Answer:
0.0164 g
Explanation:
Let's consider the reduction of silver (I) to silver that occurs in the cathode during the electroplating.
Ag⁺(aq) + 1 e⁻ → Ag(s)
We can establish the following relations.
- 1 A = 1 C/s
- The charge of 1 mole of electrons is 96,468 C (Faraday's constant)
- 1 mole of Ag(s) is deposited when 1 mole of electrons circulate.
- The molar mass of silver is 107.87 g/mol
The mass of silver deposited when a current of 0.770 A circulates during 19.0 seconds is:
English please i cant understand
Given what we know, we can confirm that the amount of heat energy that would be required in order to boil 5.05g of water is that of 11.4kJ of heat.
<h3>Why does it take this much energy to boil the water?</h3>
We arrive at this number by taking into account the energy needed to boil 1g of water to its vaporization point. This results in the use of 2260 J of heat energy. We then take this number and multiply it by the total grams of water being heated, in this case, 5.05g, which gives us our answer of 11.4 kJ of energy required.
Therefore, we can confirm that the amount of heat energy that would be required in order to boil 5.05g of water is that of 11.4kJ of heat.
To learn more about the behavior of water visit:
brainly.com/question/1416592?referrer=searchResults
The drawing is correct because there are 12 atoms of each type on each side of the arrow.
