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Kamila [148]
3 years ago
8

The crater is the bowl-shaped area around a volcano’s central vent. t/f

Chemistry
1 answer:
Lyrx [107]3 years ago
6 0
If I'm correct, the crater is actually a circular-shaped area around the volcano's central vent. My answer is false
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He total power used by humans worldwide is approximately 15 TW (terawatts). Sunlight striking Earth provides 1.336 kW per square
SCORPION-xisa [38]

Answer:

id

Explanation:

idk

3 0
3 years ago
Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston-cylinder assembly from 2 bar, 280 K to 20 bar
kaheart [24]

This question is incomplete, the complete question is;

Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston-cylinder assembly from 2 bar, 280 K to 20 bar, 520 K. If the carbon dioxide behaves as an ideal gas, determine the amount of entropy produced, in kJ/K. Assuming;

a) constant specific heats Cp = 0.939 kJ/Kg K

b) variable specific heats

Answer:

a) the amount of entropy produced is 0.731599 kJ/K

b) the amount of entropy produced is 0.69845 kJ/K

Explanation:

Given the data in the question;

5 kg of carbon dioxide (CO₂) gas undergoes a process in a well-insulated piston-cylinder assembly.

m = 5 kg

Molar mass M = 44.01 g/mol

P₁ = 2 bar, P₂ = 20

T₁ = 280 K, P₂ = 520 K

Since its insulated { q = 0 } ( kinetic and potential energy effects = 0 )

Now,

a) the amount of entropy produced, in kJ/K, Assuming constant specific heats with Cp = 0.939 kJ/Kg K

S_{Generation = m × ((Cp × In( T₂/T₁) - R × In( P₂/p₁ ))

we substitute

S_{Generation = 5 × (( 0.939  × In( 520/280) - 0.1889 × In( 20/2 ))

= 5 × ( 0.5812778 - 0.434958 )

= 5 × 0.1463198

= 0.731599 kJ/K

Therefore, the amount of entropy produced is 0.731599 kJ/K

b) the amount of entropy produced, in kJ/K, Assuming variable specific heats.

Now, from  Table A-23: Ideal Gas Properties of Selected Gases;

T₁,T₂ : s₁⁰ = 211.376 kJ/kmol-K, s₂⁰ = 236.575 kJ/kmol-K

now, s₁ = s₁⁰ / M and s₂ = s₂⁰ / M

we substitute

s₁ = s₁⁰ / M = 211.376 / 44.01  = 4.8029 kJ/kg

s₂ = s₂⁰ / M = 236.575 / 44.01 = 5.37548 kJ/kg

S_{Generation = m × (( s₂ - s₁ ) - R × In( p₂ / p₁ ))

we substitute

S_{Generation = 5 × (( 5.37548 - 4.8029  ) - 0.1880 × In( 20 / 2 ))

= 5 × ( 0.57258 - 0.432885997 )

= 5 × 0.13969

= 0.69845 kJ/K

Therefore, the amount of entropy produced is 0.69845 kJ/K

5 0
2 years ago
A sample of 42 mL of carbon dioxide gas was placed in a piston in order to maintain a constant 101 kPa of pressure.
Lilit [14]

Answer:

The answer to your question is 33.4 ml

Explanation:

Data

volume 1 = V1 = 42 ml

temperature 1 = T1 = 20°C

temperature 2 = T2 = -60°C

Volume 2 = V2 = x

Process

1.- Convert celsius to kelvin

T1 = 20 + 273 = 293°K

T2 = -60 + 273 = 233°K

2.- Use the Charles' law to solve this problem

               \frac{V1}{T1} = \frac{V2}{T2}

Solve for V2

                V2 = \frac{V1T2}{T1}

3.- Substitution

               V2 = \frac{(42)(233)}{293}

4.- Simplification

                V2 = \frac{9786}{293}

5.- Result

                V2 = 33.4ml

3 0
3 years ago
Which is caused by gravity?
MissTica
Mass wasting I think it's done
5 0
3 years ago
Argon-40 undergoes positron emission as shown:
Rudik [331]

Answer:

\frac{40}{18}Ar => \frac{40}{17}Cl + \frac{0}{1}e\\

Explanation:

3 0
3 years ago
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