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3 years ago

What varieties of matter may be the constituents in mixtures?

Chemistry

1 answer:

vodka [1.7K]3 years ago

7 0

Answer: two or more different pure substances, which may be elements or compounds.

Explanation:

The term varieties of matter is kind of ambiguos since it is not defined.

The best approach to the question is to think of matter as it can be classified into to kinds: pure substaces and mixtures.

Elements and compounds are pure substances.

Elements are pure substances constituted by only one kind of atoms. An element cannot be divided into simpler substances either by physical or chemical media.

A compound is a chemical combination of two or more different elements. A compound can be divided into simpler substances by chemical reactions, but not by physical media. The properties of compounds are different of those of the elements of which they are constituted. The composition of a compoind (ratio among its elements) is fixed.

A mixture is the physical combination of two or more pure substances (either different elements or compounds) which can be mixed in any proportion. This is, its composition is variable. The substances that form the mixture can be separated by physical media.

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3 years ago

Calculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom. What type of electr

Alecsey [184]

Answer:

The wavelength of the emitted photon will be approximately 655 nm, which corresponds to the visible spectrum.

Explanation:

In order to answer this question, we need to recall Bohr's formula for the energy of each of the orbitals in the hydrogen atom:

$E_{n} = -\frac{m_{e}e^{4}}{2(4\pi\epsilon_{0})^2\hbar^{2}}\frac{1}{n^2} = E_{1}\frac{1}{n^{2}}$ , where:

[tex]m_{e}[tex] = electron mass

e = electron charge

[tex]\epsilon_{0}[tex] = vacuum permittivity

[tex]\hbar[tex] = Planck's constant over 2pi

n = quantum number

[tex]E_{1}[tex] = hydrogen's ground state = -13.6 eV

Therefore, the energy of the emitted photon is given by the difference of the energy in the 3d orbital minus the energy in the 2nd orbital:

[tex]E_{3} - E_{2} = -13.6 eV(\frac{1}{3^{2}} - \frac{1}{2^{2}})=1.89 eV[tex]

Now, knowing the energy of the photon, we can calculate its wavelength using the equation:

[tex]E = \frac{hc}{\lambda}[tex], where:

E = Photon's energy

h = Planck's constant

c = speed of light in vacuum

[tex]\lambda[tex] = wavelength

Solving for [tex]\lambda[tex] and substituting the required values:

[tex]\lambda = \frac{hc}{E} = \frac{1.239 eV\mu m}{1.89 eV}=0.655\mu m = 655 nm[tex], which correspond to the visible spectrum (The visible spectrum includes wavelengths between 400 nm and 750 nm).

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