Answer:
Pressure,P=6×10^3Pa
Explanation:
The gas has an ideal gas behaviour and ideal gas equation
PV=NKT
T= V/N p/K ...eq1
Average transitional kinetic energy Ktr=1.8×10-23J
Ktr=3/2KT
T=2/3Ktr/K....eq2
Equating eq1 and 2
V/N p/K = 2/3Ktr/K
Cancelling K on both sides
P= 2/3N/V( Ktr)
Substituting the value of N/V and dividing by 10^-6 to convert cm^3 to m^3
P = 2/3 (5.0×10^20)/10^-6 × 1.8×10^-23
P= 6 ×10^3Pa
The answer is it will supply 1.1 x 10⁹ J of energy each second.
we can calculate this by using the following equation;
P = W/t
<span>W = P x t
</span><span>and by work energy relation;
E = W = P x t
</span>1 watt = 1j/s
1megawatt = 1000000 = 10⁶ j/s
<span>E = 1100 x 106 x 1 </span>
E = 1.1 x 10⁹ J
Answer:
Two orbitals for their electrons and six in the 2p subshell
Explanation:
Hope this helps :)
