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4 years ago

What is the composition of the mantle

2 answers:

7 0

The mantle under the crust is about 1,800 miles deep (2,890 km). It is composed mostly of silicate rocks rich inmagnesium and iron. Intense heat causes the rocks to rise. They then cool and sink back down to the core.

Sever21 [200]4 years ago

7 0

Answer:

The mantle is the middle layer of the earth, and it has a thickness of about 2900 km. It covers approximately 85% of the total earth's volume. This layer is comprised of elements such as oxygen, silicon, magnesium, aluminum, sodium, calcium, and iron. The rocks present here are mostly silicate rocks and are dominant in olivine, pyroxene, iron, and magnesium. These are denser minerals.

The mantle is again sub-divided into 2 distinct parts namely, upper mantle (includes the layer of asthenosphere) and lower mantle.

The upper mantle is comparatively less solid in nature in comparison to the lower mantle, which is marked by the presence of the asthenosphere layer, in which the rocks tend to flow in response to deformation.

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The current global population is 7.7 billion people, and the WHO estimates the minimum water need per person per day is 15 L. An

anygoal [31]

Answer:

a) The amount of freshwater available as groundwater, lakes and rivers, does not even reach one day the need for consumption for the current global population. ( t = 5.489 E-13 day )

b) The annual terrestrial precipitation, reaches to sustain the drinking water needs for the current global population

Explanation:

Let P = 7.7 billion people = 7.7 E12 person

∴ water needed for the total population for one day:

⇒ water amount = 7.7 E12 person * ( 15 L / person. day ) = 1.155 E14 L H2O /day

⇒ water amount = 1.155 E14 L H2O/day * 1 E1 Km² H2O / L H2O = 1.155 E15 Km² H2O/day * ( 365 day / year ) = 4.216 E17 Km²/year

∴ freshwater available:

freshwater = 6.34 E2 Km² H2O

how long will this water sustain the current population?

⇒ t = 6.34 E2 Km² * day / 1.155 E15 Km² = 5.489 E-13 day

this amount of freshwater does not even meet the need of the current global population.

∴ the annual terrestrial precipitation (Py) = 505000 Km³/year..........from literature

⇒ Py = 505000 Km³ H2O/year * ( 1000 m/Km )³ * ( 1000L/m³ )

⇒ Py = 5.05 E17 L H2O/year * ( 1 E1 Km² / L ) = 5.05 E18 Km² H2O/year

⇒ Py > water amount

the annual terrestrial precipitation of water, reaches to sustain the drinking water needs.

4 0

3 years ago

Draw the product formed when the structure shown below undergoes solvolysis in ch3ch2oh with heat.

mars1129 [50]

structure is attached below

7 0

3 years ago

A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The

AysviL [449]

Answer:- Heat lost by the metal is 279.45 cal.

Solution:- This type of problems are solved by using the concept, heat given = - heat taken

Metal temperature is decreasing from 45.00 degree C to 11.08 degree C. It means the heat is lost by the metal and this heat lost by metal is gained by water and the calorimeter to raise their temperature.

the equation we use is, $q=mc\Delta T$ .

where, q is the heat energy, m is mass, c is specific heat and $\Delta T$ is change in temperature.

Combined mass of calorimeter and water is 250.0 g and the specific heat is $\frac{1.035cal}{g.^0C}$ .

$\Delta T$ for calorimeter and water (combined) = 11.08 - 10.00 = 1.08 degree C

$\Delta T$ for metal = 11.08 - 45.00 = -33.92 degree C

let's plug in the values in the above equation and calculate heat gained by combined system.

q=250.0g*\frac{1.035cal}{g.^0C}*1.08^0C

q = 279.45 cal

So, the heat lost by the metal is 279.45 cal.

3 0

4 years ago

Read 2 more answers

How many grams of a 1000-gram radioactive sample are left after 3 half-lives? *

faltersainse [42]

The answer is 125 gms.

3 0

4 years ago

What is the mass of 5.7 x 1022 atoms Mn?

Svet_ta [14]

The mass of 5.2 x 10^22 manganese atoms in grams would be 635.784 grams.

What is Avogadro's constant?
Avogadro's number tells us how many particles are in one mole (or mol) of a substance. Electrons, molecules, or atoms could be among these particles. Avogadro's number has a value of approximately 6.022140857×10²³ mol⁻¹. Avogadro's number tells us how many particles are in one mole (or mol) of a substance. Electrons, molecules, or atoms could be among these particles. Avogadro's number has a value of approximately 6.022140857×10²³.

Computation of mass
Avogadro calculated that every material had 6.022 x 10^23 atoms per mole.

5.2 x 10^22 atoms of Mn would contain how many moles?
= 6.022 x 10^23 /1 x 5.2 x10^22
=11.58 moles

Mn mass equals 11.58 moles multiplied by 54.9
= 635.784 grams

You may learn more about Avogadro's constant here: brainly.com/question/15989275
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6 0

1 year ago