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3 years ago

What is the electron configuration of an element with atomic number 15 ? Please help

Chemistry

2 answers:

crimeas [40]3 years ago

7 0

If the atomic number is 15, this means this is element P in the periodic table.

the electron configuration is:

$[P]= 1S^22S^22P^63S^23P^3$

Travka [436]3 years ago

7 0

Here is the explanation. Good luck.

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tate whether the following changes are physical or chemical for rancidipication fixation of water 2 tearing of paper 3 rusting o

damaskus [11]

Answer: Physical change : tearing of paper, fixing of wtaer

Chemical change: rusting of iron , electrolysis of water, Rancidification

Explanation:

Physical change is a change in which there is no rearrangement of atoms and thus no new substance is formed. There is only change in physical state of the substance.

Example: tearing of paper, fixing of wtaer

Chemical change is a change in which there is rearrangement of atoms and thus new substance is formed. There may or may not be a change in physical state.

Example: rusting of iron , electrolysis of water, Rancidification

3 0

2 years ago

Complete this neutralization equation:<br> H2SO4 + Al(OH)3

elena-s [515]

Answer:

H2SO4 + Al(OH)3 = Al2(SO4)3 + H2O

Explanation:

4 0

3 years ago

Which statements are true about Figure I and Figure II below? (Check all that apply)

Scilla [17]

Both figures are mixtures,

Figure II is a heterogenous mixture

Figure I is a homogenous mixture

5 0

3 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

ELETE

yKpoI14uk [10]

Answer:

5 L

Explanation:

Use Charles law and rearrange formula

Change C to K

- Hope that helped! Please let me know if you need further explanation.

6 0

3 years ago