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Crazy boy [7]
3 years ago
11

True False The colors emitted depends on the number of free electrons passing through the lamp. True False When a free electron

hits an atom, the atom is always excited to the highest energy level possible. True False The kinetic energy of the free electron at the point of collision increases as the voltage of the battery increases. True False The kinetic energy of the free electron at the point of collision is higher if the atom is closer to the source of electrons. True False The only way to emit IR photons is if there are empty electronic energy levels really close to the ground state (lowest energy level). True False When atomic electrons are excited to a higher level, they always return to their lowest energy level by jumping down one level at a time.
Chemistry
1 answer:
TiliK225 [7]3 years ago
5 0

Answer:

False

False

True

False

False

False

Explanation:

The colour emitted by an excited atom depends on the emission Spectra of the atom rather than on the number of free electrons passing through the lamp.

When a free electrons hits an atom, the atom can be excited to various intermediate states other than the highest energy level depending on the amount of energy it absorbed.

Increase in voltage of the battery also increases the kinetic energy of the electron

The kinetic energy of the electron depends on the voltage difference not on the distance from the atom.

The Paschen, Brackett, and Pfund series of lines are due to transitions from higher-energy orbits to orbits with n = 3, 4, and 5, respectively; these transitions release substantially less energy, corresponding to infrared radiation.

Excited atoms can jump from a higher level to the ground state in a series of steps or directly to the ground state

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For an alloy that consists of 91.0 g copper, 111 g zinc, and 7.51 g lead, what are the concentrations (a) of Cu (in at%), (b) of
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Answer:

\% Cu=45.2\%\\\\\% Zn=53.6\%\\\\\% Pb=1.2\%

Explanation:

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In this case, given the masses of copper, zinc and lead, it is possible to compute the moles via their atomic masses first:

n_{Cu}=\frac{91.0gCu}{63.55g/mol}=1.432mol\\\\ n_{Zn}=\frac{111gZn}{65.41g/mol}=1.697mol\\\\n_{Pb}=\frac{7.51gPb}{207.2g/mol}=0.0362mol\\

Now, we compute the atomic percentages as shown below:

\% Cu=\frac{1.432}{1.432+1.697+0.0362}*100\% =45.2\%\\\\\% Zn=\frac{1.697}{1.432+1.697+0.0362}*100\%=53.6\%\\\\\% Pb=\frac{0.0362}{1.432+1.697+0.0362}*100\%=1.2\%

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