons and thus act as a lewis acid which is short of electrons.
[tex]CN^-[/tex] can readily lose electrons and thus act as a lewis base which has excess of electrons.
It is a Lewis acid/base reaction, and [tex]Fe^{3+}[/tex] is the Lewis acid.
13. In animals, nitrogenous wastes are produced mostly from the catabolism of
A. phospholipids and glycolipids
B. proteins and nucleic acids.
C. starch and cellulose
D. triglycerides and steroids
E. fatty acids and glycerol
Answer:
The answer is B
Explanation:
Calculate the volume (in mL) of 6.25 x 10-4 M ferroin solution that needs to be added to a 10.0 mL volumetric flask and diluted with deionized (DI) water in order to prepare a calibration standard solution with a concentration of 2.50 x 10-5 M ferroin. As part of your preparation for performing this experiment, repeat this calculation for each of the calibration standards you will need to prepare and record the information in your lab notebook so that you have it ready during the lab session. Group of answer choices 0.200 mL 0.400 mL 0.600 mL 0.800 mL none of the above
The volume V2 is 0.400 ml
Explanation:
The dilution equation is the product of initial values of molarity and volume which is equal to the product of final values of molarity and volume.
The dilution equation is given by
M1 [tex]\times[/tex] V1 = M2 [tex]\times[/tex] V2
where,
M represents the molarity of the solution
V represents the volume of the solution
M1 and V1 are the initial values of the solution
M2 and V2 are the final values of the solution
M1 [tex]\times[/tex] V1 = M2 [tex]\times[/tex] V2
(2.5 [tex]\times[/tex] [tex]10^{-5}[/tex]) [tex]\times[/tex] 10 = (6.25 [tex]\times 10^{-4}[/tex]) [tex]\times[/tex] V2
V2 = 2.5 [tex]\times 10^{-4}[/tex] / (6.25 [tex]\times 10^{-4}[/tex])
V2 = 0.4
The volume V2 is 0.400 ml
The volume of solution required is 0.400 mL.
In this case, we have to use the dilution formula;
C1V1 = C2 V2
In this case;
C1 = 6.25 x 10-4 M
V1 = ?
C2 = 2.50 x 10-5 M
V2 = 10.0 mL
Making V1 the subject of the formula;
V1 = C2V2/C1
V1 = 2.50 x 10-5 M x 10.0 mL/6.25 x 10-4 M
V1 = 0.400 mL
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Determine the change in entropy for 2.7 moles of an ideal gas originally placed in a container with a volume of 4.0 L when the container was expanded to a final volume of 6.0 L at constant temperature.
Answer:
The value of entropy change for the process [tex]dS = 0.009 \frac{KJ}{K}[/tex]
Explanation:
Mass of the ideal gas = 0.0027 kilo mol
Initial volume [tex]V_{1}[/tex] = 4 L
Final volume [tex]V_{2}[/tex] = 6 L
Gas constant for this ideal gas ( R ) = [tex]R_{u} M[/tex]
Where [tex]R_{u}[/tex] = Universal gas constant = 8.314 [tex]\frac{KJ}{Kmol K}[/tex]
⇒ Gas constant R = 8.314 × 0.0027 = 0.0224 [tex]\frac{KJ}{K}[/tex]
Entropy change at constant temperature is given by,
[tex]dS = R log _{e} \frac{V_{2}}{V_{1}}[/tex]
Put all the values in above formula we get,
[tex]dS = 0.0224 log _{e} [\frac{6}{4}][/tex]
[tex]dS = 0.009 \frac{KJ}{K}[/tex]
This is the value of entropy change for the process.
One way to represent this equilibrium is: N2O4(g)2NO2(g) Indicate whether each of the following statements is true, T, or false, F. AT EQUILIBRIUM we can say that: 1. The concentration of NO2 is equal to the concentration of N2O4. 2. The rate of the dissociation of N2O4 is equal to the rate of formation of N2O4. 3. The rate constant for the forward reaction is equal to the rate constant of the reverse reaction. 4. The concentration of NO2 divided by the concentration of N2O4 is equal to a constant.
Answer:
1. The concentration of [tex]NO_2[/tex] is equal to the concentration of [tex]N_2O_4[/tex] : False
2. The rate of the dissociation of [tex]N_2O_4[/tex] is equal to the rate of formation of [tex]N_2O_4[/tex]: True
3. The rate constant for the forward reaction is equal to the rate constant of the reverse reaction: False
4. The concentration of [tex]NO_2[/tex] divided by the concentration of [tex]N_2O_4[/tex] is equal to a constant : False
Explanation:
[tex]N_2O_4(g)\rightleftharpoons 2NO_2(g)[/tex]
The concentrations of reactant and product is constant and not equal.
The rate of forward and backward reactions are equal. Thus rate of the dissociation of [tex]N_2O_4[/tex] is equal to the rate of formation of [tex]N_2O_4[/tex]
For a reversible reaction, the equilibrium constant for the forward reaction is inverse of the equilibrium constant for the backward reaction and not equal.
Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.
[tex]K_c=\frac{[NO_2]^2}{[N_2O_4]}[/tex]
Final answer:
Statements 1 and 3 are False, and 2 and 4 are True. At equilibrium, the rate of formation and dissociation of N₂O₄ are equal but their concentrations are not necessarily the same. Rate constants for forward and reverse reactions are different yet define the equilibrium constant. The concentration ratio of NO₂ to N₂O₄ is constant at a given temperature.
Explanation:
When dinitrogen tetroxide (N₂O₄) is in equilibrium with nitrogen dioxide (NO₂), the following statements can be made:
1. The concentration of NO₂ is equal to the concentration of N₂O₄. False - The stoichiometry of the reaction is 1:2, so for every molecule of N₂O₄ that dissociates, two molecules of NO₂ are formed.2. The rate of the dissociation of N₂O₄ is equal to the rate of formation of N₂O₄. True - At equilibrium, the rate of the forward reaction (dissociation of N₂O₄) is equal to the rate of the reverse reaction (formation of N₂O₄).3. The rate constant for the forward reaction is equal to the rate constant of the reverse reaction. False - The rate constants for the forward and reverse reactions are generally different, but their ratio is constant at a given temperature, defining the equilibrium constant.4. The concentration of NO₂ divided by the concentration of N₂O₄ is equal to a constant. True - At equilibrium, this ratio is equal to the equilibrium constant (Keq) for the reaction at a given temperature.A chemist adds of a sodium carbonate solution to a reaction flask. Calculate the mass in grams of sodium carbonate the chemist has added to the flask. Be sure your answer has the correct number of significant digits.
The question is incomplete, here is the complete question:
A chemist adds 180.0 mL of a 1.42 M sodium carbonate solution to a reaction flask. Calculate the mass in grams of sodium carbonate the chemist has added to the flask. Be sure your answer has the correct number of significant digits.
Answer: The mass of sodium carbonate that must be added are 40.9 grams
Explanation:
To calculate the mass of solute, we use the equation used to calculate the molarity of solution:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
Molar mass of sodium carbonate = 106 g/mol
Molarity of solution = 1.42 M
Volume of solution = 180.0 mL
Putting values in above equation, we get:
[tex]1.42mol/L=\frac{\text{Mass of sodium carbonate}\times 1000}{160g/mol\times 180}\\\\\text{Mass of sodium carbonate}=\frac{160\times 1.42\times 180}{1000}=40.9g[/tex]
Hence, the mass of sodium carbonate that must be added are 40.9 grams
Be sure to answer all parts. Zinc is an amphoteric metal, meaning it reacts with both acids and bases. The standard reduction potential is −1.36 V for the following reaction: (1)Zn(OH)42−(aq) + 2e− → Zn(s) + 4OH−(aq) Calculate the formation constant Kf for the reaction: (2)Zn2+(aq) + 4OH−(aq) ⇌ Zn(OH)42−(aq)
Answer:
[tex]1.86*10^{20[/tex]
Explanation:
The equation for the reaction is:
[tex]Zn^{2+}_{(aq)} + 4OH^-_{(aq)}[/tex] ⇄ [tex]Zn(OH)^{2-}_{4(aq)}[/tex]
Oxidation can be defined as the addition of oxygen, removal of hydrogen and/or loss of electron during an electron transfer. Oxidation process occurs at the anode.
On the other hand; reduction is the removal of oxygen, addition of hydrogen and/ or the process of electron gain during an electron transfer. This process occurs at the cathode.
The oxidation-reduction process with its standard reduction potential is as follows:
[tex]Zn(OH)^{2-}_{4(aq)} + 2e^- ----->Zn_{(s)} +4OH^-_{(aq)}[/tex] [tex]E^0_{anode} = -1.36 V[/tex]
At the zinc electrode (cathode); the reduction process of the reaction with its standard reduction potential is :
[tex]Zn^{2+}_{(aq)} +2e^- -----> Zn_{(s)}[/tex] [tex]E^0_{cathode} = -0.76 V[/tex]
The standard cell potential [tex]E^0_{cell}[/tex] is given as:
[tex]E^0_{cell}=E^0_{cathode}-E^0_{anode}[/tex]
[tex]E^0_{cell}[/tex] = -0.76 V - (- 1.36 V)
[tex]E^0_{cell}[/tex] = -0.76 V + 1.36 V
[tex]E^0_{cell}[/tex] = +0.60 V
Now to determine the formation constant [tex]k_f[/tex] of the [tex]E^0_{cell}[/tex] ; we use the expression:
[tex]E^0_{cell}[/tex] = [tex]\frac{RT}{nF}Ink_f[/tex]
where;
[tex]E^0_{cell}[/tex] = +0.60 V
R = universal gas constant = 8.314 J/mol.K
T = Temperature @ 25° C = (25+273)K = 298 K
n = numbers of moles of electron transfer = 2
F = Faraday's constant = 96500 J/V.mol
[tex]+ 0.60 V = \frac{(8.314)(298)}{n(96500)} Ink_f[/tex]
[tex]+0.60 V = \frac{(0.0257)}{n}Ink_f[/tex]
[tex]+0.60V = \frac{0.0592}{n}log k_f[/tex]
[tex]logk_f = \frac{+0.60V*n}{0.0592}[/tex]
[tex]logk_f = \frac{+0.60V*2}{0.0592}[/tex]
[tex]logk_f = 20.27[/tex]
[tex]k_f= 10^{20.27}[/tex]
[tex]k_f = 1.86*10^{20}[/tex]
Therefore, the formation constant [tex]k_f[/tex] for the reaction is = [tex]1.86*10^{20[/tex]