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:
1.
work out the mean mode median and range
Explanation:
Answer:
0.681 atm
Explanation:
To solve this problem, we make use of the General gas equation.
Given:
P1 = 785 torr
V1 = 2L
T1 = 37= 37 + 273.15 = 310.15K
P2 = ?
V2 = 3.24L
T2 = 58 = 58+273.15 = 331.15K
P1V1/T1 = P2V2/T2
Now, making P2 the subject of the formula,
P2 = P1V1T2/T1V2
P2 = [785 * 2 * 331.15]/[310.15 * 3.24]
P2 = 515.715 Torr
We convert this to atm: 1 torr = 0.00132 atm
515.715 Torr = 515.715 * 0.00132 = 0.681 atm
Answer:
<em>yes</em>
Explanation:
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