*Answer:
Option A: 59.6
Explanation:
Step 1: Data given
Mass of aluminium = 4.00 kg
The applied emf = 5.00 V
watts = volts * amperes
Step 2: Calculate amperes
equivalent mass of aluminum = 27 / 3 = 9
mass of deposit = (equivalent mass x amperes x seconds) / 96500
4000 grams = (9* amperes * seconds) / 96500
amperes * seconds = 42888888.9
1 hour = 3600 seconds
amperes * hours = 42888888.9 / 3600 = 11913.6
amperes = 11913.6 / hours
Step 3: Calculate kilowatts
watts = 5 * 11913.6 / hours
watts = 59568 (per hour)
kilowatts = 59.6 (per hour)
The number of kilowatt-hours of electricity required to produce 4.00kg of aluminum from electrolysis of compounds from bauxite is 59.6 kWh when the applied emf is 5.00V
2C3H8+ 702--->6CO2+8H20
FROM Equation above 2 moles of C3H8 reacted with 7 moles of oxygen to form 6 moles of c02 plus 8 molesof H2O
the moles of c3H8 reacted is = MASS/ R.F.M
THE R.F.M =48+8=44
Number of moles is hence 0.025/44=5.68x10^-4
since ratio of C3H8 to O2 is 2:7 Therefore moles of O2 reacted is 1.989 x10^-3
mass= r.f.m x number of moles
(1.989x10^-3) x 32 =0.064g
Answer:
The answer to your question is -2855 J
Explanation:
Reaction
2C₂H₆ + 7O₂ ⇒ 4CO₂ + 6H₂O
Formula
Heat of reaction = ΔHrxn = ΣΔHrxn products - ΣΔHrxn reactants
Substitution
ΔHrxn = { 4(-393.5) + 6(-241.8)} - {2(-84.7) + 7(0)}
ΔHrxn = {-1574 -1450.8} - {-169.4}
ΔHrxn = -3024.8 + 169.4
ΔHrxn = -2855.4 J
Answer:
These substances are called amphiprotic. Water, amino acids, hydrogen carbonate ions, and hydrogen sulfate ions are common examples.
Explanation:
A substance is amphoteric (from Greek amphoteros = "each of two") if it can act as an acid or a base.
For example, aluminum hydroxide is amphoteric because it can act as a base and neutralize strong acids.
