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Svetlanka [38]
3 years ago
5

hich term names how clouds form by the collection of water vapor in the atmosphere? evaporation seepage precipitation condensati

on
Chemistry
1 answer:
SVEN [57.7K]3 years ago
5 0

Condensation

Explanation:

Condensation is a term that describes the collection of water vapor in the atmosphere into clouds.

Condensation is a phase change process that is a key component of the hydrologic cycle in which water moves in different forms in nature.

  • It occurs when tiny particles suspended in the atmosphere helps to aggregate vapor.
  • As the hot vapor rises from the surface, the temperature decreases up into the atmosphere.
  • When it comes in contact with suspended dust or some other particles, it loses its energy and settles on the dust.
  • When they aggregate, they form the clouds we see.

learn more:

Cloud brainly.com/question/8839455

#learnwithBrainly

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How many liters are in 5.98 moles of nitrogen gas at STP?
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Answer:

The correct answer is c) 134L

Explanation:

We use the formula PV =nRT. The normal conditions of temperature and pressure are 273K and 1 atm, we use the gas constant = 0, 082 l atm / K mol.

1 atm x   V = 5, 98 mol x 0, 082 l atm / K mol x 273 K

V = 5, 98 mol x 0, 082 l atm / K mol x 273 K / 1 atm

V = 133, 86828 l

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Explanation:

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Which family in the periodic table has elements that are silvery and lustrous?
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3 years ago
Read 2 more answers
A quantity of 0.225 g of a metal M (molar mass = 27.0 g/mol) liberated 0.303 L of molecular hydrogen (measured at 17°C and 741 m
Lana71 [14]

Answer:

Oxide of M is M_2O_3 and sulfate of M_2(SO_4)_3

Explanation:

0.303 L of molecular hydrogen gas measured at 17°C and 741 mmHg.

Let moles of hydrogen gas be n.

Temperature of the gas ,T= 17°C =290 K

Pressure of the gas ,P= 741 mmHg= 0.9633 atm

Volume occupied by gas , V = 0.303 L

Using an ideal gas equation:

PV=nRT

n=\frac{PV}{RT}=\frac{0.9633 atm\times 0.303 L}{0.0821 atm L/mol K\times 290 K}=0.01225 mol

Moles of hydrogen gas produced = 0.01225 mol

2M+2xHCl\rightarrow 2MCl_x+xH_2

Moles of metal =\frac{0.225 g}{27.0 g/mol}=8.3333 mol

So, 8.3333 mol of metal M gives 0.01225 mol of hydrogen gas.

\frac{8.3333}{0.01225 mol}=\frac{2}{x}

x = 2.9 ≈ 3

2M+6HCl\rightarrow 2MCl_3+3H_2

MCl_3\rightarrow M^{3+}+Cl^-

Formulas for the oxide and sulfate of M will be:

Oxide of M is M_2O_3 and sulfate of M_2(SO_4)_3.

3 0
3 years ago
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