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Alekssandra [29.7K]
3 years ago
6

It’s easier to determine the elecron configurations for the p-block elements in periods 1,2,3 than to determine the electrons co

nfigurations for the rest of the p-block elements in the periodic table beacause
Chemistry
2 answers:
Scorpion4ik [409]3 years ago
5 0

Answer:It’s easier to determine the elecron configurations for the p-block elements in periods 1,2,3 than to determine the electrons configurations for the rest of the p-block elements in the periodic table beacause from period 4, specifically from the element 31 (Ga), the atoms start to fill the d orbitals, and the energy levels of the 3d orbitals ara quite similar to the energy levels of 4p orbitals. So, for the elements Cr and Cu the right configurations do not match the configurations predicted using Aufbau method and Hund rules. Those are not the only exceptions but the two first. All is due to the proximity of the energy of the d and p orbitals and the fact that the rearrangement of the electrons result in a lower energy

Explanation:

kotegsom [21]3 years ago
3 0
<span>It’s easier to determine the elecron configurations for the p-block elements in periods 1,2,3 than to determine the electrons configurations for the rest of the p-block elements in the periodic table beacause from period 4, specifically from the element 31 (Ga), the atoms start to fill the d orbitals, and the energy levels of the 3d orbitals ara quite similar to the energy levels of 4p orbitals. So, for the elements Cr and Cu the right configurations do not match the configurations predicted using Aufbau method and Hund rules. Those are not the only exceptions but the two first. All is due to the proximity of the energy of the d and p orbitals and the fact that the rearrangement of the electrons result in a lower energy level. </span>
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Burning one gallon of gasoline releases 20 pounds of CO2 into the air. Harry’s compact car can travel 25 miles on one gallon of
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8 0
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Read 2 more answers
When a 12.8 g sample of KCL dissolves in 75.0 g of water in a calorimeter the temp. drops from 31 Celsius to 21.6 Celsius. Calcu
Delicious77 [7]

Answer:

Step 1: Calculate qsur (the surrounding is

usually the water)

qsur = ? J

m = 75.0 g water

c = 4.184 J/g

oC

ΔT = (Tfinal- Tinitial)= (21.6 – 31.0) = -9.4 oC

qsur = m · c · (ΔT)

qsur = (75.0g) (4.184 J/g

oC) (-9.4 oC)

qsur = - 2949.72 J

First, using the information we know that we

must solve for qsur, which is the water. We know

the mass for water, 75.0g, the specific heat of

the water, 4.184 j/g

o

c, and the change in

temperature, 21.6-31.0 = -9.4 oC. Plugging it

into the equation, we solve for qsur.

Step 2: Calculate qsys qsys = - (qsur)

qsys = - (- 2949.72 J)

qsys = + 2949.72

In this case, the qsur is negative, which means

that the water lost energy. Where did it go? It

went to the system. Thus, the energy of the

system is negative, opposite, the energy of the

surrounding.

Step 3: Calculate moles of the substance

that is the system

Given: 12.8 g KCl

Mol system = (g system given)

(molar mass of system)

Mol system = (12.8 g KCl)

(39.10g + 35.45g)

Mol system = 12.8 g KCl

74.55 g

Mol system = 0.172

Here, we solve for the mol in the system by

using the molar mass of the material in the

system.

Step 4: Calculate ΔH ΔH = q sys .

Mol system

ΔH= + 2949.72 J

0.172 mol

ΔH= +17179.81 J/mol or +1.72 x 104

J/mol

i hope this helps

7 0
3 years ago
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