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maksim [4K]
3 years ago
10

How many atoms are present in 3 moles of chromium?

Chemistry
2 answers:
Lena [83]3 years ago
5 0
1 mole  chromium -------------- 6.02 x 10²³ atoms
3 moles chromium ------------ ( atoms chromium )

atoms chromium = 3 x ( 6.02 x 10²³) / 1

atoms chromium = 1.806 x 10²⁴ / 1

= <span>1.806 x 10²⁴ atoms of chromium

hope this helps!</span>
DENIUS [597]3 years ago
4 0
In any element or substance, one mole of the substance is equivalent to 6.022 x10^23 atoms per mole. This is expressed as the Avogadro's number which is useful in finding out the total number of atoms in a sample. In this case, we are given with 3 moles of chromium. this is equivalent to 1.8066 x1024 atoms

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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
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

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