Answer:
Explanation:
We are to carefully sketch a curve that relates to the potential energy of two O atoms versus the distance between their nuclei.
From the diagram, O2 have higher potential energy than the N2 molecule. Because on the periodic table, the atomic size increases from left to right on across the period, thus O2 posses a larger atomic size than N2 atom.
Therefore, the bond length formation between the two O atoms will be larger compared to that of the two N atoms.
Answer:
C: The shape of the pebbles is a result of weathering and deposition
Explanation:
For the several pebbles to have a rounded shape and smooth to the touch, it will undergo weathering and deposition. This is because weathering involves breaking down of rocks and creating new sediments. This weathering could be either chemical weathering or physical weathering where Chemical weathering is the decomposition of rocks which are caused by chemical reactions and which result in formation of new compound while physical weathering is the breakdown of rocks into smaller pieces. On the other hand, deposition occurs when the agents of erosion such as wind or water deposit sediments from one spot to another which in turn changes the shape of the land.
Thus, the shape of the pebbles are as a result weathering of the parent rocks and from deposition.
Answer:
Removal of Third Electron
Explanation:
a major jump is required to remove the third electron. In general, successive ionization energies always increase because each subsequent electron is being pulled away from an increasingly more positive ion.
Ionization energy increases from bottom to top within a group, and increases from left to right within a period.
Answer:
1 electron is involved.
Explanation:
Hello,
In redox reactions, when therer's the necessity to know the involved equivalents, they equal the number of transferred electrons, in this case, since one equivalent is stated, one electron is transferred (involved).
Best regards.
Answer:
Explanation:
The half-life of K-40 (1.3 billion years) is the time it takes for half of it to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table as follows:
No. of Fraction
<u>half-lives</u> <u> t/yr </u> <u>Remaining</u>
0 0 1
1 1.3 billion ½
2 2.6 ¼
3 3.9 ⅛
We see that after 2 half-lives, ¼ of the original mass remains.
Conversely, if two half-lives have passed, the original mass must have been four times the mass we have now.
Original mass = 4 × 2.10 g =
