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Doss [256]
3 years ago
5

Order the following elements from least to greatest ionization energy.

Chemistry
1 answer:
earnstyle [38]3 years ago
5 0

Answer:

K ∠ Ca ∠ S ∠ F

Explanation:

Trend along period:

As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.

trend along group:

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased and ionization energy decrease.

Fluorine is present on right side of periodic table so its ionization energy will be high.

F = 1681 kj/mol

S = 999.6 kj/mol

Ca = 589.8 kj/mol

K = 418.8 kj/mol

while potassium is present most left of the periodic table its ionization energy is less.

The given element can be arranged in following order:

K ∠ Ca ∠ S ∠ F

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3 years ago
Explain its significance.Check all that apply. Systems with high potential energy tend to change in the direction of lower poten
KatRina [158]

Answer:

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Explain the significance of the law of conservation of energy. Check all that apply below.

1. Systems with high potential energy tend to change in the direction of lower potential energy, obtaining energy from another objects. Systems with high potential energy tend to change in the direction of lower potential energy, releasing energy into the surroundings.

2. In the exchanges, the total energy is always conserved; energy is neither created nor destroyed. In the exchanges, the total energy is always changed; energy is created and destroyed.

3. In chemical and physical changes, matter never exchanges energy with its surroundings. In chemical and physical changes, matter often exchanges energy with its surroundings.

Explanation:

1. System with high potential energy tends to change in a direction to system with low potential energy. This system with high potential energy moving towards the direction of lower potential energy will give out their potential energy to the surrounding in form of heat, as they also receive kinetic energy from the surrounding. This is a law of conservation of energy. Energy from the system was transformed from one form to another.

2. Total energy in an exchange is always conserved or transformed.

It is conserved when energy is not lost( an enclosed system). It is changed or termed destroyed when energy is lost to the surrounding in form of heat ( an open system). Energy can neither be created not destroyed but can be transformed from one form to another. This means that the energy that we think was created are the potential energy in the system, while the energy we think was destroyed, are those energy given to the surrounding in form of heat.

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5 0
3 years ago
The equilibrium constant, K, for the following reaction is 1.80x102 at 698 K. 2HI(9) =H2(g) +129) An equilibrium mixture of the
Mrrafil [7]

Explanation:

The equilibrium constant of the reaction = K_c=1.80\times 10^{-2}

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After addition 0.174 mol of HI(g)

Concentration of HI added = \frac{0.174 mol}{1 L}=0.174 M

New concentration of HI = 0.311 M + 0.174 M = 0.485 M

     2HI\rightleftharpoons H_2+I_2

Initial  concentration:

0.311 M                4.71\times 10^{-2} M           4.71\times 10^{-2} M  

At equilibrium:

(0.485 M - x)  (4.71\times 10^{-2} M+x)           (4.71\times 10^{-2} M+x)  

K_{eq}=\frac{[H_2][I_2]}{[HI]^2}

1.80\times 10^{-2}=\frac{(4.71\times 10^{-2} M+x)\times (4.71\times 10^{-2} M+x)}{ (0.485 M-x)^2}

\sqrt{1.80\times 10^{-2}}=\frac{(4.71\times 10^{-2} M+x)}{ (0.485 M-x)}

0.1342=\frac{(4.71\times 10^{-2} M+x)}{(0.485 M-x)}

0.065087 M-0.1342x=4.71\times 10^{-2} M+x

0.065087 M-4.71\times 10^{-2} M=1.1342x

x=\frac{0.017987 M}{1.3142}=0.01586 M

Equilibrium concentrations:

[HI]=0.485 M-x = 0.485 M - 0.01586 M= 0.46914 M

[H_2]=[I_2]=4.71\times 10^{-2} M+x=4.71\times 10^{-2} M+0.01586 M

[H_2]=[I_2]=0.06296 M

3 0
3 years ago
If 801 J of heat is available, what is the mass in grams of iron (specific heat = 0.45 J/g・°C) that can be heated from 22.5°C to
inysia [295]

Answer:

The correct answer will be "18.25 g".

Explanation:

The given values are:

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Heat involved,

q =  801 J

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As we know,

⇒  C = \frac{q}{m \Delta T}

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3 0
4 years ago
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Ann [662]

P2 = 54.6 kPa

Explanation:

Given:

V1 = 10.0 L. V2 = 50.0 L

P1 = 273 kPa. P2 = ?

We can use Boyle's law to solve this problem.

P1V1 = P2V2

Solving for P2,

P2 = (V1/V2)P1

= (10.0 L/50.0 L)(273 kPa)

= 54.6 kPa

5 0
3 years ago
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