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ohaa [14]
2 years ago
11

What processes in the water cycle takes water from oceans and land masses?

Chemistry
1 answer:
krek1111 [17]2 years ago
6 0

Answer:

Water Cycle

  • Earth is a truly unique in its abundance of water. Water is necessary to sustaining life on Earth, and helps tie together the Earth's lands, oceans, and atmosphere into an integrated system. Precipitation, evaporation, freezing and melting and condensation are all part of the hydrological cycle - a never-ending global process of water circulation from clouds to land, to the ocean, and back to the clouds.
  • This cycling of water is intimately linked with energy exchanges among the atmosphere, ocean, and land that determine the Earth's climate and cause much of natural climate variability.
  • The impacts of climate change and variability on the quality of human life occur primarily through changes in the water cycle. As stated in the National Research Council's report on Research Pathways for the Next Decade (NRC, 1999): "Water is at the heart of both the causes and effects of climate change."

<h2>Importance of the ocean in the water cycle</h2>

  • The ocean plays a key role in this vital cycle of water.
  • The ocean holds 97% of the total water on the planet; 78% of global precipitation occurs over the ocean, and it is the source of 86% of global evaporation.
  • Besides affecting the amount of atmospheric water vapor and hence rainfall, evaporation from the sea surface is important in the movement of heat in the climate system.
  • Water evaporates from the surface of the ocean, mostly in warm, cloud-free subtropical seas.
  • This cools the surface of the ocean, and the large amount of heat absorbed the ocean partially buffers the greenhouse effect from increasing carbon dioxide and other gases.
  • Water vapor carried by the atmosphere condenses as clouds and falls as rain, mostly in the ITCZ, far from where it evaporated, Condensing water vapor releases latent heat and this drives much of the the atmospheric circulation in the tropics.
  • This latent heat release is an important part of the Earth’s heat balance, and it couples the planet’s energy and water cycles.

  • The major physical components of the global water cycle include the evaporation from the ocean and land surfaces, the transport of water vapor by the atmosphere, precipitation onto the ocean and land surfaces, the net atmospheric transport of water from land areas to ocean, and the return flow of fresh water from the land back into the ocean.
  • . The additional components of oceanic water transport are few, including the mixing of fresh water through the oceanic boundary layer, transport by ocean currents, and sea ice processes.
  • On land the situation is considerably more complex, and includes the deposition of rain and snow on land; water flow in runoff; infiltration of water into the soil and groundwater; storage of water in soil, lakes and streams, and groundwater; polar and glacial ice; and use of water in vegetation and human activities.
  • Illustration of the water cycle showing the ocean, land, mountains, and rivers returning to the ocean.
  • Processes labeled include: precipitation, condensation, evaporation, evaportranspiration (from tree into atmosphere), radiative exchange, surface runoff, ground water and stream flow, infiltration, percolation and soil.
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n unknown metal is either aluminum, iron or lead. If 150. g of this metal at 150.0 °C was placed in a calorimeter that contains
Nitella [24]

Answer : The metal used was iron (the specific heat capacity is 0.449J/g^oC).

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of unknown metal = ?

c_2 = specific heat of water = 4.184J/g^oC

m_1 = mass of unknown metal = 150 g

m_2 = mass of water = 200 g

T_f = final temperature of water = 34.3^oC

T_1 = initial temperature of unknown metal = 150.0^oC

T_2 = initial temperature of water = 25.0^oC

Now put all the given values in the above formula, we get

150g\times c_1\times (34.3-150.0)^oC=-200g\times 4.184J/g^oC\times (34.3-25.0)^oC

c_1=0.449J/g^oC

Form the value of specific heat of unknown metal, we conclude that the metal used in this was iron (Fe).

Therefore, the metal used was iron (the specific heat capacity is 0.449J/g^oC).

6 0
3 years ago
Cl2(aq) + H2O H+(aq) + Cl–(aq) + HOCl(aq)
dimulka [17.4K]
44. (a) N2O3 (b) SF4 (c) AlCl3 (d) Li2CO3 46. H Br δ+ δ− 48. The metallic potassium atoms lose one electron and form +1 cations, and the nonmetallic fluorine atoms gain one electron and form –1 anions. K → K+ + e– 19p/19e– 19p/18e– F + e– → F– 9p/9e– 9p/10e– The ionic bonds are the attractions between K+ cations and F– anions. 50. See Figure 3.6. 52. (a) covalent…nonmetal-nonmetal (b) ionic…metal-nonmetal 54. (a) all nonmetallic atoms - molecular (b) metal-nonmetal - ionic 56. (a) 7 (b) 4 58. Each of the following answers is based on the assumption that nonmetallic atoms tend to form covalent bonds in order to get an octet (8) of electrons around each atom, like the very stable noble gases (other than helium). Covalent bonds (represented by lines in Lewis structures) and lone pairs each contribute two electrons to the octet. (a) oxygen, O If oxygen atoms form two covalent bonds, they will have an octet of electrons around them. Water is an example: H O H (b) fluorine, F If fluorine atoms form one covalent bond, they will have an octet of electrons around them. Hydrogen fluoride, HF, is an example: H F (c) carbon, C If carbon atoms form four covalent bonds, they will have an octet of electrons around them. Methane, CH4, is an example: H H H H C (d) phosphorus, P If phosphorus atoms form three covalent bonds, they will have an octet 
6 0
3 years ago
Which type of wave is a sound wave?
Vesna [10]

Answer:

longitudinal wave

Explanation:

hope it helps you and Mark me down as brainlist

3 0
2 years ago
The constant volume heat capacity of a gas can be measured by observing the decrease in temperature when it expands adiabaticall
miv72 [106K]

Answer:

The value is  C_p  = 42. 8 J/K\cdot mol

Explanation:

From the question we are told that  

     \gamma = \frac{C_p }{C_v}

The  initial volume of the  fluorocarbon gas is  V_1 = V

 The final  volume of the fluorocarbon gas isV_2 = 2V

  The initial  temperature of the fluorocarbon gas is  T_1  =  298.15 K

  The final  temperature of the fluorocarbon gas is T_2  =  248.44 K

   The initial  pressure is P_1  = 202.94\  kPa

    The final   pressure is  P_2  =  81.840\  kPa

Generally the equation for  adiabatically reversible expansion is mathematically represented as

       T_2 =  T_1  * [ \frac{V_1}{V_2} ]^{\frac{R}{C_v} }

Here R is the ideal gas constant with the value  

        R =  8.314\  J/K \cdot mol

So  

   248.44 =   298.15  * [ \frac{V}{2V} ]^{\frac{8.314}{C_v} }

=> C_v  =  31.54 J/K\cdot mol

Generally adiabatic reversible expansion can also be mathematically expressed as

    P_2 V_2^{\gamma} = P_1 V_1^{\gamma}

=>[ 81.840 *10^3] [2V]^{\gamma} = [202.94 *10^3] V^{\gamma}    

=>  2^{\gamma} =  2.56

=>    \gamma =  1.356

So

     \gamma  =  \frac{C_p}{C_v} \equiv  1.356 = \frac{C_p}{31.54}

=>    C_p  = 42. 8 J/K\cdot mol

3 0
3 years ago
ammonium phosphate is an important ingredient in many solid fertilizers. It can be made by reacting aqueous phosphoric acid with
Kaylis [27]

Answer:

0.57 moles (NH4)3PO4 (2 sig. figs.)

Explanation:

To quote, J.R.

"Note:  liquid ammonia (NH3) is actually aqueous ammonium hydroxide (NH4OH) because NH3 + H2O -> NH4OH.

H3PO4(aq) + 3NH4OH(aq) ==> (NH4)3PO4 + 3H2O  

Assuming that H3PO4 is not limiting, i.e. it is present in excess

1.7 mol NH4OH x 1 mole (NH4)3PO4/3 moles NH4OH = 0.567 moles = 0.57 moles (NH4)3PO4 (2 sig. figs.)"

3 0
3 years ago
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