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3 years ago

What is the best explanation for why solid sodium chloride CANNOT conduct electricity and why molten sodium chloride can?

Chemistry

1 answer:

ira [324]3 years ago

6 0

Answer: See explanation

Explanation:

The explanation for why solid sodium chloride can't conduct electricity while molten sodium chloride can is explained below:

Ionic compounds that are in their solid state like sodium chloride have their ions fixed in position. Due to this reason, the able to move, therefore we can say that the solid ionic compounds cannot be able to conduct electricity.

On the other hand, ionic compounds in their molten state, are free to flow unlike when they're in their solid state and therefore we can say that molten sodium chloride can be able to conduct electricity.

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Magnesium chloride is an important coagulant used in the preparation of tofu from soy milk. Its solubility in water at 200C is 5

Gennadij [26K]

Answer:

(a) Homogeneous. 4.7 g of MgCl₂.

(b) 9.1 g

Explanation:

(a)

At 200°C, we can dissolve 54.6g of MgCl₂ in 100 g of water. The mass that we could dissolve in 38.2 g of water is:

$38.2gWater.\frac{54.6gMgCl_{2}}{100gWater} =20.9g$

Since we can dissolve up to 20.9 g of MgCl₂ and we added only 16.2 g, the mixture is homogeneous and we could add 20.9 g -16.2 g = 4.7 g of solute to make it saturated.

(b)

At 800°C, we can dissolve 66.1 g of MgCl₂ in 100 g of water. The mass that we could dissolve in 38.2 g of water is:

$38.2gWater.\frac{66.1gMgCl_{2}}{100gWater} =25.3g$

Since we can dissolve up to 25.3 g of MgCl₂ and we added only 16.2 g, we could add 25.3 g - 16.2 g = 9.1 g of solute to make it saturated.

7 0

3 years ago

An object at 20∘C absorbs 25.0 J of heat. What is the change in entropy ΔS of the object? Express your answer numerically in jou

nevsk [136]

Question:

Part D) An object at 20∘C absorbs 25.0 J of heat. What is the change in entropy ΔS of the object? Express your answer numerically in joules per kelvin.

Part E) An object at 500 K dissipates 25.0 kJ of heat into the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.

Part F) An object at 400 K absorbs 25.0 kJ of heat from the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.

Part G) Two objects form a closed system. One object, which is at 400 K, absorbs 25.0 kJ of heat from the other object,which is at 500 K. What is the net change in entropy ΔSsys of the system? Assume that the temperatures of the objects do not change appreciably in the process. Express your answer numerically in joules per kelvin.

Answer:

D) 85 J/K

E) - 50 J/K

F) 62.5 J/K

G) 12.5 J/K

Explanation:

Let's make use of the entropy equation: ΔS = $\frac{Q}{T}$

Part D)

Given:

T = 20°C = 20 +273 = 293K

Q = 25.0 kJ

Entropy change will be:

ΔS = $\frac{25*1000}{293}$

= 85 J/K

Part E)

Given:

T = 500K

Q = -25.0 kJ

Entropy change will be:

ΔS = $\frac{-25*1000}{500}$

= - 50 J/K

Part F)

Given:

T = 400K

Q = 25.0 kJ

Entropy change will be:

ΔS = $\frac{25*1000}{400}$

= 62.5 J/K

Part G:

Given:

T1 = 400K

T2 = 500K

Q = 25.0 kJ

The net entropy change will be:

ΔS = $(\frac{25*1000}{400}) + (\frac{-25*1000}{500}$

= 12.5 J/K

7 0

3 years ago

Which of these would remove water from a river?

mariarad [96]

Hello, I don’t see the options

5 0

4 years ago

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creativ13 [48]

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3 years ago

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Aleksandr-060686 [28]

Answer:

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Explanation:

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3 years ago