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andrezito [222]
2 years ago
11

What is the electron configuration for 08 16

Chemistry
1 answer:
Elis [28]2 years ago
3 0

The electron configuration for Oxygen : [He] 2s²2p⁴

<h3>Further explanation</h3>

Writing electron configurations starts from the lowest to the highest sub-shell energy level. There are 4 sub-shells in the shell of an atom, namely s, p, d, and f. The maximum number of electrons for each sub-shell is  

• s: 2 electrons  

• p: 6 electrons  

• d: 10 electrons and  

• f: 14 electrons  

Charging electrons in the sub-shell uses the following sequence:  

<em>1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc.  </em>

The element is Oxgen, with symbol O, and :

the atomic number=8=number of electron

the atomic mass=16

The electron configuration based on the number of electrons(for Oxygen=8), so the configuration :

\tt _8^{16}O:1s^22s^22p^4 or we can write with noble gas [He] 2s²2p⁴

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A nuclear fission reaction and a nuclear fusion reaction are similar because both reactions
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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
Law Incorporation [45]

<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

7 0
3 years ago
What is the full number for 5.7x10^4
dsp73

Answer:

5.7*10^4 is equal to 57,000.

Explanation:

First, we must multiply 10 by its power, 4. That would be 10 4 times.

10*10*10*10 = 10,000.

Then, we multiply it by 5.7.

5.7*10,000 = 57,000.

Regards!

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