Data Given:
Time = t = ?
Current = I = 10 A
Faradays Constant = F = 96500
Chemical equivalent = e = 107.86/1 = 107.86 g
Amount Deposited = W = 17.3 g
Solution:
According to Faraday's Law,
W = I t e / F
Solving for t,
t = W F / I e
Putting values,
t = (17.3 g × 96500) ÷ (10 A × 107.86 g)
t = 1547.79 s
t = 1.54 × 10³ s
Given that
Mass of water = 65.34 g
Amount of heat = mass of water * specific heat (temperature change
)
= 65.34 g * 4.184 J / g-C ( 21.75-18.43 )C
= 907.63 J
= 0.908 KJ
And
1 cal = 4.186798 J
907.63 J * 1 cal / 4.186798 J =216.78 cal
Or0.218 kcal
Answer:
0.0613 L
Explanation:
Given data
- Initial pressure (P₁): 1.00 atm
- Initial volume (V₁): 1.84 L
- Final pressure (P₂): 30.0 atm
Since we are dealing with an ideal gas, we can calculate the final volume using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁ / P₂
V₂ = 1.00 atm × 1.84 L / 30.0 atm
V₂ = 0.0613 L
<span>oxidizing substance removes electrons from another substance, which are then added to itself, the oxidizing substance becomes “reduced” (more negative). And because it “accepts” electrons .</span>