Answer: 406 hours
Explanation:

where Q= quantity of electricity in coloumbs
I = current in amperes = 39.5 A
t= time in seconds = ?
The deposition of copper at cathode is represented by:

Coloumb of electricity deposits 1 mole of copper
i.e. 63.5 g of copper is deposited by = 193000 Coloumb
Thus 19.0 kg or 19000 g of copper is deposited by =
Coloumb

(1hour=3600s)
Thus it will take 406 hours to plate 19.0 kg of copper onto the cathode if the current passed through the cell is held constant at 39.5 A
Answer:H2=11.4g
CH4=28.6g
Explanation:The complete combustion of the two gases can be represented by a balanced reaction below
1. CH4 +2O2___CO2+2H2O
2.2H2+O2___2H2O
Combining the two we have CH4 +2H2+3O2___
CO2+4H2O
Since the mixture contains 40gof CH4 and 2, therefore 20g of CH4 and 8g of H2 combines.
Calculated from their molecular Mass i.e CH4 12+4×2)=20 and 2H2= 2×2×2=8g
Mass of CH4=20/28×40=28.6g
2H2=8/28×40=11.4g
Answer:
destructive interference?
Explanation:
Answer:
3853 g
Step-by-step explanation:
M_r: 107.87
16Ag + S₈ ⟶ 8Ag₂S; ΔH°f = -31.8 kJ·mol⁻¹
1. Calculate the moles of Ag₂S
Moles of Ag₂S = 567.9 kJ × 1 mol Ag₂S/31.8kJ = 17.858 mol Ag₂S
2. Calculate the moles of Ag
Moles of Ag = 17.86 mol Ag₂S × (16 mol Ag/8 mol Ag₂S) = 35.717 mol Ag
3. Calculate the mass of Ag
Mass of g = 35.717 mol Ag × (107.87 g Ag/1 mol Ag) = 3853 g Ag
You must react 3853 g of Ag to produce 567.9 kJ of heat
Answer:
THE FINAL TEMPREATURE OF IRON SWORD IS
1331°C