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levacccp [35]
3 years ago
15

The photon of light that is emitted as an electron drops back to its original orbit is:

Chemistry
1 answer:
Alex73 [517]3 years ago
3 0

Answer:

energy

Explanation:

The photon of light that is emitted as an electron drops back to its original orbit is energy and this energy is released during de-excitation process.

The electron is jumped into higher level and back into lower level by absorbing and releasing the energy.

The process is called excitation and de-excitation.

Excitation:

When the energy is provided to the atom the electrons by absorbing the energy jump to the higher energy levels. This process is called excitation. The amount of energy absorbed by the electron is exactly equal to the energy difference of orbits.  For example if electron jumped from K to L it must absorbed the energy which is equal the energy difference of these two level. The excited electron thus move back to lower energy level which is K by releasing the energy because electron can not stay longer in higher energy level and comes to ground state.

De-excitation:

When the excited electron fall back to the lower energy levels the energy is released in the form of radiations. this energy is exactly equal to the energy difference between the orbits. The characteristics bright colors are due to the these emitted radiations. These emitted radiations can be seen if they are fall in the visible region of spectrum

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When 1.04g of cyclopropane was burnt in excess oxygen in a bomb calorimeter, the temperature rose by 3.69K. The total heat capac
STatiana [176]

Answer:

\Delta _{comb}H=-2,093\frac{kJ}{mol}

Explanation:

Hello!

In this case, since these calorimetry problems are characterized by the fact that the calorimeter absorbs the heat released by the combustion of the substance, we can write:

Q_{rxn}+Q_{cal}=0

Thus, given the temperature change and the total heat capacity, we obtain the following total heat of reaction:

Q_{rxn}=-14.01kJ/K*3.69K\\\\Q_{rxn}=-51.70kJ

Now, by dividing by the moles in 1.04 g of cyclopropane (42.09 g/mol) we obtain the enthalpy of combustion of this fuel:

n=\frac{1.04g}{42.09g/mol}=0.0247mol\\\\\Delta _{comb}H=\frac{Q_{rxn}}{n}\\\\  \Delta _{comb}H=-2,093\frac{kJ}{mol}

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4 0
3 years ago
Give atleast 5 differences in cell structures of plant and animal cells
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The shape, the cell wall, chlorophyll, the components of the nucleus, cytoplasm
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2 years ago
Which of these molecules has the same number of shared electron pairs as unshared electron pairs? (a) hcl, (b) h2s, (c) pf3, (d)
Paul [167]
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4 years ago
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The bond angles in SCl2 are expected to be Multiple Choice a little more than 109.5°. 109.5°. 120°. a little less than 109.5°. 1
nalin [4]

Answer:

a little less than 109.5°

Explanation:

SCl2 has four regions of electron density around the central atom of the molecule. This implies that it has a tetrahedral electron domain geometry with an expected bond angle of 109.5° according to valence shell electron pair repulsion theory.

However, there are two lone pair of electrons on the central atom of the molecule which decreases the bond angle a little less than 109.5° owing to repulsion between electron pairs.

4 0
3 years ago
what is an empirical formulaWhat is the percent composition by mass of nitrogen in (NH4)2CO3 (gram-formula mass = 96.0 g/mol)?
Margarita [4]

Answer:

Percentage composition = 14.583%

Explanation:

In chemistry, the emprical formular of a compound is the simplest formular a compound can have. It shows the simplest ratio in which the elements are combined in the compound.

Percentage composition by mass of Nitrogen

Nitrogen = 14g/mol

In one mole of the compound;

Mass of Nitrogen = 1 mol * 14g/mol = 14g

Mass of compound = 1 mol * 96.0 g/mol = 96

Percentage composition of Nitrogen = (Mass of Nitrogen /  Mass of compound) * 100

percentage composition = 14/96   * 100

Percentage composition = 0.14583 * 100

Percentage composition = 14.583%

3 0
3 years ago
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