When solid surfaces slide over each other, the kind of friction that occurs is called <u>Sliding Friction.</u>
Answer:
Rank in increasing order of effective nuclear charge:
Explanation:
This explains the meaning of effective nuclear charge, Zeff, how to determine it, and the calculations for a valence electron of each of the five given elements: F, Li, Be, B, and N.
<u>1) Effective nuclear charge definitions</u>
- While the total positive charge of the atom nucleus (Z) is equal to the number of protons, the electrons farther away from the nucleus experience an effective nuclear charge (Zeff) less than the total nuclear charge, due to the fact that electrons in between the nucleus and the outer electrons partially cancel the atraction from the nucleus.
- Such effect on on a valence electron is estimated as the atomic number less the number of electrons closer to the nucleus than the electron whose effective nuclear charge is being determined: Zeff = Z - S.
<u><em>2) Z eff for a F valence electron:</em></u>
- F's atomic number: Z = 9
- Total number of electrons: 9 (same numer of protons)
- Period: 17 (search in the periodic table or do the electron configuration)
- Number of valence electrons: 7 (equal to the last digit of the period's number)
- Number of electrons closer to the nucleus than a valence electron: S = 9 - 7 = 2
- Zeff = Z - S = 9 - 2 = 7
<u><em>3) Z eff for a Li valence eletron:</em></u>
- Li's atomic number: Z = 3
- Total number of electrons: 3 (same number of protons)
- Period: 1 (search on the periodic table or do the electron configuration)
- Number of valence electrons: 1 (equal to the last digit of the period's number)
- Number of electrons closer to the nucleus than a valence electron: S = 3 - 1 = 2
- Z eff = Z - S = 3 - 2 = 1.
<em>4) Z eff for a Be valence eletron:</em>
- Be's atomic number: Z = 4
- Total number of electrons: 4 (same number of protons)
- Period: 2 (search on the periodic table or do the electron configuration)
- Number of valence electrons: 2 (equal to the last digit of the period's number)
- Number of electrons closer to the nucleus than a valence electron: S = 4 - 2 = 2
- Z eff = Z - S = 4 - 2 = 2
<u><em>5) Z eff for a B valence eletron:</em></u>
- B's atomic number: Z = 5
- Total number of electrons: 5 (same number of protons)
- Period: 13 (search on the periodic table or do the electron configuration)
- Number of valence electrons: 3 (equal to the last digit of the period's number)
- Number of electrons closer to the nucleus than a valence electron: S = 5 - 3 = 2
- Z eff = Z - S = 5 - 2 = 3
<u><em>6) Z eff for a N valence eletron:</em></u>
- N's atomic number: Z = 7
- Total number of electrons: 7 (same number of protons)
- Period: 15 (search on the periodic table or do the electron configuration)
- Number of valence electrons: 5 (equal to the last digit of the period's number)
- Number of electrons closer to the nucleus than a valence electron: S = 7 - 5 = 2
- Z eff = Z - S = 7 - 2 = 5
<u><em>7) Summary (order):</em></u>
Atom Zeff for a valence electron
- <u>Conclusion</u>: the order is Li < Be < B < N < F
Answer:
Bromine (Br) loses an electron, so it is the reducing agent.
Explanation:
A reducing agent also called a reducer, is known to be an electron donor. A reducing agent is oxidized, because it loses electrons in the redox reaction.
A oxidising agent also called a oxidant or oxidiser, is known to be an electron acceptor. A oxidising agent is reduced, because it gains electrons in the redox reaction.
Cl2(aq) + 2Br-(aq) --> 2Cl-(aq) + Br2(aq)
Half ionic equations,
Cl2(aq) + 2e- --> 2Cl-(aq)
2Br-(aq) --> Br2(aq) + 2e-
Reducing agent = Br-
Oxidizing agent = Cl2
Answer:
See attached picture.
Explanation:
Hello!
In this case, since C2H3Cl is an organic compound we need a central C-C parent chain to which the three hydrogen atoms and one chlorine atom provides the electrons to get all the octets except for H as given on the statement.
In such a way, on the attached picture you can find the required Lewis dot structure without formal charges and with all the unshared electron pairs, considering there is a double bond binding the central carbon atoms in order to compete their octets.
Best regards!
