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tiny-mole [99]
3 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]3 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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Can cells that are haploid (single set of chromosomes ) undergo meiosis
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3 years ago
What is the balanced symbol equation for magnesium chloride + sodium carbonate --> magnesium carbonate + sodium chloride
Semenov [28]

Answer:

MgCl₂+ Na₂CO₃ ==> MgCO₃ + NaCl

From a quick observation

You see that the right hand side of the eqn is deficient of Sodium and Chlorine

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MgCl₂ + Na₂CO₃ ==> MgCO₃ + 2NaCl.✅

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How do disease causing organism reach from one organism to other​
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Identify how many of the original parent nuclei remain after three half-lives
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3 0
3 years ago
A 1.52 g sample of KCIO3 is reacted according to the balanced equation below. How many liters of O2 is produced at a pressure of
ipn [44]

Answer:

0.486 L

Explanation:

Step 1: Write the balanced reaction

2 KCIO₃(s) ⇒ 2 KCI (s) + 3 O₂(g)

Step 2: Calculate the moles corresponding to 1.52 g of KCIO₃

The molar mass of KCIO₃ is 122.55 g/mol.

1.52 g × 1 mol/122.55 g = 0.0124 mol

Step 3: Calculate the moles of O₂ produced from 0.0124 moles of KCIO₃

The molar ratio of KCIO₃ to O₂ is 2:3. The moles of O₂ produced are 3/2 × 0.0124 mol = 0.0186 mol

Step 4: Calculate the volume corresponding to 0.0186 moles of O₂

0.0186 moles of O₂ are at 37 °C (310 K) and 0.974 atm. We can calculate the volume of oxygen using the ideal gas equation.

P × V = n × R × T

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7 0
3 years ago
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