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Nataly_w [17]
3 years ago
15

What are energy sublevels?

Chemistry
2 answers:
MaRussiya [10]3 years ago
5 0
The answer is B: multiple orbitals of specific magnitude within an energy level
sdas [7]3 years ago
3 0

Answer:

Option B

Explanation:

First, we need to know what is an energy level.

An energy level corresponds to the row of the periodic table of element. So, if you see the attached picture, you can see that we have 7 rows where the elements are distributed, so, we have 7 energy levels for all those elements.

Knowing that, the sub levels are the caps where the electrons of the atoms are carried. These sublevels or caps, are called orbitals, these can be of several types

s orbytal: can hold 2 electrons

p orbytal: can hold up to 6 electrons.

d orbytal: can hold up to 10 electrons

f orbytal: can hold up to 14 electrons

g orbytal: can hold up to 18 electrons.

Depending on the row (or energy level) and the atom, we can know how many electrons can carry an element, in which period or row is, and the sub levels. For example, the Chlorine, with an atomic number of 17, can carry up to 7 electrons in it's outer level and it's on the third row (two energy levels). This can be known with it's electronic configuration:

[Cl] = 1s^2 2s^2 2p^6 3s^2 3p^5

The last energy level is 3, so it's the third period, and the electrons of those sub level are 2 and 5, 7 electrons.

Hope this can help better

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If the mass is 44g and the volume is 10ml, what is the density of the object
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Answer:44/10

Explanation:

Because D= G/ml

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At a given temperature the vapor pressure of pure liquid benzene and toluene are 745 torr and 290 torr, respectively. A solution
IRISSAK [1]

Answer:

Vapour pressure of benzene over the solution is 253 torr

Explanation:

According to Raoult's law for a mixture of two liquid component A and B-

vapour pressure of a component (A) in solution = x_{A}\times P_{A}^{0}

vapour pressure of a component (B) in solution = x_{B}\times P_{B}^{0}

Where x_{A},x_{B} are mole fraction of component A and B in solution respectively

P_{A}^{0},P_{B}^{0} are vapour pressure of pure A and pure B respectively

Here mole fraction of benzene in solution is 0.340 and vapour pressure of pure benzene is 745 torr

So, vapour pressure of benzene in solution = 0.340\times 745 torr

                                                                         = 253 torr

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Which model is the best for identifying geometric isomers?
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A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli
Diano4ka-milaya [45]

Answer:

\large \boxed{109.17 \, ^{\circ}\text{C}}

Explanation:

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50/50 ethylene glycol (EG):water

V = 4.70 gal

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ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

\Delta T_{b} = iK_{b}b

i is the van’t Hoff factor —  the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}

2. Kilograms of water

m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1  gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}

3. Molal concentration of EG

b =  \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}

4. Increase in boiling point

\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}

5. Boiling point

\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}

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The answer to this is t<span>he atom is mostly empty space.</span>
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