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2 years ago

Use electron configurations to account for the stability of the lanthanide ions Ce⁴⁺ and Eu²⁺.

Chemistry

1 answer:

Citrus2011 [14]2 years ago

7 0

Answer:

Explanation:

The Ce metal has electronic configuration as follows

[Xe] 4f¹5d¹6s²

After losing 4 electrons , it gains noble gas configuration ,. So Ce ⁺⁴ is stable.

Eu has electronic configuration as follows

[ Xe ] 4 f ⁷6s²

[ Xe ] 4 f ⁷

Its outermost orbit contains 2 electrons so Eu²⁺ is stable. Its +3 oxidation state is also stable.

Ce⁺²

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A sample of neon gas at a pressure of 1.08 atm fills a flask with a volume of 250 mL at a temperature of 24.0 °C. If the gas is

musickatia [10]

Answer:

124.91mL

Explanation:

Given parameters:

P₁ = 1.08atm

V₁ = 250mL

T₁ = 24°C

P₂ = 2.25atm

T₂ = 37.2°C

V₂ = ?

Solution:

To solve this problem, we are going to apply the combined gas law;

$\frac{P_{1} V_{1} }{T_{1} } = \frac{P_{2} V_{2} }{T_{2} }$

P, V and T represents pressure, volume and temperature

1 and 2 delineates initial and final states

Convert the temperature to kelvin;

T₁ = 24°C, T₁ = 24 + 273 = 297K

T₂ = 37.2°C , T₂ = 37.2 + 273 = 310.2K

Input the variables and solve for V₂

$\frac{1.08 x 250}{298} = \frac{2.25 x V_{2} }{310.2}$

V₂ = 124.91mL

6 0

2 years ago

A 1.0 g sample of a cashew was burned in a calorimeter containing 1000. g of water, and the temperature of the water changed fro

Savatey [412]

Answer:

The correct answer is option C.

Explanation:

1.0 g sample of a cashew :

Heat released on combustion of 1.0 gram of cashew = -Q

We have mass of water = m = 1000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water : Q

$Q=1000 g\times 4.184 J/g^oC\times 5^oC=20,920 J$

Heat released on combustion of 1.0 gram of cashew is -20,920 J.

3.0 g sample of a marshmallows :

Heat released on combustion of 3.0 g sample of a marshmallows = -Q'

We have mass of water = m = 2000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water : Q'

$Q'=2000 g\times 4.184 J/g^oC\times 5^oC=41,840 J$

Heat released on 3.0 g sample of a marshmallows= -Q' = -41,840 J

Heat released on 1.0 g sample of a marshmallows : q

$q =\frac{-Q'}{3} = \frac{-41,840 J}{3}=-13,946.67 J$

Heat released on combustion of 1.0 gram of marshmallows -13,946.67 J.

-20,920 J. > -13,946.67 J

The combustion of 1.0 g of cashew releases more energy than the combustion of 1.0 g of marshmallow.

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2 years ago

At 10 K Cp,m(Hg(s)) = 4.64 J K−1 mol−1. Between 10 K and the melting point of Hg(s), 234.3 K, heat capacity measurements indicat

umka21 [38]

Answer:

S°m,298K = 85.184 J/Kmol

Explanation:

∴ T = 10 K ⇒ Cp,m(Hg(s)) = 4.64 J/Kmol

∴ 10 K to 234.3 K ⇒ ΔS = 57.74 J/Kmol

∴ T = 234.3 K ⇒ ΔHf = 2322 J/mol

∴ 234.3 K to 298.0 K ⇒ ΔS = 6.85 J/Kmol

⇒ S°m,298K = S°m,0K + ∫CpdT/T(10K) + ΔS(10-234.3) + ΔHf/T(234.3K) + ΔS(234.3-298)

⇒ S°m,298K = 0 + 10.684 J/Kmol + 57.74 J/Kmol + 9.9104 J/Kmol + 6.85 J/kmol

⇒ S°m,298K = 85.184 J/Kmol

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