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:
6.9 Sleps 1. (8)(13) = 15 2. 104/15 = 15/15 3. 6.93333333 4. 6.9 Sleps
Answer:
34,6g of (NH₄)₂SO₄
Explanation:
The boiling-point elevation describes the phenomenon in which the boiling point of a liquid increases with the addition of a compound. The formula is:
ΔT = kb×m
Where ΔT is Tsolution - T solvent; kb is ebullioscopic constant and m is molality of ions in solution.
For the problem:
ΔT = 109,7°C-108,3°C = 1,4°C
kb = 1.07 °C kg/mol
Solving:
m = 1,31 mol/kg
As mass of X = 600g = 0,600kg:
1,31mol/kg×0,600kg = 0,785 moles of ions. As (NH₄)₂SO₄ has three ions:
0,785 moles of ions×
= 0,262 moles of (NH₄)₂SO₄
As molar mass of (NH₄)₂SO₄ is 132,14g/mol:
0,262 moles of (NH₄)₂SO₄×
= <em>34,6g of (NH₄)₂SO₄</em>
<em></em>
I hope it helps!
Answer:
32 g Cu
Explanation:
1 mol Cu -> 63.5 g
0.5 mol Cu ->x
x=(0.5 mol *63.5 g)/1 mol x= 32 g Cu
Carbon dioxide and oxygen are removed from the air.
Explanation:
When air is passed through aqueous sodium hydroxide solution the carbon dioxide is removed from the air.
First the carbon dioxide will dissolve and react with water to form carbonic acid ( H₂CO₃) :
CO₂ + H₂O → H₂CO₃
The the carbonic acid will react with sodium hydroxide to form sodium carbonate (Na₂CO₃):
H₂CO₃ + 2 NaOH → Na₂CO₃ + 2 H₂O
After this by passing the air over heated cooper the oxygen is removed.
2 Cu + O₂ → 2 CuO
Learn more about:
neutralization reaction
brainly.com/question/2632201
#learnwithBrainly
