PV=nRT
P=nRT/V
P=[(0.650mol)(0.08206)(298K)]/(0.750L)=21.2atm
Well it's an alkali metal if that's what you're asking<span />
*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
Answer:
Be,Mg,Ra etc
Explanation:
It should be palced in group 2A because as it reacts with chlorine in ratio of 1:2 . It's valancy is 2 and is metal as it react with non metal donating two electrons .
one more thing it fits there orderly
Answer:
Energy decreases and wavelength gets longer.
Explanation:
The energy of a wave is given by :
Where
h is Planck's constant
f is the frequency of the wave
As frequency decreases, the energy of the wave also decreases and wavelength gets longer. Hence, the option is (d).
