Answer:
stay the same.
Explanation: Period 3 consists of the full 1s, 2s, and 2p electron orbitals, plus the 3s and 3p valence orbitals, which are filled with a total of 8 more electrons as we move from left (Na) to the far right (Ar):
Na: 1s2 2s2 2p6 3s1
Ar: s2 2s2 2p6 3s2 3p6
As we move from left to right, and ignoring the already-filled 1s, 2s, and 2p orbitals, the period three starting and ending elements have the following:
Na: 3s1
Ar: 3s2, 3p6
All the new electrons electrons filled the third energy level (3s and 3p). So the energy level does not change, just the orbitals.
<h2><u>
Answer:</u></h2>
(These are not rounded to the correct decimal)
130.94 atm
13,266.6 kPa
99,571.4 mmHg
<h2><u>
Explanation:</u></h2>
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PV = nRT
V = 245L
P = ?
R = 0.08206 (atm) , 8.314 (kPa) , 62.4 (mmHg)
T = 273.15 + 27 = 300.15K
n = 1302.5 moles
How I found (n).
5.21kg x 1000g/1kg x 1 mole/4.0g = 1302.5 moles
Now, plug all the numbers into the equation.
Pressure in atm = (1302.5)(0.08206)(300.15) / 245 = 130.94 atm (not rounded to the correct decimal)
Pressure in kPa = (1302.5)(8.314)(300.15) / 245 = 13,266.6 kPa (not rounded to the correct decimal)
Pressure in mmHg = (1302.5)(62.4)(300.15) / 245 = 99,571.4 mmHg (not rounded to the correct decimal)
Answer:
30 neutrons
Explanation:
Since the mass of the iron nuclide is 56 , there must be 56−26=30 neutrons, 30 massive, neutral particles in this iron nucleus.
I think the answer is -12.7
