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Lesechka [4]
3 years ago
7

What are the basic forces that drive the water cycle? Explain what they are.

Chemistry
2 answers:
weqwewe [10]3 years ago
5 0

Answer:

kcgiobb jkffiob hffyio hiiddhoo hhioiv

alukav5142 [94]3 years ago
5 0

Answer: Have a blessed day

Explanation:

Water Cycle: Component # 1. Evaporation:

Evaporation is the process where a liquid changes from its liquid state to a gaseous state. Liquid water becomes water vapour. Temperature is the primary factor for evaporation.

During the water cycle the water of the oceans and freshwater bodies, such as lakes and rivers, is warmed by the sun and evaporates. During the process of evaporation, impurities in the water are left behind. As a result, the water that goes into the atmosphere is cleaner than it was on Earth.

Water Cycle: Component # 2. Condensation:

Condensation is the opposite of evaporation. Condensation is the process where a gas is changed into a liquid. Condensation occurs when the temperature of the vapour decreases.

The water droplets formed from condensation are very small and they remain suspended in the atmosphere. There millions of droplets of suspended water form clouds in the sky or fog at ground level. Water condenses into droplets only when there are small dusts particles present around.

Water Cycle: Component # 3. Precipitation:

When the temperature and atmospheric pressure are right, the small droplets of water in clouds form larger droplets or raindrops and it is called precipitation. The raindrops fall to earth.

As a result of evaporation, condensation and precipitation, water travels from the surface of the Earth goes into the atmosphere, and returns to Earth again.

Water Cycle: Component # 4. Surface Runoff:

Much of the water that returns to Earth as precipitation runs off the surface of the land and flows down-hill into streams, rivers, ponds and lakes. Small streams flow into the ocean. Surface runoff is an important part of the water cycle because, through surface runoff, much of the water returns again to the oceans, where a great deal of evaporation occurs.

Water Cycle: Component # 5. Infiltration:

ADVERTISEMENTS:

Infiltration is an important process where rain water soaks into the ground, through the soil and underlying rock layers. Some of this water ultimately returns to the surface at springs or in low spots downhill. Some of the water remains underground and is called groundwater.

As the water infiltrates through the soil and rock layers, many of the impurities in the water are filtered out. This filtering process helps in cleaning of the water.

Water Cycle: Component # 6. Transpiration or Evapotranspiration:

This is performed by plants. As plants absorb water from the soil, the water moves from the roots through the stems to the leaves. Once the water reaches the leaves, some of it evaporates from the leaves, adding to the amount of water vapour in the air. This process of evaporation through plant leaves is called transpiration. In large forests, an enormous amount of water will transpire through leaves.

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Help 4 me???? ples????
frutty [35]

Answer:

its the first one.

Explanation:

If sodium loses an electron, it now has 11 protons, 11 neutrons, and only 10 electrons, leaving it with an overall charge of +1

3 0
3 years ago
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Calculate the number of grams of CO that can react with 0.400 kg of Fe2O3.
aleksandr82 [10.1K]

Answer:

mass of CO = 210.42 g

mass in three significant figures = 210. g

Explanation:

Given data:

mass of Fe2O3 = 0.400 Kg

mass of CO= ?

Solution:

chemical equation:

Fe2O3 + 3CO → 2Fe + 3CO2

Now we will calculate the molar mass of  Fe2O3 and CO.

Molar mass of  Fe2O3 = (55.845 × 2) + (16 × 3) = 159.69 g/mol

Molar mass of CO = 12+ 16 = 28 g/mol

now we will convert the kg of Fe2O3 in g.

mass of Fe2O3 = 0.400 kg × 1000 = 400 g

number of moles of Fe2O3  = 400 g/ 159.69 g/mol = 2.505 mol

mass of CO = moles of Fe2O3 × 3( molar mass of CO)

mass of CO = 2.505 mol × 84 g/mol

mass of CO = 210.42 g

mass in three significant figures = 210. g

5 0
4 years ago
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What changes in color occur when bromine reacts with an alkene​
Gala2k [10]

Answer:

Explanation: Bromine, the dark red color disappears quickly as the atoms of bromine bond with the atoms of carbon in the double bond.

7 0
4 years ago
2Al + 6HCl → 2AlCl3 + 3H2
Nadya [2.5K]

Answer: The reaction produces 2.93 g H₂.

M_r:                        133.34  2.016

       2Al + 6HCl → 2AlCl₃ + 3H₂

Moles of AlCl₃ = 129 g AlCl₃ × (1 mol AlCl₃/133.34 g AlCl₃) = 0.9675 mol AlCl₃

Moles of H₂ = 0.9675 mol AlCl₃ × (3 mol H₂/2 mol AlCl₃) = 1.451 mol H₂

Mass of H₂ = 1.451 mol H₂ × (2.016 g H₂/1 mol H₂) = 2.93 g H₂

Explanation:

6 0
3 years ago
A chemist fills a reaction vessel with 9.20 atm nitrogen monoxide (NO) gas, 9.15 atm chlorine (CI) gas, and 7.70 atm nitrosyl ch
ivanzaharov [21]

Answer:

The reactions free energy \Delta G = -49.36 kJ

Explanation:

From the question we are told that

      The pressure of (NO) is P_{NO} = 9.20 \ atm

      The  pressure of  (Cl) gas is  P_{Cl} = 9.15 \ atm

       The  pressure of nitrosly chloride (NOCl) is P_{(NOCl)} = 7.70 \ atm

The reaction is

              2NO_{(g)} + Cl_2 (g)    ⇆   2 NOCl_{(g)}

 From the reaction we can  mathematically evaluate the \Delta G^o (Standard state  free energy ) as

                    \Delta G^o = 2 \Delta G^o _{NOCl} -   \Delta G^o _{Cl_2}  - 2 \Delta G^o _{NO}

The Standard state  free energy for NO is  constant with a value  

                 \Delta G^o _{NO} = 86.55 kJ/mol

 The Standard state  free energy for Cl_2 is  constant with a value                  

             \Delta G^o _{Cl_2} = 0kJ/mol

 The Standard state  free energy for NOCl is  constant with a value

         \Delta G^o _{NOCl} =66.1kJ/mol

Now substituting this into the equation

        \Delta G^o = 2 * 66.1 - 0 - 2 * 87.6

                = -43 kJ/mol

The pressure constant is evaluated as

         Q =  \frac{Pressure \ of  \ product }{ Pressure  \ of \ reactant }

Substituting  values  

        Q = \frac{(7.7)^2 }{(9.2)^2 (9.15) } = \frac{59.29}{774.456}

           = 0.0765

The free energy for this reaction is evaluated as

           \Delta  G  =  \Delta  G^o  + RT ln Q

Where R is gas constant with a value  of  R = 8.314 J / K \cdot mol

          T is temperature in K  with a given value of  T = 25+273 = 298 K

   Substituting value

                \Delta  G  = -43 *10^{3} + 8.314 *298 * ln [0.0765]

                       = -43-6.36

                      \Delta G = -49.36 kJ

4 0
3 years ago
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