Which response is a positive way to manage stress?

How might the oceans on early earth have formed?

baherus [9]

Gradual leakage of water stored in hydrate minerals of Earth's rocks could have formed a portion of its water. Water may also have come from volcanism: water vapor in the atmosphere that originated in volcanic eruptions may have condensed to form rain, slowly filling Earth's oceanic basins. So, the answer would be C.

I hope this helped! Good luck :)

3 0

3 years ago

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Be sure to answer all parts. Enter your answers in scientific notation.

Andrews [41]

Answer:

a) f = 3.02x10¹⁵ s⁻¹, and λ = 99.4 nm.

b) 99.4 nm

Explanation:

a) The energy of radiation is given by:

E = h*f

Where h is the Planck constant (6.626x10⁻³⁴ J.s), and f is the frequency. To have the highest frequency, the energy must be the highest too, because they're directly proportional. So we must use E = -E1 = 20x10⁻¹⁹ J

20x10⁻¹⁹ = 6.626x10⁻³⁴xf

f = 3.02x10¹⁵ s⁻¹

The wavelenght is the velocity of light (3.00x10⁸ m/s) divided by the frequency:

λ = 3.00x10⁸/3.02x10¹⁵

λ = 9.94x10⁻⁸ m = 99.4 nm

b) To have the shortest wavelength, it must be the highest energy and frequency, so it would be the same as the letter a) 99.4 nm.

8 0

3 years ago

How many sulfur atoms are in H2SO4?

NemiM [27]

Answer: 1

Explanation:

6 0

3 years ago

How many significate figures does 0.003010 have

guajiro [1.7K]

The answer is C 4 3 4 3 hope it helps

4 0

4 years ago

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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

7 0

3 years ago