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natulia [17]
2 years ago
11

I’ll mark brainliest please

Chemistry
1 answer:
DaniilM [7]2 years ago
5 0
I’m assuming your just writing the formula? If so
Potassium chloride: KCL
Potassium nitride: KNO2
Potassium sulfide: K2S
calcium chloride: CaCl2
Calcium nitride: Ca3N2
Calcium sulfide: CaS
Silver chloride: AgCl
Silver nitride: Ag3N
Silver sulfide: Ag2S
Manganese (||) chloride: MnCl2
Manganese (||) nitride: Mn3N2
Manganese (||) sulfide: MnS
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Write an equation that shows the formation of the chloride ion from a neutral chlorine atom
jeka57 [31]

The gaining of electron by an atom results in the formation of anion shown by the negative charge on the atom whereas lose of electron results in the formation of cation shown by positive charge on the atom. The atom lose or gain electron to complete their octet and get stable in nature.

The chlorine atom will gain an electron and form chloride anion with one negative charge on it. The chloride ion is more stable in nature compared to the chlorine atom due to complete octet of chloride ion by gaining of electron.

Electronic configuration of chlorine atom is:

[Ne]3s^{2}3p^{5}

By gaining of one electron, electronic configuration of chloride ion is:

[Ne]3s^{2}3p^{6}

Thus, the equation that shows the formation of the chloride ion from a neutral chlorine atom is:

Cl+1e^{-}\rightarrow Cl^{-}

4 0
3 years ago
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Explanation:

7 0
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An atom (not a hydrogen atom) absorbs a photon whose associated wavelength is 300 nm and then immediately emits photon whose ass
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Answer:

The net energy is 2.196 eV

Explanation:

Basically, the energy of an atom increases when it absorbs a photon. In addition, the wavelength of the emitted photon is longer such that the atom absorbed a net energy in the process.

Using:

ΔE = h*c*(1/λ_{1} - 1/λ_{2})

where:

ΔE is the net energy in eV (electron-volt). 1 eV is equivalent to 1.602*10^{-19} J.

h = 4.135*10^{-15} eVs

c = 3*10^{8} m/s

λ_{1} = 300 nm = 300*10^{-9} m

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Thus:

ΔE = 4.135*10^{-15} eVs*3*10^{8} m/s*(\frac{1}{300*10^{-9}m } }-\frac{1}{640*10^{-9}m })

ΔE = 4.135*10^{-15}*3*10^{8}*1.77*10^{6} eV = 2.196 eV

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