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dlinn [17]
3 years ago
9

Which phenomena naturally warms earth lower atmosphere and surface?

Chemistry
1 answer:
Alenkinab [10]3 years ago
6 0

The phenomenon that naturally warms the earth's lower atmosphere and surface is called the greenhouse effect.


Put me as Brainliest!!!

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The final temperature and pressure in a container is 234K at 210 kPa.
andre [41]

Answer:

Initial pressure = 157 kpa (Approx)

Explanation:

Given:

final temperature = 234 K

final pressure = 210 kpa

Initial temperature = 175 K

Find:

Initial pressure

Computation:

Initial pressure / Initial temperature = final pressure / final temperature

Initial pressure / 175 = 210 / 234

Initial pressure = 157 kpa (Approx)

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2 years ago
What is the frequency of a wave?
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Divide velocity by the wavelength.
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3 years ago
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How is kinetic energy related to the Kelvin scale?
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The Kelvin temperature of a substance is directly equal to the avg kinetic energy of particles of a substance. Hope that is what your looking for
8 0
3 years ago
Potassium chlorate (used in fireworks, flares, and safety matches) forms oxygen and potassium chloride when heated. KClO3(s) → K
Zina [86]

Answer:

10.3 g of oxygen are formed when 26.4 g of potassium chlorate is heated

Explanation:

This is the balanced equation:

2KClO₃(s) → 2KCl(s) + 3O₂(g)

Ratio beteween the salt and oxygen is 2:3

Molar mass of KClO₃ = 122.55 g/m

Let's find out the moles of salt

Mass / Molar mass

26.4 g /122.55 g/m = 0.215 moles

So, this is the final rule of three:

If 2 moles of KClO₃ make 3 moles of oxygen

0.215 moles of KClO₃ make  (0.215  .3) /2 = 0.323 moles of O₂ are produced

Molar mass O₂ = 32 g/m

Moles . molar mass = mass

0.323 m  . 32g/m = 10.3 g

4 0
3 years ago
What volume of lead (of density 11.3 g/cm3 ) has the same mass as 395 cm3 of a piece of redwood (of density 0.38 g/cm3 )? Answer
Zepler [3.9K]
d_{1}=\frac{m}{V_{1}}\\\\
V_{1}=395cm^{3}\\
d_{1}=0,38\frac{g}{cm^{3}} \ \ \ \ \Rightarrow \ \ \ \ m=395cm^{3}*0,38\frac{g}{cm^{3}}=150,1g\\\\\\
d_{2}=\frac{m}{V_{2}}\\\\
m=150,1g\\
d_{2}=11,3\frac{g}{cm^{3}} \ \ \ \ \Rightarrow \ \ \ \ V_{2}=\frac{150,1g}{11,3\frac{g}{cm^{3}}}\approx13,28cm^{3}
8 0
3 years ago
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