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ohaa [14]
3 years ago
11

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

Chemistry
1 answer:
krek1111 [17]3 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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When using 100ml or 50ml graduated cylinder to what decimal place can your volume be estimated?<br>​
Olegator [25]

Answer:

I know that the 100-mL graduated cylinders are always read to 1 decimal place.

I think for 50 mL graduated cylinders, it lets you measure volumes up to 50.0 mL to the nearest 0.1 or 0.2 mL, depending on your exact cylinder.

3 0
3 years ago
aspirin C6H4(CO2)(CO2CH3),can be prepared in the chemistry laboratory by the reactions of salicylic acid, C6H4(CO2H)(OH),with ac
Ket [755]
262mol 1=kg
g=1000
from stoichoimetry

x=102*1000/360
x=102000/360
x=283.33

density =m/v
=283.33/1.082
=262mol

8 0
3 years ago
When rolling a die, find the probability of rolling an even number or rolling a
frozen [14]
<h3><em><u>ᎪꪀsωꫀᏒ</u></em></h3>

even no = 3/6 = 1/2

no. less than 5 = 4/6 = 2/3

3 0
2 years ago
The density of an aqueous solution of nitric acid is 1.64 g/mL and the concentration is 1.85 M. What is the concentration of thi
galina1969 [7]

Answer:

Mass % of the solution = 7.1067 %

Explanation:

Given :

Molarity of nitric acid solution = 1.85 M

Density of the solution = 1.64 g/mL

<u>Molarity of a solution is defined as the number of moles of solute present in 1 liter of the solution.</u>

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Lets, consider the volume of the solution = 1 L

Thus,

Moles of nitric acid present in the solution:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Moles of Nitric acid=Molarity \times {Volume\ of\ the\ solution}

So,

Moles of Nitric acid  = 1.85 moles

Molar mass of nitric acid = 63 g/mol

The mass of Nitric acid can be find out by using mole formula as:

moles=\frac{Mass\ taken}{Molar\ mass}

Thus,  

Mass\ of\ Nitric\ acid=Moles \times Molar mass}

Mass\ of\ Nitric\ acid=1.85 g \times 63 g/mol}

<u>Mass of Nitric acid = 116.55 g</u>

Also,

Density=\frac{Mass}{Volume}

Given : Density = 1.64 g/mL

Also, 1 L = 10³ mL

Volume of the solution is 1000 mL

So, mass of the solution:

Mass\ of\ the\ solution=Density \times {Volume\ of\ the\ solution}

Mass\ of\ the\ solution=1.64 g/mL \times {1000 mL}

<u>Mass of the solution  = 1640 g</u>

Mass % is defined as the mass of solute in 100 g of the solution. The formula for the calculation of mass % is shown below:

Mass \% =\frac{Mass\ of\ the\ solute}{Mass\ of\ the\ solution} \times {100}

So,

Mass \%=\frac{116.55}{1640} \times {100}

<u>Mass % = 7.1067 %</u>

6 0
3 years ago
A solution of 2-propanol and 1-octanol behaves ideally. Calculate the chemical potential of 2-propanol in solution relative to t
andrew-mc [135]

Answer:

The chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.    

Explanation:

The chemical potential of 2-propanol in solution relative to that of pure 2-propanol can be calculated using the following equation:

\mu (l) = \mu ^{\circ} (l) + R*T*ln(x)

<u>Where:</u>

<em>μ (l): is the chemical potential of 2-propanol in solution    </em>

<em>μ° (l): is the chemical potential of pure 2-propanol   </em>

<em>R: is the gas constant = 8.314 J K⁻¹ mol⁻¹ </em>

<em>T: is the temperature = 82.3 °C = 355.3 K </em>

<em>x: is the mole fraction of 2-propanol = 0.41 </em>

\mu (l) = \mu ^{\circ} (l) + 8.314 \frac{J}{K*mol}*355.3 K*ln(0.41)

\mu (l) = \mu ^{\circ} (l) - 2.63 \cdot 10^{3} J*mol^{-1}

\mu (l) - \mu ^{\circ} (l) = - 2.63 \cdot 10^{3} J*mol^{-1}  

Therefore, the chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.    

I hope it helps you!    

8 0
3 years ago
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