The voltage generated by the zinc concentration cell described by, zn(s)|zn2 (aq, 0.100 m)||zn2 (aq, ? m)|zn(s),is 16.0 mv at 25 °c. calculate the concentration of the zn2 (aq) ion at the cathode.

Answers

Answer 1

Answer: The concentration cell is:
Zn(s) \ Zn²⁺(aq,0.100 M) // Zn²⁺(aq, x M) \ Zn(s)
voltage = 16 mV x (1V / 10³ mV) = 16 x 10⁻³ V
- In the cell notation, the concentration on the left is that of the anode and that on the right is that of the cathode.
- Oxidation takes place at the anode and reduction takes place at the cathode.
so [Zn²⁺]oxidation = 0.100 M
[Zn²⁺] reduction = x M
From Nernst equation:
Ecell = -0.0592 / n log [Zn²⁺] oxidation / [Zn²⁺]reduction
Number of electrons, n = 2. Substitute and solve for x:
16 x 10⁻³ V = - 0.0592 / 2 log (0.100 /x)
log 0.100 / x = - 0.54
0.100 / x = 0.288
x = 0.347
So the concentration of Zn²⁺ at the cathode = 0.406

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A chemist designs a galvanic cell that uses these two half-reactions:O2 (g) + 4H+(aq) + 4e− → 2H2O (l) Eo =+1.23V
Zn+2 (aq) + 2e− → Zn(s) Eo=−0.763V

Answer the following questions about this cell.

Write a balanced equation for the half-reaction that happens at the cathode.
Write a balanced equation for the half-reaction that happens at the anode.
Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous as written. Do you have enough information to calculate the cell voltage under standard conditions

Answers

Answer: The reaction is spontaneous and there is not enough information to calculate the cell voltage.

Explanation:

The substance having highest positive $E^o$ reduction potential will always get reduced and will undergo reduction reaction.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

For a:

The half reactions for the cell occurring at cathode follows:

$O_2(g)+4H^+(aq)+4e^-\rightarrow H_2O(l);E^o_(cathode)=+1.23V$

For b:

The half reactions for the cell occurring at anode follows:

$Zn(s)\rightarrow Zn^(2+)+2e^-;E^o_(anode)=-0.763V$ ( × 2)

For c:

The balanced equation for the overall reaction of the cell follows:

$O_2(g)+4H^+(aq)+2Zn(s)\rightarrow H_2O(l)+2Zn^(2+)(aq)$

For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative.

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_(cell)$

For a reaction to be spontaneous, the standard electrode potential must be positive.

To calculate the $E^o_(cell)$ of the reaction, we use the equation:

$E^o_(cell)=E^o_(cathode)-E^o_(anode)$

Putting values in above equation, we get:

$E^o_(cell)=1.23-(-0.763)=1.993V$

As, the standard electrode potential of the cell is coming out to be positive, the reaction is spontaneous in nature.

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_(cell)=E^o_(cell)-(0.059)/(n)\log ([Zn^(2+)]^2)/([H^(+)]^4* p_(O_2))$

As, the concentrations and partial pressures are not given. So, there is not enough information to calculate the cell voltage.

Hence, the reaction is spontaneous and there is not enough information to calculate the cell voltage.

If sodium arsenite is Na3AsO3, the formula for calcium arsenite would be

Answers

Answer:

Ca₃(AsO₃)₂

Explanation:

Sodium arsenite, with the chemical formula Na₃AsO₃, is formed by the cation Na⁺ and the anion AsO₃³⁻. For the molecule to be neutral, 3 cations Na⁺ and 1 anion AsO₃³⁻ are required.

Calcium arsenite would be formed by the cation Ca²⁺ and the anion AsO₃³⁻. For the molecule to be neutral, we require 3 cations Ca²⁺ and 2 anions AsO₃³⁻. The resulting chemical formula is Ca₃(AsO₃)₂.

Please help me I will give a brainleist to the. first person to answer

Answers

Answer:

Explanation:

The Agricultural Revolution gave Britain the most productive agriculture in Europe, with 19th-century yields as much as 80% higher than the Continental average. ... By the 19th century, marketing was nationwide and the vast majority of agricultural production was for the market rather than for the farmer and his family.

hope this helps :)

Calculate the molar mass of the following:(a)the anesthetic halothane, C2HBrClF3
(b)the herbicide paraquat, C12H14N2CL2
(c)caffein, C8H10N4O2
(d)urea, CO(NH2)2
(e)a typical soap,C17H35CO2Na

Answers

Answer:

a)C2HBrClF3 = 197.35 g/mol

b)C12H14N2CL2 = 229.06g/mol

c)C8H10N4O2 = 194.22g/mol

d) CO(NH2)2=60.07 g/mol

e)C17H35CO2Na = 306.52 g/mol

Explanation:

Molar mass of a compound is equal to the sum of the atomic masses of the constituent elements.

a) C2HBrClF3

$Molar\ mass = 2(At. mass C)+1(at.mass H) +1(At. mass Br) + 1(At.mass Cl) + 3(At.mass F)\n=2(12.01 g/mol) + 1(1.01g/mol)+1(79.90 g/mol) +1(35.45g/mol)+3(18.99g/mol)=197.35g/mol$

b) C12H14N2CL2

$Molar\ mass = 12(C) + 14(H) + 2(N) + 2(Cl)\n\n=12(12.01) + 14(1.01) + 2(14.01) + 2(35.45) = 229.06g/mol$

c) C8H10N4O2

$Molar\ mass = 8(C) + 10(H) + 4(N) + 2(O)\n\n=8(12.01) + 10(1.01) + 4(14.01) + 2(16.00) =194.22g/mol$

d) CO(NH2)2

$Molar\ mass = 1(C) + 1(O) + 2(N) + 4(H)\n\n=1(12.01) + 1(16.00) + 2(14.01)+4(1.01) =60.07 g/mol$

e) C17H35CO2Na

$Molar Mass = 18(C) + H(35) +2(O) + 1(Na)\n\n=18(12.01) + 35(1.01) + 2(16.00) + 1(22.99) =306.52 g/mol$

The combustion of ethane (C2H6) produces carbon dioxide and steam. 2C2H6(g)+7O2(g)⟶4CO2(g)+6H2O(g) How many moles of CO2 are produced when 5.90 mol of ethane is burned in an excess of oxygen?

Answers

Final answer:

The combustion of ethane yields carbon dioxide, and with 5.90 moles of ethane being reacted, it results in the production of 11.8 moles of CO2.

Explanation:

The question pertains to the concept of stoichiometry in chemistry, and the chemical reaction in question is a combustion reaction involving ethane (C2H6). From the balanced reaction, it is evident that 2 moles of ethane (C2H6) produce 4 moles of carbon dioxide (CO2). Therefore, if we have 5.90 moles of ethane reacting, it's a straightforward calculation to determine that this would yield twice that many moles of CO2. We simply multiply the moles of ethane by the stoichiometric ratio (4/2) to get the moles of CO2.

Example Calculation: 5.90 moles of ethane x (4 moles CO2 / 2 moles C2H6) = 11.8 moles CO2

So, when 5.90 moles of ethane are burned in an excess of oxygen, 11.8 moles of CO2 are produced.

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Final answer:

In the combustion of ethane, for every mole of ethane burned, two moles of carbon dioxide are produced. Hence, when 5.90 moles of ethane are burned, 11.8 moles of carbon dioxide are produced.

Explanation:

The chemical reaction given, 2C2H6(g) + 7O2(g) ⟶ 4CO2(g) + 6H2O(g), states that 2 moles of ethane (C2H6) produce 4 moles of carbon dioxide (CO2). Thus, the mole-to-mole ratio of ethane to carbon dioxide is 2:4, or simplified, 1:2. So, for every mole of ethane burned, two moles of carbon dioxide are produced.

Given that 5.90 moles of ethane are burned, we can calculate the quantity of carbon dioxide produced by multiplying 5.90 moles by 2. Hence, when 5.90 moles of ethane are burned in an excess of oxygen, 11.8 moles of carbon dioxide are produced.

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Which are correct ???

Answers

Answer:

last one

Explanation:

endothermic reactions release energy

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