*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
The correct answer is approximately 11.73 grams of sulfuric acid.
The theoretical yield of water from Al(OH)3 is lower than that of H₂SO₄. As a consequence, Al(OH)3 is the limiting reactant, H₂SO₄ is in excess.
The balanced equation is:
2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O
Each mole of Al(OH)3 corresponds to 3/2 moles of H₂SO₄. The molecular mass of Al(OH)3 is 78.003 g/mol. There are 15/78.003 = 0.19230 moles of Al(OH)3 in the five grams of Al(OH)3 available. Al(OH)3 is in limiting, which means that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.
The molar mass of H₂SO₄ is 98.706 g/mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.706 = 28.289 g
40 grams of sulfuric acid is available, out of which 28.289 grams is consumed. The remaining 40-28.289 = 11.711 g is in excess, which is closest to the first option, that is, 11.73 grams of H₂SO₄.
"Indoor environments are much less polluted than outdoor <span>environments" is the best option since indoor environments often rely on separate air and ventilation systems. </span>
The maximum number of covalent bonds formed by a single carbon atom is 4
Answer:
Distance = 200 km
Distance = 204 km
Speed = 77 km/h
Time = 21.42 h
Explanation:
Given:
A.
Speed = 100 km/h , Time = 2 h
Find:
Distance
B.
Speed = 68 km/h , Time = 3 h
Find:
Distance
C.
Distance = 154 km , Time = 2 h
Find:
Speed
D.
Distance = 1500 km speed = 70 km/h
Find:
Time
Computation:
Speed = distance / time
A.
Distance = 100 x 2
Distance = 200 km
B.
Distance = 68 x 3
Distance = 204 km
C.
Speed = 154 / 2
Speed = 77 km/h
D.
Time = 1500 / 70
Time = 21.42 h
