The molar concentration of phosphate (present as dihydrogen phosphate) in seawater can be calculated by converting the given mass concentration of phosphate (0.07 ppm) into moles using the molar mass of phosphate. After this conversion, the phosphate concentration is approximately 7.14 x 10^-4 mol/L.
Explanation:The question asks us to calculate the molar concentration of the dihydrogen phosphate ion in seawater. Remember that molar concentration is the amount of a solute (in this case phosphorus as dihydrogen phosphate) divided by the volume of the solution (seawater). We are given that the concentration of phosphorus is 0.07 ppm by mass, which, converted into grams per liter (g/L), yields 0.07 g/L. Additionally, the molar mass of H₂PO₄⁻ is approximately 97.994 g/mol.
Step-by-step calculation:
First, convert the given mass concentration of phosphorus into moles by using the molar mass of phosphorus: (0.07 g/L) / (97.994 g/mol) = 7.14 x 10⁻⁴ mol/L.Since the phosphorus was assumed to be in the form of dihydrogen phosphate, we don't need any further calculations as the molar concentration of dihydrogen phosphate is equal to that of phosphorus.Therefore, the molar concentration of phosphate in seawater is approximately 7.14 x 10⁻⁴ mol/L.
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For the reaction X2 + Y + Z -> XY + XZ it is found that doubling the concentration of X2 doubles the reaction rate. triplingthe concentration on triples the rate, and doubling the concentration of 2 has no effect, (a) What is the rate law for thisreaction? (b) Why is it that the change in the concentration of 2 has no effect on the rate? (c) Suggest a mechanism for thereaction that is consistent with the rate law.
Answer:
(a) R=k[X2][Y] (b) reaction is zero order wrt Z (c) X2 + Y --- XY + X (slow step)
X + Z --- XZ (fast step)
Explanation:
(a) Suppose the reaction rate R with respect to a component X2 with concentration [X2] is generally expressed as follows:
R = k [X2]^n (1)
Where k is the rate constant and n is the order of reaction with respect to X2.
When the reaction rate is found to double by doubling the concentration of X2, the following equation could be developed:
2R = k *(2[X2])^n (2)
Dividing equation (2) by (1) yields
2R/R= k *(2[X2])^n / k *[X2]^n
2=2^n , n = 1
Thus, the reaction is first order with respect to X2.
In the same manner,
Tripling the concentration of Y triples the rate,
R = k [Y]^m (3)
Where k is the rate constant and m is the order of reaction with respect to Y.
When the reaction rate is found to triple by tripling the concentration of Y, the following equation could be developed:
2R = k *(3[Y])^m (4)
Dividing equation (4) by (3) yields
3R/R= k *(3[Y])^m / k *[Y]^m
3=3^m , m = 1
Thus, the reaction is first order with respect to Y.
Therefore,the rate law for this reaction is R = k [X2] [Y]
(b)
Provided that doubling the concentration of Z has no effect, we perform similar analysis as above with p representing order of reaction wrt Z:
R = k [Z]^p (5)
R = k (2*[Z]^p (6)
R/R= k *(2[Z])^p / k *[Z]^p
1=2^p , p = 0
Thus, the reaction is zero order wrt Z. This explains why the change in concentration has no effect on the rate.
(c) A suggested mechanism for the reaction that is consistent with the rate law will therefore be
X2 + Y ---- XY + X (slow step)
X + Z ---- XZ (fast step)
The rate law is determined by the slow step, and this is very consistent with the experimentally observed data.
Besides, addition of the two step yields the overall reaction, and both steps are reasonable.
Final answer:
The rate law is Rate = k[X2][Y] indicating first order dependencies on X2 and Y. Z has no effect on the rate, suggesting it's not involved in the rate-determining step, which implies a reaction mechanism with an initial slow step not involving Z.
Explanation:
Rate Law and Reaction Mechanism
Based on the given information, we can deduce the rate law for the reaction. As the rate doubles when the concentration of X2 is doubled, and it triples when the concentration of Y is tripled, this suggests first order dependencies with respect to X2 and Y. There is no effect on the rate when the concentration of Z is doubled, which means Z is zero order in the rate law. Therefore, the rate law can be expressed as:
Rate = k[X2][Y]
The change in concentration of Z has no effect on the rate because it is not involved in the rate-determining step of the mechanism. The reaction mechanism likely involves a slow initial step that does not include Z, followed by a rapid step that includes it. Thus, the overall rate of the reaction is controlled by the first, slower step in the mechanism.
Two changes are described below. a. Ice melts to form water. b. Sugar cubes dissolve in hot coffee. Which statement is true about the two changes?a. Both are chemical changes because there is a change in the composition of ice and sugar.b. Both are physical changes because there is a change in the physical states of ice and sugar.c. A is a physical change due to a change of state, but B is a chemical change because new molecules are formed.d. A is a chemical change due to a change of state, but B is a physical change because new molecules are formed.
Answer: Option (b) is the correct answer.
Explanation:
A change that does not lead to any difference in chemical composition of a substance is known as a physical change.
For example, shape, size, mass, volume, density, boiling point, etc of a substance are all physical properties.
As ice melts to form water shows that only the state of matter is changing. Hence, it is a physical change. Similarly, sugar cubes dissolve in hot coffee is also a physical change as no new compound has formed.
On the other hand, changes that lead to bring change in chemical composition of a substance is known as a chemical change.
For example, exploding dynamite, rotting cheese etc are all chemical changes.
Thus, we can conclude that both are physical changes because there is a change in the physical states of ice and sugar.
Of the atoms listed below, which one will have at least one electron in its d orbital?
A. Mg
B. K
C. S
D. Cr
E. none of the above
Answer:
D. Cr
Explanation:
In order to determine which atom has at least one electron in its d orbital, we have to write their theoretical electron configurations.
₁₂Mg 1s² 2s² 2p⁶ 3s²₁₉K 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹₁₆S 1s² 2s² 2p⁶ 3s² 3p⁴₂₄Cr 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁴Cr has 4 electrons in d orbitals. Cr belongs to the d-block in the periodic table.
Dinitrogen tetroxide partially decomposes according to the following equilibrium: N2O4 (g) → 2 NO2 (g)
A 1.000-L flask is charged with 3.00 × 10-2 mol of N2O4. At equilibrium, 2.36 × 10-2 mol of N2O4 remains.
Keq for this reaction is __________.
Answer:
Keq for this reaction is 6.94x10⁻³
Explanation:
The equilibrium equation is this one:
N₂O₄ (g) ⇄ 2NO₂ (g)
Initially we have 0.03 moles from the dinitrogen tetroxide and nothing from the dioxide.
In the reaction, some amount of compound (x) has reacted.
As ratio is 1:2, we have double x in products.
Finally in equilibrium we have:
N₂O₄ (g) ⇄ 2NO₂ (g)
0.03 - x 2x
And we know [N₂O₄] in equilibrium so:
0.03 - x = 0.0236
x = 0.03 - 0.0236 → 6.4x10⁻³
As this is the amount that has reacted, in equilibrium I have produced:
6.4x10⁻³ .2 = 0.0128 moles of NO₂
This is the expression for K,
[NO₂] ² / [N₂O₄]
0.0128² / 0.0236 = 6.94x10⁻³
Answer:
6.94x10-3
Explanation:
What element is being oxidized in the following redox reaction?
MnO4 (aq) + H2C2O4(aq) → Mn2+(aq) + CO2(g)
Answer:
C is the element thats has been oxidized.
Explanation:
MnO₄⁻ (aq) + H₂C₂O₄ (aq) → Mn²⁺ (aq) + CO₂(g)
This is a reaction where the manganese from the permanganate, it's reduced to Mn²⁺.
In the oxalic acid, this are the oxidation states:
H: +1
C: +3
O: -2
In the product side, in CO₂ the oxidation states are:
C: +4
O: -2
Carbon from the oxalate has increased the oxidation state, so it has been oxidized.
The zinc within a copper-plated penny will dissolve in hydrochloric acid if the copper coating is filed down in several spots (so that the hydrochloric acid can get to the zinc). The reaction between the acid and the zinc is as follows: 2H+(aq)+Zn(s)→H2(g)+Zn2+(aq). When the zinc in a certain penny dissolves, the total volume of gas collected over water at 25 ∘C was 0.949 L at a total pressure of 758 mmHg .
Answer:
Mass of hydrogen gas evolved is 0.0749 grams.
Explanation:
Total pressure of the gases = p = 758 mmHg
Vapor pressure of water = 23.78 mmHg
Pressure of hydrogen gas ,P = p - 23.78 mmHg = 758 mmHg - 23.78 mmHg
P = 734.22 mmHg = [tex]\frac{734.22}{760} atm=0.966 atm[/tex]
Temperature of of hydrogen gas ,T= 25°C =298.15 K
Volume of hydrogen gas = V = 0.949 L
Moles of hydrogen gas =n
PV = nRT (Ideal gas equation )
[tex]n=\frac{PV}{RT}=\frac{0.966 atm\times 0.949 L}{0.0821 atm L/mol K\times 298.15 K}[/tex]
n = 0.03745 mol
Moles of hydrogen gas = 0.03745 mol
Mass of 0.03745 moles of hydrogen gas = 0.03745 mol × 2 g/mol = 0.0749 g
Mass of hydrogen gas evolved is 0.0749 grams.
The human body obtains 1062 kJ from a candy bar.If this energy were used to vaporize water at 100 ∘C, how much water in liters could be vaporized? (Assume that the density of water is 1.0 g/mL.) The heat of vaporization of water at 100 ∘C is 40.7 kJ/mole.
Answer:
The Answer Of this question is 0.4052 L ...
The van Deemter equation describes plate height in terms of constants A, B, C, and the linear velocity, ux. Which statement is NOT true for the van Deemter equation?
A) A takes into account multiple pathways through the column. The value of A is column specific and independent of linear flow.
B) B takes into account longitudinal diffusion of the analyte in the mobile phase. The value of B is column specific and the impact of B on the plate height is inversely proportional to the linear velocity. x x + Cu u B H » A +
C) C takes into account equilibrium time between the stationary and mobile phase. The value of C is column specific and the impact of C on plate height is proportional to the linear velocity.
D) To minimize plate height, the optimal flow rate is the maxima for the plot of versus flow rate.
E) Linear velocity is the flow rate of the mobile phase.
Answer:
D) To minimize plate height, the optimal flow rate is the maxima for the plot of plate height versus flow rate.
Explanation:
Van Deemter equation in chromatography, relates the variance per unit length of a separation column to the linear mobile phase velocity by considering physical, kinetic, and thermodynamic properties of a separation.
The formula is:
H = A + B/u + C×u
Where:
H = HETP (plate height)
A = Eddy diffusion term .
B = Longitudinal diffusion.
C = Resistance against mass transfer.
u = Linear velocity .
A) A takes into account multiple pathways through the column. The value of A is column specific and independent of linear flow. TRUE. Eddy difussion term depends of column packing
B) B takes into account longitudinal diffusion of the analyte in the mobile phase. The value of B is column specific and the impact of B on the plate height is inversely proportional to the linear velocity. TRUE. Longitudinal difussion is another term that is specific to a column. Also, as the formula is H = A + B/u + C×u, the impact of B on the plate height is inversely proportional to the linear velocity.
C) C takes into account equilibrium time between the stationary and mobile phase. The value of C is column specific and the impact of C on plate height is proportional to the linear velocity. TRUE. C is the time that system needs to equilibirum. Based on the formula, the impact of C on plate height is proportional to the linear velocity
D) To minimize plate height, the optimal flow rate is the maxima for the plot of plate height versus flow rate. FALSE. The optimal flow is the minimum of the graph
E) Linear velocity is the flow rate of the mobile phase. TRUE. The Van Deemter equation describes u as the linear velocity of the mobile phase.
I hope it helps!
Final answer:
The correct statement that is NOT true for the van Deemter equation is option D) To minimize plate height, the optimal flow rate is the maxima for the plot of versus flow rate.
Explanation:
The correct statement that is NOT true for the van Deemter equation is option D) To minimize plate height, the optimal flow rate is the maxima for the plot of versus flow rate.
This means that in order to minimize the plate height, the optimal flow rate is not at the maximum point on the plot of plate height versus flow rate. The optimal flow rate is usually at a point of minimum plate height.
Express the equilibrium constant for the following reaction.P(g) + 3/2 Cl2(g) ↔ PCl3(g)K = [PCl3]^2/[P]^2[Cl2]^3K = [PCl3]/[P][Cl2]^3/2K = [P][Cl2]^3/2/[PCl3]K = [P]^1/2[Cl2]^1/3/[PCl3]^1/2K = [PCl3]^2/[P]2[Cl2]3
The correct expression for the equilibrium constant (K) is:
K = [PCl₃]² / [P] × [Cl₂](3/2)
The correct expression for the equilibrium constant (K) of the given reaction is:
K = [PCl₃]² / [P] × [Cl₂](3/2)
In this expression:
[PCl₃] represents the molar concentration of PCl₃ (gaseous product).
[P] represents the molar concentration of P (reactant P in its gaseous state).
[Cl₂] represents the molar concentration of Cl₂ (reactant Cl₂ in its gaseous state).
The exponents in the expression correspond to the stoichiometric coefficients in the balanced chemical equation:
P(g) + 3/2 Cl₂(g) ↔ PCl₃(g)
The stoichiometric coefficients of P and Cl₂ are both 1, and the coefficient of PCl₃ is 1. To make the equation balanced, we need to multiply Cl₂ by 3/2 (1.5).
So, the correct expression for the equilibrium constant (K) is:
K = [PCl₃]² / [P] × [Cl₂](3/2)
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The following (unbalanced) reaction is one of the steps to producing acid rain [as H2SO4(aq)] from sulfur-containing coal. In a study of this reaction, a flask contains an equilibrium concentration of the three substances SO2(g), O2(g) and SO3(g) held at a constant temperature.
SO2 (g) + O2 (g) <-----> SO3 (g)
The concentrations at equilibrium are found to be: [SO2] = 3.61 x 10-3M [O2] = 6.11 x 10-4 M [SO3] = 1.01 x 10-2 M
(a) Write the Equilibrium Constant expression, Kc for this specific reaction.
(b) Calculate the value of Kc for this temperature.
(c) Would you classify this as a product-favored reaction?
Answer:
a) kc= [SO3 ]/([SO2 ][O2 ])
b) kc= 2.27*10⁶ M⁻¹
v) the reaction is product-favored
Explanation:
for the reaction, the equilibrium constant is
SO2 (g) + O2 (g) <-----> SO3 (g)
he equilibrum constant is
kc= [SO3 ]/([SO2 ]*[O2 ])
replacing values
kc= [SO3 ]/([SO2 ]*[O2 ]) = 1.01*10⁻² M/(3.61*10⁻³M*6.11 x 10⁻⁴ M) = 2.27*10⁶ M⁻¹
since kc>>1 the reaction is product-favored
The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In the first step of the Ostwald process, ammonia is reacted with oxygen gas to produce nitric oxide and water. What is the maximum mass of H 2 O H2O that can be produced by combining 52.3 g 52.3 g of each reactant? 4 NH 3 ( g ) + 5 O 2 ( g ) ⟶ 4 NO ( g ) + 6 H 2 O ( g )
Answer:
35.3124 g is the maximum mass of [tex]H_2O[/tex] that can be produced.
Explanation:
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
For [tex]NH_3[/tex] :-
Mass of [tex]NH_3[/tex] = 52.3 g
Molar mass of [tex]NH_3[/tex] = 17.031 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{52.3\ g}{17.031\ g/mol}[/tex]
[tex]Moles\ of\ NH_3= 3.0709\ mol[/tex]
For [tex]O_2[/tex] :-
Given mass of [tex]O_2[/tex]= 52.3 g
Molar mass of [tex]O_2[/tex] = 31.9898 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{52.3\ g}{31.9898\ g/mol}[/tex]
[tex]Moles\ of\ O_2=1.6349\ mol[/tex]
According to the given reaction:
[tex]4NH_3+5O_2\rightarrow 4NO_4+6H_2O[/tex]
4 moles of [tex]NH_3[/tex] reacts with 5 moles of [tex]O_2[/tex]
1 mole of [tex]NH_3[/tex] reacts with 5/4 moles of [tex]O_2[/tex]
Also,
3.0709 moles of [tex]NH_3[/tex] reacts with [tex]\frac{5}{4}\times 3.0709[/tex] moles of [tex]O_2[/tex]
Moles of [tex]O_2[/tex] = 3.8386 moles
Available moles of [tex]O_2[/tex] = 1.6349 moles
Limiting reagent is the one which is present in small amount. Thus, [tex]O_2[/tex] is limiting reagent.
The formation of the product is governed by the limiting reagent. So,
5 moles of [tex]O_2[/tex] on reaction forms 6 moles of [tex]H_2O[/tex]
1 mole of [tex]O_2[/tex] on reaction forms 6/5 moles of [tex]H_2O[/tex]
Thus,
1.6349 mole of [tex]O_2[/tex] on reaction forms [tex]\frac{6}{5}\times 1.6349[/tex] moles of [tex]H_2O[/tex]
Moles of [tex]H_2O[/tex] = 1.9618 moles
Molar mass of [tex]H_2O[/tex] = 18 g/mol
Mass of sodium sulfate = Moles × Molar mass = 1.9618 × 18 g = 35.3124 g
35.3124 g is the maximum mass of [tex]H_2O[/tex] that can be produced.
Answer:
35.3 g
Explanation:
From the balanced equation given we can say:
4 moles of NH3 reacts with 5 moles of O2 to give 6 moles of H2O.
4*17 g of NH3 reacts with 5*32 g of O2 to give 6*18 g of H2O.
68 g of NH3 reacts with 160 g of O2 to give 108 g of H2O.
Here the limiting reagent is O2 and excess reagent is NH3.
52.3 g of O2 will react with [tex]\frac{68}{160}\times52.3=22.23\ g\ of\ NH_{3}[/tex] to give :
[tex]\frac{108}{160}*52.3=35.3\ g\ of\ H_{2}O[/tex]
Hence the maximum mass of H2O that can be produced by 52.3 g of reactants is 35.3 g.
Research and draw (on paper) the isomers of another coordination compound that has octahedral geometry.
Answer:
Sulfur hexafluoride SF6
Explanation:
In chemistry, the shape of the compounds in which six ligands (atoms, molecules or ions) are arranged around a central atom or ion, defining the vertices of an octahedron, is called octahedral molecular geometry or Oh. It is a very common structure, and it is very studied for its importance in the coordination chemistry of transition metals. From it, other important molecular geometries are derived by continuous deformation, such as the elongated octahedron, the flat octahedron, the square-based pyramid and the flat square. Indirectly, it is also related to tetrahedral molecular geometry.
The concept of octahedral coordination geometry was developed by Alfred Werner to explain the stoichiometry and isomeries in the coordination compounds. An example of a strictly octahedral compound is SF6 sulfur hexafluoride, but chemists use the term in a lax form, so that it is applied to compounds that are not mathematically octahedra, such as cobalt hexaamine (III).
Isomers are molecules that have the same molecular formula but different structure. It is classified as structural isomers and stereoisomers. Structural isomers differ in the way of joining their atoms and are classified into chain, position and function isomers.
Follow the Carbon in Citric Acid Cycle: 1. How many cycles does it take to convert Acetyl CoA to CO2
Each round of Citric Acid Cycle produce 2 molecules of carbon dioxide. So one cycle of TCA cycle is enough to convert Acetyl CoA to carbon dioxide.
Explanation:Citric Acid Cycle or Kreb's Cycle or the TCA cycle is the 1st dedicated step towards the aerobic respiration. The end product of glycolysis is Pyruvate which is a three carbon compound. It's acted upon by Pyruvate Decarboxylase to produce a 2 carbon compound Acetyl CoA and a molecule of carbon dioxide. This Acetyl CoA now reacts with oxaloacetate to produce Citric Acid which is the 1st step of Citric Acid Cycle. This now produce several intermediates and a lot of reduced electron carriers along with 2 molecules of carbon dioxide and ends up being oxaloacetate again. So one cycle of Citric Acid Cycle is necessary to convert Acetyl CoA to CO2.
At 10 K Cp,m(Hg(s)) = 4.64 J K−1 mol−1. Between 10 K and the melting point of Hg(s), 234.3 K, heat capacity measurements indicate that the entropy increases by 57.74 J K−1 mol−1. The standard enthalpy of fusion of Hg(s) is 2322 J mol−1 at 234.3 K. Between the melting point and 298.0 K, heat capacity measurements indicate that the entropy increases by 6.85 J K−1 mol−1. Determine the Third-Law standard molar entropy of Hg(l) at 298 K.
Answer:
S°m,298K = 85.184 J/Kmol
Explanation:
∴ T = 10 K ⇒ Cp,m(Hg(s)) = 4.64 J/Kmol
∴ 10 K to 234.3 K ⇒ ΔS = 57.74 J/Kmol
∴ T = 234.3 K ⇒ ΔHf = 2322 J/mol
∴ 234.3 K to 298.0 K ⇒ ΔS = 6.85 J/Kmol
⇒ S°m,298K = S°m,0K + ∫CpdT/T(10K) + ΔS(10-234.3) + ΔHf/T(234.3K) + ΔS(234.3-298)
⇒ S°m,298K = 0 + 10.684 J/Kmol + 57.74 J/Kmol + 9.9104 J/Kmol + 6.85 J/kmol
⇒ S°m,298K = 85.184 J/Kmol
To find the Third-Law standard molar entropy of Hg(l) at 298 K, we sum up the entropy changes between 10K and 234.3 K, the entropy change due to fusion, and then the entropy change between the melting point and 298K. The sum gives us a final value of 74.51 J K−1 mol−1.
Explanation:The Third-Law standard molar entropy of Hg(l) at 298 K can be calculated by summing up the entropy changes that occur from 10K to 298K.
First calculate the entropy up to the melting point from 10 K which can be determined using the equation ΔS = ∫(Cp,mdT)/T.
This integral can be approximated as a rectangle from 10K to 234.3 K, hence ΔS₁ = (234.3-10)(4.64 J K−1 mol−1)/10 = 57.74 J K−1 mol−1.
Next, calculate the entropy change associated with the fusion process using the equation, ΔS = ΔH/T, giving ΔS₂ = 2322 J mol−1 / 234.3 K = 9.92 J K−1 mol−1.
Finally, add the entropy increase from the melting point to 298.0 K, which is given as 6.85 J K−1 mol−1.
Summing these values gives the Third-Law standard molar entropy of Hg(l) at 298 K: 57.74 J K−1 mol−1 + 9.92 J K−1 mol−1 + 6.85 J K−1 mol−1 = 74.51 J K−1 mol−1.
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Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an electrolytic cell. How many hours will it take to plate 15.0 kg of copper onto the cathode if the current passed through the cell is held constant at 34.5 A ? Assume the copper in the electrolytic solution is present as Cu 2 + .
Final answer:
To electroplate 15.0 kg of copper onto the cathode using 34.5 A, first calculate the moles of copper needed then convert this to the required charge using Faraday's constant. Finally, divide the total charge by the current to find the time in seconds and convert to hours, resulting in approximately 367 hours.
Explanation:
To calculate the time required to electroplate 15.0 kg of copper using a current of 34.5 A, we need to use Faraday's laws of electrolysis. First, we need to determine the number of moles of copper to be plated. The molar mass of copper is approximately 63.55 g/mol, so:
15,000 g / 63.55 g/mol = 236.025 mol Cu
Since each copper ion (Cu2+) requires two electrons to be reduced to copper metal, the number of moles of electrons needed is twice the number of moles of copper:
2 × 236.025 mol = 472.05 mol e-
Each mole of electrons corresponds to a charge of 96,485 coulombs (Faraday's constant), so:
472.05 mol e- × 96,485 C/mol = 45,562,240.25 C
Now, we can calculate the time required to deliver this charge at a rate of 34.5 A (since 1 A = 1 C/s), using:
Time (s) = Total Charge (C) / Current (A)
Time (s) = 45,562,240.25 C / 34.5 A = 1,320,643 s
Converting seconds to hours:
1,320,643 s / (60 s/min) / (60 min/h) ≈ 367 h
Therefore, it will take approximately 367 hours to electroplate 15.0 kg of copper onto the cathode with a constant current of 34.5 A.
Part A Find ΔS∘ for the reaction between nitrogen gas and hydrogen gas to form ammonia: 12N2(g)+32H2(g)→NH3(g) Express your answer using one decimal place. ΔS∘Δ S ∘ = J/K Previous Answer Request Answer Incorrect; Try Again; 3 attempts remaining Part B Complete previous part(s) Provide Feedback
Answer:
[tex]\Delta S^{0}[/tex] for the given reaction is -99.4 J/K
Explanation:
Balanced reaction: [tex]\frac{1}{2}N_{2}(g)+\frac{3}{2}H_{2}(g)\rightarrow NH_{3}(g)[/tex]
[tex]\Delta S^{0}=[1mol\times S^{0}(NH_{3})_{g}]-[\frac{1}{2}mol\times S^{0}(N_{2})_{g}]-[\frac{3}{2}mol\times S^{0}(H_{2})_{g}][/tex]
where [tex]S^{0}[/tex] represents standard entropy.
Plug in all the standard entropy values from available literature in the above equation:
[tex]\Delta S^{0}=[1mol\times 192.45\frac{J}{mol.K}]-[\frac{1}{2}mol\times 191.61\frac{J}{mol.K}]-[\frac{3}{2}mol\times 130.684\frac{J}{mol.K}]=-99.4J/K[/tex]
So, [tex]\Delta S^{0}[/tex] for the given reaction is -99.4 J/K
The change in entropy in the reaction forming ammonia from nitrogen and hydrogen gas can be calculated using the stoichiometry of the balanced equation and the entropy values of the reactants and products.
Explanation:The question is, 'Find ΔS∘ for the reaction between nitrogen gas and hydrogen gas to form ammonia: 1/2N2(g) + 3/2H2(g) → NH3(g)' To calculate this, we need to consider the stoichiometry of the balanced chemical equation, which tells us that one mole of N₂ will react with three moles of H₂ to form two moles of NH3. The stoichiometric factors derived from this equation can be used to determine the change in entropy (ΔS∘) for the reaction.
The change in entropy for a reaction can be determined using the equation ΔS∘ = ΣS∘(products) - ΣS∘(reactants). As per the balanced chemical equation N₂(g) + 3H₂(g) → 2NH3(g), ΔS∘ would be equal to the sum of the absolute entropies of 2 moles of NH3 (products) minus the sum of the absolute entropies of 1/2 moles of N2 and 3/2 moles of H2 (reactants). The specific entropy values would depend on data provided in a standard thermodynamic table.
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Compare and contrast the three different types of friction. Explain one way they are all alike, and at least one way they are different.
Answer:
The correct answer is static, sliding and rolling friction.
Explanation:
There are three different types of friction. Static, sliding and rolling friction occur between solid surfaces. Static friction is strongest, followed by sliding friction and then rolling friction, which is the weakest of the three. The equality of these three types of friction is that they produce heat and make movement difficult. The difference is in their magnitude and in the conditions that produce each type of friction. Static is produced when a body at rest begins to move, sliding is produced when this body is already in motion and rolling is produced when a body rolls on a surface, deforming one or both of them.
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A solid sample of Rb2SO3 weighing 6.24 g reacts with 1.38 L gaseous HBr, measured at 75°C and 0.953 atm pressure. The solid RbBr, extracted from the reaction mixture and purified, has a mass of 7.32 g.
(a) What is the limiting reactant?
(b) What is the theoretical yield of RbBr, assuming com- plete reaction?
(c) What is the actual percentage yield of product?
Answer:
(a) HBr;
(b) 7.61 g;
(c) 96.2 %
Explanation:
Firstly, write the balanced chemical equation:
[tex]Rb_2SO_3 (aq) + 2 HBr (aq)\rightarrow 2 RbBr (aq) + SO_2 (g) + H_2O (l)[/tex]
(a) Find moles of each reactant dividing the mass by the molar mass of rubidium sulfite, then applying the ideal gas law for HBr:
[tex]n_{Rb_2SO_3}=\frac{6.24 g}{251.00 g/mol} = 0.02486 mol[/tex]
[tex]pV_{HBr}=n_{HBr}RT[/tex]
[tex]\therefore n_{HBr} = \frac{pV_{HBr}}{RT} = \frac{0.953 atm\cdot 1.38 L}{0.08206 \frac{L atm}{mol K}\cdot 348.15 K} = 0.04603 mol[/tex]
Find the limiting reactant by dividing each moles by the stoichiometric coefficients and comparing the two numbers:
[tex]eq._{Rb_2SO_3} = \frac{0.02486 mol}{1} = 0.02486 mol[/tex]
[tex]eq._{HBr} = \frac{0.04603 mol}{2} = 0.02302 mol[/tex]
That said, the equivalent of HBr is lower, so it's the limiting reactant.
(b) According to the balanced equation, the moles of HBr are equal to the moles of RbBr, so moles of RbBr theoretically are equal to:
[tex]n_{RbBr} = 0.04603 mol[/tex]
Using the molar mass of RbBr, convert this into mass:
[tex]m_{RbBr} = 0.04603 mol\cdot 165.372 g/mol = 7.61 g[/tex]
(c) To find the percent yield, divide the actual mass produced by the theoretical mass calculated in (b) and multiply by 100 %:
[tex]\%_{yield} =\frac{7.32 g}{7.61 g}\cdot 100\% = 96.2 \%[/tex]
My grandmother left me some silver years ago. I opened the box containing the silver and it had tarnished. Using the equation below, if I had 300 g of silver that had tarnished and 175g H2S, which would be the limiting reactant? 2 Ag(s) + H2S(g) → Ag2S(s) + H2(g) A. hydrogen B. silver C. There is no limiting reactant in this case D. hydrogen sulfide E. silver sulfide
Answer:
B
Explanation:
The limiting reagent in a chemical reaction is the one that is consumed completely in the course of the reaction. It dictates the extent to which the reaction would proceed as the completion of the reaction is based solely on it.
There are several ways to determine the limiting reagent. The easiest way to do this that works is to divide the number of moles of each reagent by their stoichiometric coefficient in the balanced equation. The reactant with the least value is the limiting reagent.
Thus, we need to calculate the number of moles of silver and hydrogen sulphide. To do this, we simply divide the masses by the relative atomic masses or molecular masses.
The atomic mass of silver is 108.
The number of moles of silver is =
300/108 = 2.8
We now divide this by the stoichiometric coefficient: 2.8/2 = 1.4
We do same for hydrogen sulphide. The molar mass of hydrogen sulphide is 34g/mol
The number of moles is thus 175/34 = 5.15
We now divide this by stoichiometric coefficient = 5.15/1 = 5.15
We can see that silver has less number of the value and hence it is the limiting reactant.
How many grams of coffee must evaporate from 350 g of coffee in a 100-g glass cup to cool the coffee and the cup from 95.0°C to 45.0°C ? Assume the coffee has the same thermal properties as water and that the average heat of vaporization is 2340 kJ/kg (560 kcal/g). Neglect heat losses through processes other than evaporation, as well as the change in mass of the coffee as it cools. Do the latter two assumptions cause your answer to be higher or lower than the true answe
Answer:
31.3 g
The answer is higher than the true answer.
Explanation:
By neglecting the heat lost by other processes, the energy conservation states that:
Qcooling + Qevaporate = 0
The cooling process happens without phase change, so the heat can be calculated by:
Qcooling = m*c*ΔT
Where m is the mass, c is the heat capacity (cwater = 4184 J/kg.K), and ΔT is the temperature variation (final - initial).
The evaporate process happen without changing of temperature (pure substance), and the heat can be calculated by:
Qevaporate = m*L
Where m is the mass evaporated and L is the heat of evaporation (2340000 J/kg).
0.350*4184*(45 - 95) + m*2340000 = 0
2340000m = 73220
m = 0.0313 kg
m = 31.3 g
Because of the assumptions made, the real mass is not that was calculated. There'll be changing mass when the coffee is cooling, and there'll be heat loses by other processes because the system is not isolated. Also, the substance is not pure. So, there'll be more factors at the energy equation, thus, the answer is higher than the true answer.
To cool 350 g of coffee in a 100-g glass cup from 95.0°C to 45.0°C, 33.2 grams of coffee must evaporate.
To solve this problem, we first need to calculate the total heat that needs to be removed from the coffee and the cup.
Steps to Calculate:
Calculate the heat loss needed to cool the coffee:Thus, 33.2 grams of coffee must evaporate to cool the coffee and the cup from 95.0°C to 45.0°C. Neglecting other heat losses means this answer is slightly larger than the true answer.
A thermos bottle (Dewar vessel) has an evacuated space between its inner and outer walls to diminish the rate of transfer of thermal energy to or from the bottle’s contents. For good insulation, the mean free path of the residual gas (air; average molecular mass 5 29) should be at least 10 times the distance between the inner and outer walls, which is about 1.0 cm. What should be the maximum residual gas pressure in the evacuated space if T 5 300 K? Take an average diameter of d 5 3.1 × 10210 m for the molecules in the air.
There are quite a bunch of typo errors in the question; here is the correct question below:
A thermos bottle (Dewar vessel) has an evacuated space between its inner and outer walls to diminish the rate of transfer of thermal energy to or from the bottle’s contents. For good insulation, the mean free path of the residual gas (air; average molecular mass = 29) should be at least 10 times the distance between the inner and outer walls, which is about 1.0 cm. What should be the maximum residual gas pressure in the evacuated space if T = 300 K? Take an average diameter of d = 3.1 × 10⁻¹⁰ m for the molecules in the air.
Answer:
9.57 × 10⁻⁷ atm
Explanation:
The mean free path ( λ ) can be illustrated by the equation:
λ = [tex]\frac{1}{\sqrt{2} \pi d{^2}N/V }[/tex] ---------- (1)
N/V = [tex]\frac{N_AP}{RT}[/tex] ------------- (2)
From the above relation, we can deduce that;
P= [tex]\frac{RT}{\sqrt{2}\pi d{^2}N{_A}λ }[/tex] -------------(3)
let I= λ
From the above equations;
d= diameter of the atom
[tex]{N_A}[/tex] = avogadro's constant
P= pressure
R= universal rate constant which is given by 0.08206 L atm mol⁻¹ k⁻¹
T= temperature
From the question, we are given that the mean free path of the residual air molecules ( d = 3.1 × 10⁻¹⁰ m) is equal to 10cm = 0.1m
Therefore, we can determine the pressure using equation (3)
i.e
P= [tex]\frac{RT}{\sqrt{2}\pi d{^2}N{_A}λ }[/tex]
= [tex]\frac{(8.314J{K^-^1)(300K)}}{\sqrt{2}(3.14)(3.1*10^{-10}m)^2(6.022*10^{23}mol^{-1})(0.1m) }[/tex]
=97.06 × 10⁻³ Pa × [tex]\frac{1atm}{1.01325*10^5Pa}[/tex]
=9.57 × 10⁻⁷ atm
Therefore, the maximum residual gas pressure in the calculated space is; 9.57 × 10⁻⁷ atm
Predict the shape, state the hybridizationof the central atom,and give the ideal bond angle(s) and any expected deviations ineach of the following.
a) BiF52-
Shape:
Hybridization:
Ideal bond angle(s):
Deviation from ideal angle(s):
b) BrO3-
Shape:
Hybridization:
Ideal bond angle(s):
Deviation from ideal angle(s):
c) IF4+
Shape:
Hybridization:
Ideal bond angle(s):
Deviation from ideal angle(s):
Answer:
BiF5
Shape- trigonal bipyramid
Hybridisation-sp3d
Ideal bond angle-120 and 90
BrO3-
Shape- tetrahedral
Hybridisation-sp3
Ideal bond angle 109o28'
IF4+
Shape- trigonal bipyramid
Hybridisation-sp3d
Ideal bond angle-120o and 90o
Explanation:
The shape adopted by a molecule according to VSEPR is such as minimizes repulsion between electron pairs. The shape of a molecule depends on electron pairs present on the outermost shell of the central atom. The bond angles are such that electrons are positioned as far apart in space as possible, given the number of electron pairs present.
BiF52- has a square pyramidal shape, sp3d2 hybridization, with ideal bond angles of 90° and 120°. BrO3- has a trigonal pyramidal shape, sp3 hybridization, with an ideal bond angle of 109.5°. IF4+ has a square planar shape, sp3d2 hybridization, with ideal bond angles of 90° and 180°.
Explanation:The shape, hybridization of the central atom, ideal bond angle(s), and any expected deviations for the molecules are as follows:
BiF52-: The shape is square pyramidal. The hybridization of the central atom, Bi, is sp3d2. The ideal bond angles are 90° and 120°. Due to the presence of lone pairs, there can be deviations from these angles.BrO3-: This molecule has a trigonal pyramidal shape. The hybridization of the central Br atom is sp3. The ideal bond angle is 109.5°. Since there is one lone pair of electrons, deviations from the ideal angle occur.IF4+: The shape of this molecule is square planar. The central atom I is in sp3d2 hybridization. The ideal bond angles are 90° and 180°. As it is a symmetrical molecule with no lone pairs, there aren't deviations from this angle.Learn more about Molecule Structures here:https://brainly.com/question/31544427
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When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezing point of pure . On the other hand, when of iron(III) nitrate are dissolved in the same mass of , the freezing point of the solution is lower than the freezing point of pure . Calculate the van't Hoff factor for iron(III) nitrate in . Be sure your answer has a unit symbol, if necessary, and round your answer to significant digits.
Answer:
i = 3,5
Explanation:
There are missing the following values:
132 g Alanine
1150g of X
4,4°C the first freezing point dercreasing
132g of Iron(III) nitrate
5,6°C the second freezing point decreasing.
The freezing point depression is a colligative property that describes the decrease of the freezing point of a solvent on the addition of a non-volatile solute.
The formula is:
ΔT = i kf mb
Where ΔT is freezing point decreasing, i is Van't Hoff factor, kf, is cryoscopic constant and mb is molality of solution.
For alanine Van't Hoff factor is 1 (Ratio between particles in dissolution and before dissolution), molality is:
132g×(1mol/89,09g) = 1,48mol / 1,150kg = 1,29 mol/kg
Replacing:
4,4K = 1 kf 1,29mol/kg
kf = 3,41 K·kg/mol
Now, for Iron(III) nitrate molality is:
132g×(1mol/241,86g) = 0,546mol / 1,150kg = 0,475 mol/kg
Replacing:
5,6K = i×3,41 K·kg/mol×0,475 mol/kg
i = 3,5
I hope it helps!
vWould radiation with a wavelength 0.91 nm or a frequency of 5.9×1011 s−1 be detected by an X-ray detector? Would radiation with a wavelength 0.91 or a frequency of 5.9×1011 be detected by an X-ray detector? Only the radiation with a wavelength 0.91 nm can be observed by an X-ray detector. Only the radiation with a frequency of 5.9×1011 s−1 can be observed by an X-ray detector. Both types of radiation can be observed by an X-ray detector. Neither type of radiation can be observed by an X-ray detector.
Answer:
Only the radiation with a wavelength 0.91 nm can be observed by an X-ray detector.
Explanation:
To answer this question we need to consult the ranges in which x rays are in the electromagnetic spectrum:
The X radiation in the electromagnetic spectrum fall in the region of:
frequency: 3 x 10¹⁶ Hz to 3 x 10¹⁹ Hz (1Hz = 1s⁻¹)
wavelengt: 1 pm to 10 nm
Comparing the values in our question,
0.91 nm will be detected
5.9 x 10¹¹ Hz will not be detected.
The radiation with a wavelength 0.91 or a frequency of [tex]5.9*10^{11}[/tex] will have:
A. Only the radiation with a wavelength 0.91 nm can be observed by an X-ray detector.
Electromagnetic spectrum:X-rays are both types of high energy (high frequency) electromagnetic radiation. They are packets of energy that have no charge or mass (weight).
The X radiation in the electromagnetic spectrum fall in the region of:
Frequency : 3 x 10¹⁶ Hz to 3 x 10¹⁹ Hz (1Hz = 1s⁻¹)
Wavelength : 1 pm to 10 nm
On comparing to the values given in the question:
0.91 nm will be detected
5.9 x 10¹¹ Hz will not be detected.
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Which of the following are characteristics of reverse-phase chromatography?I The stationary phase is polar.II The mobile phase is more polar than the stationary phase.III Less polar mobile phase has a lower eluent strength.IV More polar mobile phase has a higher eluent strength.V The stationary phase is nonpolar.
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The correct options would be:
The mobile phase is more polar than the stationary phase
The stationary phase is nonpolar
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The characteristics of reversed phase chromatography are The mobile phase is more polar than the stationary phase and The stationary phase is non polar.
Hence, option II and V are correct.
What is RPC?RPC is Reversed Phase Chromatography. In reversed phase chromatography the stationary phase is non polar [hydrophobic] and the mobile phase is very polar [hydrophilic].
Now lets check all the option one by one:
Option (I): The stationary phase is non polar in reverse phase chromatography.
So it is incorrect option.
Option (II): The mobile phase is more polar than the stationary phase because in polar mobile phase has high affinity towards the polar solute.
So, it is correct option.
Option (III): Less polar mobile phase has higher eluent strength not lower eluent strength.
So, it is incorrect option.
Option (IV): More polar mobile phase has less eluent strength.
So, it is incorrect option.
Option (V): The stationary phase is non polar [hydrophobic] is reverse phase chromatography.
So, it is correct option.
Thus, from above conclusion we can say that the The mobile phase is more polar than the stationary phase and The stationary phase is non polar are the characteristics of reverse phase chromatography.
Hence option II and V are correct.
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Select the set of quantum numbers that represents each electron in a ground‑state Be atom. n = 2 , ???? = 0 , m ???? = 0 , m s = − 1 2 n = 1 , ???? = 0 , m ???? = 0 , m s = + 1 2 n = 2 , ???? = 0 , m ???? = 0 , m s = + 1 2 n = 2 , ???? = 1 , m ???? = 0 , m s = − 1 2 n = 1 , ???? = 0 , m ???? = − 1 , m s = + 1 2 n = 1 , ???? = 0 , m ???? = 0 , m s = − 1 2 n = 2 , ???? = 0 , m ???? = 0 , m s = − 1 n = 2 , ???? = 1 , m ???? = 1 , m s = + 1 2
Answer:
n: 1, ℓ: 0, ml: 0, ms:+1/2
n: 1, ℓ: 0, ml: 0, ms:-1/2
n: 2, ℓ: 0, ml: 0, ms:+1/2
n: 2, ℓ: 0, ml: 0, ms:-1/2
Explanation:
Beryllium has 4 electrons and its electron configuration is 1s² 2s².
The principal quantum number (n) describes the level of energy. Then, the first two electrons have n = 1, and the second 2 electrons have n = 2.
The azimuthal number (ℓ) describes the subshell of energy. All the 4 electrons are in s subshells, which correspond to ℓ = 0.
The magnetic quantum number (ml) describes the orbital of the subshell. The s subshell has only 1 s orbital, so the only possible value for ml is 0.
The spin quantum number (ms) describes the spin of the electron and can take 2 values: +1/2 or -1/2.
Considering these rules, the quantum numbers for these 4 electrons are:
n: 1, ℓ: 0, ml: 0, ms:+1/2
n: 1, ℓ: 0, ml: 0, ms:-1/2
n: 2, ℓ: 0, ml: 0, ms:+1/2
n: 2, ℓ: 0, ml: 0, ms:-1/2
Concerning the 10.0 mL of 0.50 M NaCl to 100 mL of solution:
Does dilution change the concentration?
Yes
No
Answer:
Yes
Explanation:
Before dilution:
Volume of NaCl = 10 mL
Concentration of NaCl = 0.5 M
Number of moles = Molarity*Volume = 0.5*10 = 5 millimoles.
Note that number of moles of NaCl does not change on dilution as we are only adding water.
After dilution:
Volume of NaCl = 100 mL
Number of moles = 5 millimoles (no change)
New Concentration = Number of moles per volume in litres = [tex]\frac{5\times10^{-3}}{100\times10^{-3}}[/tex]= 0.05 Molar
Hence the concentration became one-tenth of the initial concentration after dilution.
Final answer:
Yes, dilution does change the concentration of a solution. By diluting a 0.50 M NaCl solution to a total volume of 100 mL, the concentration is reduced to 0.050 M, because the same amount of NaCl is spread out in a larger volume of water.
Explanation:
Concerning dilution, yes, it does change the concentration of the solution. When you add more solvent to a solution, the concentration of the solute decreases. For example, if you have a 0.50 M NaCl solution and add enough water to make the total volume 100 mL, the concentration of NaCl changes because the same amount of solute (NaCl) is now dispersed in a greater volume of solvent (water).
Step-by-step Calculation:
Determine the initial amount of NaCl in moles by multiplying the initial volume by the concentration: Moles of NaCl = 0.50 M × 0.010 L = 0.005 moles.
Keep in mind that the amount of NaCl does not change during dilution.
Calculate the final concentration by dividing the moles of NaCl by the final volume of the solution after dilution: Final concentration = 0.005 moles / 0.100 L = 0.050 M NaCl.
A chemist prepares a solution of copper(II) sulfate CuSO4 by measuring out 79.g of copper(II) sulfate into a 500.mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in /molL of the chemist's copper(II) sulfate solution. Be sure your answer has the correct number of significant digits.
Answer:
The concentration of the copper(II) sulfate solution is 0.99 M
Explanation:
Step 1: Data given
Mass of copper(II) sulfate = 79 grams
Volume of the flask = 500 mL
Molar mass copper(II) sulfate = 159.61 g/mol
Step 2: Calculate moles copper(II) sulfate
Moles CuSO4 = Mass CuSO4 / molar mass CuSO4
Moles CuSO4 = 79.0 grams / 159.61 g/mol
Moles CuSO4 = 0.495 moles
Step 3: Calculate concentration
Concentration CuSO4 = moles / volume
Concentration CuSO4 = 0.495 moles / 0.5 L
Concentration = 0.99 M
The concentration of the copper(II) sulfate solution is 0.99 M
A fixed amount of gas at 25.0°C occupies a volume of 10.0 L when the pressure is 667 torr. Use Boyle's law to calculate the pressure (torr) when the volume is reduced to 7.88 L at a constant temperature of 25.0°C.a. 846 torrb. 526 torrc. 0.118 torrd. 1.11 torre. 5.26 × 104 torr
Answer:
The pressure, when the volume is reduced to 7.88L, is 846 torr (option A)
Explanation:
Step 1: Data given
The temperature of a gas = 25.0°C
AT 25 °C the gas occupies a volume of 10.0L and a pressure of 667 torr.
The volume reduces to 7.88 L but the temperature stays constant.
Step 2: Boyle's law
(P1*V1)/T1 = (P2*V2)/T2
⇒ Since the temperature stays constant, we can simplify to:
P1*V1 = P2*V2
⇒ with P1 = the initial pressure 667 torr
⇒ with V1 = the initial volume = 10.0 L
⇒ with P2 = the final pressure = TO BE DETERMINED
⇒ with V2 = the final volume = 7.88L
P2 = (P1*V1)/V2
P2 = (667*10.0)/7.88
P2 = 846 torr
The pressure, when the volume is reduced to 7.88L, is 846 torr (option A)
Final answer:
Using Boyle's law, the new pressure of the gas when its volume decreases from 10.0 L to 7.88 L (with initial pressure 667 torr) is calculated to be 846 torr.
Explanation:
To calculate the new pressure of a gas when its volume changes, we can use Boyle's law, which states that for a fixed amount of gas at constant temperature, the pressure of the gas is inversely proportional to its volume.
This means if the volume decreases, the pressure increases, and vice versa, as long as the temperature and the amount of gas remain unchanged.
When the volume of the gas decreases from 10.0 L to 7.88 L and the initial pressure is 667 torr, we set up Boyle's law as follows: P1 * V1 = P2 * V2
Substituting the known values gives: 667 torr * 10.0 L = P2 * 7.88 L
Assuming no typographical errors, to solve for P2 (the new pressure), we rearrange the equation:
P2 = (667 torr * 10.0 L) / 7.88 L
After performing the division, we find: P2 = 846 torr.
Therefore, the answer is (a) 846 torr, which aligns with Boyle's expectation that a decrease in volume leads to an increase in pressure.
After the stannous chloride solution has been prepared, a few pieces of tin metal, Sn, are dropped into the bottle, to prevent oxidation by air.
Determine the standard cell potential for the following reaction: Sn(s)+Sn4+(aq)→2Sn2+(aq)
Express your answer with the appropriate units.
Answer:
E° = 0.29 V
Explanation:
Let's consider the following redox reaction.
Sn(s) + Sn⁴⁺(aq) → 2 Sn²⁺(aq)
We can identify both half-reactions:
Reduction (cathode): Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V
Oxidation (anode): Sn(s) → Sn²⁺(aq) + 2 e⁻ Ered = -0.14 V
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.15 V - (-0.14 V) = 0.29 V