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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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The densities of gases a, b, and c at stp are 1.25 g/l, 2.86 g/l and 0.714 g/l, respectively. calculate the molar mass of each s

kicyunya [14]

The molar mass of a, b and c at STP is calculated as below

At STP T is always= 273 Kelvin and ,P= 1.0 atm

by use of ideal gas equation that is PV =nRT
n(number of moles) = mass/molar mass therefore replace n in the ideal gas equation

that is Pv = (mass/molar mass)RT
multiply both side by molar mass and then divide by Pv to make molar mass the subject of the formula

that is molar mass = (mass x RT)/ PV

density is always = mass/volume

therefore by replacing mass/volume in the equation by density the equation
molar mass=( density xRT)/P where R = 0.082 L.atm/mol.K

the molar mass for a
= (1.25 g/l x0.082 L.atm/mol.k x273k)/1.0atm = 28g/mol

the molar mass of b
=(2.86g/l x0.082L.atm/mol.k x273 k) /1.0 atm = 64 g/mol

the molar mass of c

=0.714g/l x0.082 L.atm/mol.K x273 K) 1.0atm= 16 g/mol

therefore the
gas a is nitrogen N2 since 14 x2= 28 g/mol
gas b =SO2 since 32 +(16x2)= 64g/mol
gas c = methaneCH4 since 12+(1x4) = 16 g/mol

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3 years ago

Complete combustion of 2.0 metric tons of coal to gaseous carbon dioxide releases 6.6X10¹⁰ J of heat. Convert this energy to

Jobisdone [24]

6.6 × $10^{10}$ J = 6.2 × $10^{7}$ British thermal units

First of all, we should know that 1 British thermal unit = 1055 Joules.

From question, heat = 6.6 × $10^{10}$ J

1 BTU = 1055 J

6.6 × $10^{10}$ J = $\frac{1}{1055}$ × 6.6 × $10^{10}$ British thermal units

= 6.2 × $10^{7}$ British thermal units

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1 year ago

Lets say you were doing an experiment on two different chemicals in two different test tubes. One chemical is giving off oxygen

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Answer:

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IgorLugansk [536]

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4 years ago

Read 2 more answers

You add 100.0 g of water at 52.0 °C to 100.0 g of ice at 0.00 °C. Some of the ice melts and cools the water to 0.00 °C. When the

lara31 [8.8K]

Answer:

$m_{ice} = 65.336\,g$

Explanation:

Accoding to the First Law of Thermodynamics, the heat released by the water melts a portion of ice. That is to say:

$Q_{water} = Q_{ice}$

$(100\,g)\cdot \left(4.184\,\frac{J}{kg\cdot ^{\textdegree}C}\right)\cdot (52\,^{\textdegree}C - 0\,^{\textdegree}C) = m_{ice}\cdot \left(333\,\frac{J}{g} \right)$

The amount of ice that is melt is:

$m_{ice} = 65.336\,g$

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