Answer:
Xe:[Kr]4d¹⁰5(sp³d³)₆⁺² => Octahedral Geometry (AX₆)⁺²
Explanation:
Xe:[Kr]4d¹⁰(5s²5p₋₁²p₀²p₁²5d₋₂d₋₁d₀)⁺² => Xe[Kr]5(sp³d³)₆²
Ca. #Valence e⁻ = Xe + 6F - 2e⁻ = 1(8) + 6(7) - 2 = 48
Ca. #Substrate e⁻ = 6F = 6(8) = 48
#Nonbonded free pairs e⁻ = (V - S)/2 = (48 - 48)/2 = 0 free pairs
#Bonded pairs e⁻ = 6F substrates = 6 bonded pairs
BPr + NBPr = 6 + 0 = 6 e⁻ pairs => Geometry => [AX₆]⁺² => Octahedron
Xe:[Kr]4d¹⁰(5s²5p₋₁²p₀²p₁²5d₋₂d₋₁d₀)⁺² => Xe[Kr]5(sp³d³)₆⁺²
XeF₆⁺² => 6(sp³d³) hybrid orbitals => Octahedral Geometry (AX₆)
Answers:
1st: 189.6 g/mol
2nd: 0.1357 L
3rd: 1.41 M
Explanation:
Finding Molar Mass:
SnCl2 = <u>Tin(II) Chloride</u>
Tin has a molar mass of <u>118.71 g/mol</u>
Chloride has a molar mass of <u>35.453 g/mol</u>
Chloride*2 = <u>70.906</u>
<u>118.71 + 70.906 ≈ 189.6 g/mol</u>
Finding Liters of Solution:
L = mL/1000
135.7 mL / 1000 = <u>0.1357</u>
Finding Molarity:
molarity = <u>moles of solute / liters of solution</u>
M = (36.4g / 189.6g) / 0.1357 L = <u>1.41 M</u>
Hope this helped ;)
Answer: p2 = 1.06p1
Explanation: pressure increases with temperature increase.
According to Gass law
P1/T1 = P2/T2
T1 = 20°c = 20 +273 = 293k
T2 = 40°c = 40 +373 = 313k
Therefore
P2 = P1T2/T1 = 313P2/293
P2 = 1.06P1
Answer:
Average atomic mass = 14.0067 amu.
Explanation:
Given data:
Abundance of 1st isotope = 99.63%
Atomic mass of 1st isotope = 14.003 amu
Abundance of 2nd isotope = 0.37%
Atomic mass of 2nd isotope = 15.000 amu
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (99.63×14.003)+(0.37×15.000) /100
Average atomic mass = 1395.119 + 5.55 / 100
Average atomic mass = 1400.67 / 100
Average atomic mass = 14.0067 amu.
The question above may be solved using the overall mass and component balances. If we let x be the number of liters of peroxide that should be added, the final mixture will have a quantity of x+12 L. The peroxide balance before and after mixing is,
x(1) + 12(0.08) = (x + 12)(0.16)
The value of x is 1.143. Thus, approximately 1.143 L of peroxide should be added to the original solution.
