Which is NOT true of a mixture?

Chemistry

1 answer:

Reika [66]3 years ago

6 0

Answer:

OPTUON C,CHEMICALLY BONDED

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What is the wavenumber of the radiation emitted when a hydrogen

LUCKY_DIMON [66]

Answer: Wavenumber of the radiation emitted is $0.08\times 10^{8}m^{-1}$

Explanation:

The relationship between wavelength and energy of the wave follows the equation:

$E=\frac{hc}{\lambda}$

where,

E = energy of the radiation = $1.634\times 10^{-18}J$

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

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

$\lambda$ = wavelength of radiation = ?

Putting values in above equation, we get:

$1.634\times 10^{-18}J=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{\lambda}\\\\\lambda=12.16\times 10^{-8}m$

$\bar {\nu}=\frac{1}{\lambda}=\frac{1}{12.16\times 10^{-8}}=0.08\times 10^{8}m^{-1}$

Thus wavenumber of the radiation emitted is $0.08\times 10^{8}m^{-1}$

8 0

3 years ago

One of the few xenon compounds that form is cesium xenon heptafluoride (CsXeF7). How many moles of CsXeF7 can be produced from t

larisa [96]

The balanced chemical equation is written as:

CsF(s) + XeF6(s) ------> CsXeF7(s)

We are given the amount of cesium fluoride and xenon hexafluoride used for the reaction. We need to determine first the limiting reactant to proceed with the calculation. From the equation and the amounts, we can say that the limiting reactant would be cesium fluoride. We calculate as follows:

11.0 mol CsF ( 1 mol CsXeF7 / 1 mol CsF ) = 11.0 mol CsXeF7

6 0

3 years ago

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What is the molarity of a solution prepared by dissolving 10.7 g NaI in 0.250 L?0.0714

hjlf

Answer:0.286M

Explanation:

10.7g / 149.89g/mol×0.25L

4 0

3 years ago

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