Given what we know, we can confirm that in a voltaic cell, the anode loses electrons and is oxidized, meanwhile, the cathode is reduced by gaining electrons.
<h3 /><h3>What is a voltaic cell?</h3>
- It is described as an electrochemical cell.
- These cells use chemical reactions to produce electrical energy.
- During this reaction, an anode loses electrons, thus oxidizing.
- Meanwhile, the cathode gains electrons and is reduced.
Therefore, given the nature of the voltaic cell, we can confirm that during its reaction, the anode is oxidized by losing electrons while the cathode becomes reduced by gaining them.
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The correct answer is D:
it involves breaking molecular bonds between copper compounds
.
The explanation :
-when we melt a copper it is a physical change because the substance is still copper and have the same shape.
- but for example Burning a copper it is a chemical change. Fire activates a chemical reaction between copper and oxygen.
-The oxygen in the air reacts with the copper and the chemical bonds are broken.
- the chemical change is changing the other compound bonded to the copper atoms.
So, the correct answer is D
Answer:
Percentage of copper = 88%
Explanation:
Given data:
Mass of copper = 51.2 g
Mass of tin = 6.84 g
Percentage of copper = ?
Solution:
Formula:
Percentage of copper = mass of copper / total mass × 100
Now we will determine the total mass:
Total mass = mass of copper + mass of tin
Total mass = 51.2 g + 6.84 g
Total mass = 58.04 g
Now we will calculate the percentage of copper.
Percentage of copper = 51.2 g / 58.04 g × 100
Percentage of copper = 0.88 × 100
Percentage of copper = 88%
Answer:
105 grams PbI₂
Explanation:
Pb(NO₃)₂ + 2KI => 2KNO₃ + PbI₂(s)
moles Pb(NO₃)₂ = 0.265L(1.2M) = 0.318 mole
moles KI = 0.293(1.55M) = 0.454 mole => Limiting Reactant
moles PbI₂ from mole KI in excess Pb(NO₃)₂ = 1/2(0.454 mole) = 0.227 mol PbI₂
grams PbI₂ = 0.227 mol PbI₂ x 461 g/mole = 104.68 g ≈ 105 g PbI₂(s)
Answer:
the acceleration of the racecar is 6.5 m/s²
Explanation:
Given;
initial velocity of the racecar, u = 0
final velocity of the racecar, v = 39 m/s
time of motion, t = 6.0 s
The acceleration of the racecar is calculated as;
Therefore, the acceleration of the racecar is 6.5 m/s²
