Answer:
The change in enthalpy is +210 kJ, so it is an endothermic reaction.
Explanation:
- The change in enthalpy shows that the reaction is exothermic or endothermic.
The change in enthalpy = The energy of products - the energy of reactants.
- If ΔH is positive value, the reaction is endothermic, the energy of the products is higher than that of the reactants.
- If ΔH is negative value, the reaction is exothermic, the energy of the products is lower than that of the reactants.
<em>For this reaction: </em>The change in enthalpy = The energy of products - the energy of reactants = 370 kJ - 160 kJ = + 210 kJ.
<em>So, the right choice is: The change in enthalpy is +210 kJ, so it is an endothermic reaction.</em>
When carbon undergoes sp2 hybridization it forms methane ? I believe.
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:
Mass is Universal and weight is tied to the strength of the gravitational force of the planet .
Explanation:
The reason is because without mass you have no weight witch means if you have a certain amount of mass you will have a certain weight .
Given:
Concentration of Fluoride ions = 0.100 M
Concentration of Hydrogen Fluoride = 0.126 M
Asked: Concentration of fluoride ions after the addition of 5ml of 0.0100 M HCl to 25 mL of the solution
Assume: 50:50 ratio of fluoride ions and HF
12.5ml*0.1mol/L *1L/1000mL + 12.5*0.126mol/L * 1L/1000mL = 2.825x10^-3 moles F-
5ml * 0.01 mol/L *1L/1000mL = 5x10^-5 moles
Assume: Volume additive
Final concentration = 2.825x10^-3 + 5x10^-5 moles/ 30 ml * 1000ml/L =0.0958 M
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