Answer: The mass of deposited cadmium is 23.9g
Explanation:
According to Faraday Law of Electrolysis, the mass of substance deposited at the electrode is directly proportional to the quantity of electricity passed through the electrolyte.
Faraday has found that the amount of electricity needed to liberate one gm eq. of substance from an electrolyte is 96500C.
Given that
time (t) =1.55min *60 =93seconds
Current (I) =221A
Q= It = 221 * 93 =20553C
Molar mass of CdS04 =112.41
96500C will liberate 112.41g
20553C will liberate Xg
Xg= (20553*112.41)/96500
=23.9g
Therefore , the amount of Cd deposited at the electrode is 23.9g
The answer of your question is True
Answer:
13.8 moles
Explanation:
You know that in a mole you have 6.022x10²³ something in this case of molecules. By knowing that by each mole you have 6.022x10²³ molecules, then you can establish the factor of conversion:
8.32x10²⁴molec of CO2 x 1 mol/6.022x10²³ molec of CO2
You divide 8.32x10²⁴/6.022x10²³ and you get 13.81 moles.
Answer:
- <u>Ten (10) molecules of water</u><em> are bonded to sodium sulfate decahydrate.</em>
Explanation:
A hydrated compound is one that contains water molecules chemically bonded. So, the correct question is how many waters are <u><em>bonded</em></u> to sodium suilfate?
The latin prefix<em> deca</em> means ten (10), and the word <em>hydrate </em>means that contains water; then, <em>decahydrate</em> is used in chemistry to designate a compound that is bonded to 10 parts (molecules) of water.
Hence, sodium sulfate decahydrate means that each molecule of the compound is bonded to 10 molecules of water.
In the chemical formula, the water molecules of a hydrated compound are shown after the chemical formula of the compound using a period and a coefficient (the number of water molecules bonded) before the water molecules.
Then, given that the chemical formula for sodium sulfate is Na₂SO₄, the chemical formula for sodium sulfate decahydrate is Na₂SO₄·10H₂O.
According to Henry's law, solubility is proportional to the external pressure. Since both cases in the problem have the same temperature, only the pressure varies from 1.0 atm to 2.5 atm. This means that just as the pressure is multiplied by a factor of 2.5, we can multiply the solubility by a factor of 2.5 as well. This gives us 0.041 g/L * 2.5 = 0.1025 g/L.