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tiny-mole [99]
4 years ago
6

An electrochemical cell is constructed using two half-cells: Al(s) in Al(NO2)3(aq) and Cu(s) in Cu(NO3)2(aq). The two half cells

are connected by a KNO3 salt bridge and two copper wires from the electrodes to a voltmeter. Based on their respective standard reduction potentials, which half-cell is the cathode?
Chemistry
1 answer:
wlad13 [49]4 years ago
4 0

Answer:

Cu(s) in Cu(NO₃)₂(aq)

Explanation:

The standard reduction potential (E°) is the energy necessary to reduce the atom in a redox reaction. When an atom reduces it gains electrons from other than oxides. As higher is E°, easily it will reduce. The substance that reduces is at the cathode of a cell, where the electrons go to, and the other that oxides are at the anode of the cell.

The standard reduction potentials from Al(s) and Cu(s) are, respectively, -1.66V and +0.15V, so the half-cell of Cu(s) in Cu(NO₃)₂(aq) is the cathode.

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React it hydrogen bromide  

Step-by-step explanation:

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CH₂=CH₂ + HBr ⟶ CH₃CH₂Br

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Why will the crucible and its contents gain mass when heated?
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4 years ago
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A soft drink is made by dissolving CO2 at 3.00 atm in a flavored solution and sealing the solution in an aluminum can at 20oC. W
erastovalidia [21]

Answer:

Volume is 1.065L

Explanation:

Hello,

We can easily solve this problem by using general gas equation.

PV / T = K

P1V1/T1 = P2V2/T2

Data;

P1 = 3.0atm

P2 = 1.0atm

T1 = 20°C = (20 + 273.15)K = 293.15K

T2 = 20°C = (20 + 273.15)K = 293.15K

V1 = 355mL = 0.355L

V2 = ?

From the data given, we can substitute it into the equation,

(P1 × V1) / T1 = (P2 × V2) / T2

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3 0
4 years ago
Calculate the pH of each of the following aqueous solutions. (Enter your answers to two decimal places.) (a) 10.0 mL deionized w
Anna007 [38]

Answer:

a. pH = 7.0

b. pH = 12.52

c. pH = 12.70

d. pH = 12.78

Explanation:

a. Deionized water has the [H⁺] of pure water = 1x10⁻⁷ (Kw = 1x10⁻¹⁴ = [H⁺][OH⁻] - [H⁺] = [OH⁻ -)

pH = -log[H⁺] = 7

b. Moles NaOH = 5x10⁻³L * (0.10mol / L) = 5x10⁻⁴moles OH⁻ / 0.015L = 0.0333M = [OH⁻]

<em>-Total volume = 10mL+5mL = 15mL = 0.015L</em>

pOH = -log[OH⁻] = 1.48

pH = 14-pOH

pH = 12.52

c. Moles NaOH = 0.010L * (0.10mol / L) = 1x10⁻³moles OH⁻ / 0.020L = 0.0500M = [OH⁻]

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d. Moles NaOH = 0.015L * (0.10mol / L) = 1.5x10⁻³moles OH⁻ / 0.025L = 0.060M = [OH⁻]

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4 0
3 years ago
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melisa1 [442]
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