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3 years ago

What are the Common compounds of niobium in which it is found

Chemistry

1 answer:

blagie [28]3 years ago

7 0

Explanation:

Common Compounds of Niobium Nb

[Nb+3, Nb+5]

Compound NameFormulaMolar Mass

Niobium(III) OxideNb2O3233.811

Niobium(III) SulfateNb2(SO4)3474.0006

Niobium(V) PhosphateNb3(PO4)5753.576

Niobium(V) BromideNbBr5492.4264

Niobium(V) PentoxideNb2O5265.8098

Niobium OxychlorideNbOCl3215.2648

Niobium(V) IodideNbI5727.4287

Niobium(V) PentachlorideNbCl5270.1714

Niobium(V) ThiocyanateNb(SCN)5383.3184

Niobium(III) ChlorideNbCl3199.2654

Niobium(V) Dihydrogen PhosphateNb(H2PO4)5577.84259

Niobium NitrideNbN106.9131

Niobium(III) DichromateNb2(Cr2O7)3833.77676

Niobium HydroxideNb(OH)3143.9284

Niobium(V) PerchlorateNb(ClO4)5590.15938

Niobium(V) HypochloriteNb(ClO)5350.16838

Niobium(III) HypochloriteNb(ClO)3247.26358

Niobium(V) CarbonateNb2(CO3)5485.85726

Niobium(III) TartrateNb2(C4H4O6)3630.02564

Niobium(III) ChromateNb2(CrO4)3533.79386

Niobium(V) HexafluorosilicateNb2(SiF6)5896.192356

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2 years ago

Microwave radiation has a wavelength on the order of 1.0 cm. Calculate the frequency and the energy of a single photon of this r

denis23 [38]

Answer :

(1) The frequency of photon is, $3\times 10^{10}Hz$

(2) The energy of a single photon of this radiation is $1.988\times 10^{-23}J/photon$

(3) The energy of an Avogadro's number of photons of this radiation is, 11.97 J/mol

Explanation : Given,

Wavelength of photon = $1.0cm=0.01m$ (1 m = 100 cm)

(1) Now we have to calculate the frequency of photon.

Formula used :

$\nu=\frac{c}{\lambda}$

where,

$\nu$ = frequency of photon

$\lambda$ = wavelength of photon

c = speed of light = $3\times 10^8m/s$

Now put all the given values in the above formula, we get:

$\nu=\frac{3\times 10^8m/s}{0.01m}$

$\nu=3\times 10^{10}s^{-1}=3\times 10^{10}Hz$ $(1Hz=1s^{-1})$

The frequency of photon is, $3\times 10^{10}Hz$

(2) Now we have to calculate the energy of photon.

Formula used :

$E=h\times \nu$

where,

$\nu$ = frequency of photon

h = Planck's constant = $6.626\times 10^{-34}Js$

Now put all the given values in the above formula, we get:

$E=(6.626\times 10^{-34}Js)\times (3\times 10^{10}s^{-1})$

$E=1.988\times 10^{-23}J/photon$

The energy of a single photon of this radiation is $1.988\times 10^{-23}J/photon$

(3) Now we have to calculate the energy in J/mol.

$E=1.988\times 10^{-23}J/photon$

$E=(1.988\times 10^{-23}J/photon)\times (6.022\times 10^{23}photon/mol)$

$E=11.97J/mol$

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Give all possible ml values for orbitals that have each of the following:<br> (a) l = 3;

Angelina_Jolie [31]

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What are quantum numbers?

Quantum numbers are used to describe where around a nucleus a particular electron can be found.
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The values that each number can be are based on a set of rules.

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It describes the orientation of the orbitals.
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The value of this quantum number ranges from (-l to +l).
When l = 2, the value of will be -2, -1, 0, +1, +2.

To know more about magnetic quantum numbers, refer:

brainly.com/question/14650566

#SPJ4

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