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3 years ago

How can I tell how many electrons in each energy level? When my question states "third sublevel" does this mean sublevel P?

1 answer:

6 0

Sorry for the delay! My internet is a bit bad.
P is the third sublevel. Each sublevel (the angular momentum quantum number), has its own number:
<span>s = 1, p =3, d = 5, f = 7</span>
The number of electrons for each is:
s-2
p-6
d-10
f-14
It's easier to just memorize these numbers, but the equation for determining the sublevel number is 2n (n = the principal quantum number). The principal quantum number is based on the period the element is in.

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A physical change _____ change the composition or identity of the substance. will will not 2. An example of a physical change is

Sav [38]

Answer:

not sure

Explanation:

a physical change (can)change.,....

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3 years ago

When a person looks at a bright light, tiny muscles in the eye contract so less light can enter the eye.

bagirrra123 [75]

Answer:

involuntary, attached to the eyeball, nonstriated.

Explanation:

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3 years ago

Read 2 more answers

1. Mr. Brown weighs 670 newtons and he climbed a ladder that was 5 meters

ziro4ka [17]

Work = force*distance
Work = 670 * 5
Work = 3350 Nm

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2 years ago

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:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

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 $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$

7 0

3 years ago

student mixes vinegar and baking soda inside a sealed plastic bag would the reaction be the same if baking soda and vinegar were

pychu [463]

Yes, because nothing has changed. The same ingredients are used and if one reaction happened in a a bag then why wouldn’t it happen in a container?

3 0

3 years ago