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jek_recluse [69]
3 years ago
13

Use the periodic table to determine the electron configuration for dysprosium (Dy) and americium (Am) in noble-gas notation

Chemistry
2 answers:
sergij07 [2.7K]3 years ago
3 0

Answer:

Dysprosium [Dy]=[Xe]4f^{10}6s^2

Americium [Am]=[Rn]5f^77s^2

Dysprosium is a chemical element with symbol Dy and atomic number of 66. It is a rare earth metal and as it contains partially filled f sub shells, it belongs to f block. Xe is the nearest noble gas and has atomic number of 54.

Americium is a chemical element with symbol Am and atomic number of 95. It is a rare earth metal and as it contains half filled f sub shells, it belongs to f block. Radon is the nearest noble gas and has atomic number of 86.

Alex3 years ago
3 0

Dy: [Xe]6s24f10

Am: [Rn]7s25f 7 :)

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The equilibrium constant in terms of pressure. Kp for the following process is 0.179 at 50 °C. It increases to 0.669 at 86 °C. C
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Answer:

a) ΔHvap=35.3395 kJ/mol

b) Tb=98.62 °C

Explanation:

Given the reaction:

C₇H₁₆ (l) ⇔ C₇H₁₆ (g)

Kp=P(C₇H₁₆) since the concentration ratio for a pure liquid is equal to 1.

When

T₁=50°C=323.15K ⇒P₁=0.179

T₂=86°C=359.15K ⇒P₂=0.669

The Clasius-Clapeyron equation is:

ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})

ln(\frac{0.669}{0.179}) =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{359.15K}-\frac{1}{323.15K})

1.3184 =-\frac{AH_{vap}}{8.3145 J.mol^{-1}K^{-1}} (-3.10186*10^{-4}K{^-1})

ΔHvap=35339.5 J/mol=35.3395 KJ/mol

Normal boiling point ⇒ P=1 atm

Hence, we find the normal boiling point where:

T₁=323.15K

P₁=0.179 atm

P₂=1 atm

ln(\frac{P_2}{P_1}) =-\frac{AH_{vap}}{R} (\frac{1}{T_2}-\frac{1}{T_1})

ln(\frac{1atm}{0.179atm}) =-\frac{35339.5 J/mol}{8.3145 J.mol^{-1}K^{-1}} (\frac{1}{T_2}-\frac{1}{323.15K})

1.7203=-4250.34 (\frac{1}{T_2}-\frac{1}{323.15K})

T₂=371.77 K= 98.62 °C

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3 years ago
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