Answer:
D
Explanation:
Ocean breezes keep coastal galveston cooler than Del Rio, which is inland exposed to southerly winds.
Answer: The average valence electron energy (AVEE) of this element =
1014.2 KJ/ mol or 1.0142mJ/mol.
Explanation:
The average valence electron energy = (number of electrons in s subshell x Ionization energy of that subshell) + (number of electrons in p subshell x Ionization energy of that subshell) / total number of electrons in both subshells of the valence shells.
The 5A elements are non-metals like Nitrogen and Phosphorus with the metallic character increasing as you go down the group, So a new 5A element will have characteristics of its group with 5 valence electron in its outermost shell represented as ns2 np3
Therefore the average valence electron energy (AVEE) of this element will be calculated as
The average valence electron energy = (2 x 1370 kJ/mol + 3 x 777 kJ/mol.) / 5
2740+2331/ 5 =5071/5
=1014.2 KJ/ mol or 1.0142mJ/mol.
Answer:
0.057 M
Explanation:
Step 1: Given data
Solubility product constant (Ksp) for HgBr₂: 2.8 × 10⁻⁴
Concentration of mercury (II) ion: 0.085 M
Step 2: Write the reaction for the solution of HgBr₂
HgBr₂(s) ⇄ Hg²⁺(aq) + 2 Br⁻
Step 3: Calculate the bromide concentration needed for a precipitate to occur
The Ksp is:
Ksp = 2.8 × 10⁻⁴ = [Hg²⁺] × [Br⁻]²
[Br⁻] = √(2.8 × 10⁻⁴/0.085) = 0.057 M
Answer:
.081 g of O2
Explanation:
4Cr + 3O2 -----> 2Cr2O3
.175 g Cr x [1 mole / 52.0 g] x [2 moles Cr2O3 / 4 moles Cr] x [152 g / 1 mole] = .256 g of Cr2O3
.175 g Cr x [1 mole / 52.0 g] x [3 moles O2 / 4 moles Cr] x [32 g / 1 mole] = .081 g of O2
