Answer:
A jump occurs when a core electron is removed.
Explanation:
A jump in ionization energy occurs when a core electron is removed. A large jump in the ionization energy easily be seen from the electronic configuration of an element.
For Beryllium, the electronic configuration of is 1s2 2s2.
There are two valence electrons in the outermost shell hence the ionization energy data for beryllium will show a sudden jump or increase in going from the second to the third ionization energy owing to the removal of a core electron
The electronic configuration for Nitrogen is 1s2 2s2 2p3. Five valence electrons are found in the outermost shell so the ionization energy data for nitrogen will show a sudden jump or increase in going from the fifth to sixth ionization energy because of the removal of a core electron
The electronic configuration of oxygen is 1s2 2s2 2p4. There are six valence electrons hence ionization energy for oxygen atom will show a sudden jump or increase in going from the sixth to the seventh ionization energy because of the removal of a core electron
The electronic configuration of Lithium is 1s2 2s1
There is one valence electron in its outermost shell so its ionization energy data will show a sudden jump or increase in going from the first to the second ionization energy because of the removal of a core electron.
Answer:
Second element(Titanium); [Ar] 3d2 4s2
Third element(Vanadium):Ar 3d3 4s2
Explanation:
Given that there are only three d orbitals in universe L instead of five, the electronic configuration of the second and third elements in the first transition series will now look thus;
Second element(Titanium); [Ar] 3d2 4s2
Third transition element(Vanadium):Ar 3d3 4s2
Hence, the electronic configuration of Titanium and Vanadium in universe L is just the same as what it is on earth.
Answer:
Viewing systems from multiple perspectives.
Discovering causes and effects using model tractability.
Improving system understanding through visual analysis.
Explanation:
Got this from google, lol. But, I put three here just in case you could get extra credit for more than two.
The height of the described cylinder is 0.85 cm
Mass= 98g
Density= 8.9g/cm^3
Density= (mass/volume)
Substitute the values
8.94= 98/ volume
Volume= 
Volume= 10.97
Volume of a cylinder can be found with (V = π r 2 h.)
h= height of cylinder ?
radius= 2cm
Substitute the values
10.97= π × 
× h
h= 10.97/12.57
h= 0.85 cm
Therefore, height of the described cylinder is 0.85 cm
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