When particles collide with the surface of the solid.
First let us determine the electronic configuration of
Bromine (Br). This is written as:
Br = [Ar] 3d10 4s2 4p5
Then we must recall that the greatest effective nuclear
charge (also referred to as shielding) greatly increases as distance of the
orbital to the nucleus also increases. So therefore the electron in the
farthest shell will experience the greatest nuclear charge hence the answer is:
<span>4p orbital</span>
Answer:
New pressure P2 = 4.95 atm
Explanation:
Given:
Old volume V1 = 1.50 L
New volume V2 = 0.50 L
Old pressure P1 = 1.65 atm
Find:
New pressure P2
Computation:
P1V1 = P2V2
So,
(1.50)(1.65) = (0.50)(P2)
New pressure P2 = 4.95 atm
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<u>Given:</u>
Change in internal energy = ΔU = -5084.1 kJ
Change in enthalpy = ΔH = -5074.3 kJ
<u>To determine:</u>
The work done, W
<u>Explanation:</u>
Based on the first law of thermodynamics,
ΔH = ΔU + PΔV
the work done by a gas is given as:
W = -PΔV
Therefore:
ΔH = ΔU - W
W = ΔU-ΔH = -5084.1 -(-5074.3) = -9.8 kJ
Ans: Work done is -9.8 kJ
