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Natalija [7]
3 years ago
14

Determine the mass of an object that, when places in a 25-mL Graduated cylinder containing 14 mL of water, causes the level of t

he water to rise to 19 mL. The object has a density of 3.2 g/mL.
How do you set up the equation?
Chemistry
2 answers:
Alja [10]3 years ago
8 0

Answer: The mass of the object is 16 g.

Explanation:

Density is defined as the mass contained per unit volume.

Density=\frac{mass}{Volume}

Given : Mass of object = ?

Density of the object = 3.2g/ml

Volume of the object = Volume of water displaced =(19-14) ml = 5 ml

Putting in the values we get:

3.2g/ml=\frac{Mass}{5ml}

Mass=16g

Thus the mass of the object is 16 g.

lakkis [162]3 years ago
7 0
Mass = Density x Volume
Mass = 3.2g/mL x 5 mL
Mass = 16g
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The production of water proceeds according to the following equation.
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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
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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
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