Answer:
97 000 g Na
Explanation:
The absortion (or liberation) of energy in form of heat is expressed by:
q=m*Cp*ΔT
The information we have:
q=1.30MJ= 1.30*10^6 J
ΔT = 10.0°C = 10.0 K (ΔT is the same in °C than in K)
Cp=30.8 J/(K mol Na)
If you notice, the Cp in the question is in relation with mol of Na. Before using the q equation, we can find the Cp in relation to the grams of Na.
To do so, we use the molar mass of Na= 22.99g/mol
Now, we are able to solve for m:
=97 000 g Na
Answer:
yes
Explanation:
Usually, it would not affect the crucible, but depending on the temperature of the flame the enamel of the crucible may begin to melt and stick to the metal object being used to handle the crucible. This tiny amount that is melted off can cause very small changes in the original mass of the crucible, which although it is almost unnoticeable it is still there. Therefore, the answer to this question would be yes.
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
Molal or molar? There is a difference