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Anni [7]
4 years ago
13

Consider hydrogen atom in an excited state of 3s1. What is the energy of its electron

Chemistry
1 answer:
SVEN [57.7K]4 years ago
8 0

<u>Given:</u>

Excited state electron configuration of H atom = 3s¹

<u>To determine:</u>

Energy of the electron in this state

<u>Explanation:</u>

The energy 'E' of an electron in a hydrogen atom occupying a level, 'n' is given by the formula:

En = -E₀/n²

where, E₀ is the ground state energy = 13.6 eV

n = principle quantum number i.e. n = 1,2,3...

Therefore,

En = -13.6/n² eV

In this case for a 3s¹ configuration, n = 3

E = -13.6/(3)² = -1.51 eV

Ans: Energy of the electron in a 3s1 state is -1.51 eV

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A container of oxygen with a volume of 60 L is heated from 300 K to 400 k, What is the new volume?
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Answer:

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Explanation:

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V2 = V1 T2/T1

T1 = 300K

V1 = 60L

T2 = 400K

V2 = ?

V2 = V1 T2/T1

V2 = (60L)(400K) / (300K)

V2 = 80L

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What does homogeneous mixture of elements mean
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Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32
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<u>Answer:</u>

<u>For a:</u> The standard Gibbs free energy of the reaction is -347.4 kJ

<u>For b:</u> The standard Gibbs free energy of the reaction is 746.91 kJ

<u>Explanation:</u>

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}           ............(1)

  • <u>For a:</u>

The given chemical equation follows:

2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)

<u>Oxidation half reaction:</u>   Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-       ( × 2)

<u>Reduction half reaction:</u>   3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)

We are given:

n=2\\E^o_{cell}=+1.08V\\F=96500

Putting values in equation 1, we get:

\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

  • <u>For b:</u>

The given chemical equation follows:

6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)

<u>Oxidation half reaction:</u>   Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-       ( × 6)

<u>Reduction half reaction:</u>   2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)

We are given:

n=6\\E^o_{cell}=-1.29V\\F=96500

Putting values in equation 1, we get:

\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

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