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Whitepunk [10]
3 years ago
14

Soil erosion is a secondary environmental problem because it only degrades the land and does not affect water or air quality.

Chemistry
1 answer:
Anastaziya [24]3 years ago
3 0

The correct answer is - False.

The soil erosion is a primary environment problem. It latter has effects that fall into the category of secondary environmental issues. The erosion of the soil is a very serious environmental issue, as it is the prime natural resources that enables the growth of plants, thus food for the living organisms, including the humans. Also, with high rates of erosion, the process of desertification significantly increases, as the soil is stripped of its top layers. The sediments that have been stripped of the soil can cause pollution of the air, though the stronger winds and storms that carry them, or with sediments getting to the water bodies and making them all murky, muddy.

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1.) What are the chemical properties of matter?
denis23 [38]

Answer:

Chemical properties are properties that can be measured or observed only when matter undergoes a change to become an entirely different kind of matter. They include reactivity, flammability, and the ability to rust. Reactivity is the ability of matter to react chemically with other substances.

Explanation:

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Whats is a chemical change of a paper bag
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Combustion will cause a chemical change in paper.

This is because the carbon compounds are being oxidized into carbon dioxide and water vapour (or h2o, same thing).



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3 years ago
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Which organisms are the best example of producers in a food web? A. zooplankton B. krill C. killer whales D. marine phytoplankto
notsponge [240]
D. Marine phytoplankton
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3 years ago
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How do many types of cells, like bacteria, reproduce?
umka2103 [35]

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cells like bacteria are bisexual so they split and their offspring is 100 percent like the parent and this process happens over and over

Explanation:

8 0
2 years ago
Titanium has an HCP crystal structure, a c/a ratio of 1.669, an atomic weight of 47.87 g/mol, and a density of 4.51 g/cm3. Compu
IrinaK [193]

Answer : The atomic radius for Ti is, 1.45\times 10^{-8}cm

Explanation :

Atomic weight = 47.87 g/mole

Avogadro's number (N_{A})=6.022\times 10^{23} mol^{-1}

First we have to calculate the volume of HCP crystal structure.

Formula used :  

\rho=\frac{Z\times M}{N_{A}\times V} .............(1)

where,

\rho = density  = 4.51g/cm^3

Z = number of atom in unit cell (for HCP = 6)

M = atomic mass  = 47.87 g/mole

(N_{A}) = Avogadro's number  

V = volume of HCP crystal structure = ?

Now put all the values in above formula (1), we get

4.51g/cm^3=\frac{6\times (47.87g/mol)}{(6.022\times 10^{23}mol^{-1}) \times V}

V=1.06\times 10^{-22}cm^3

Now we have to calculate the atomic radius for Ti.

Formula used :

V=6R^2c\sqrt{3}

Given:

c/a ratio = 1.669 that means,  c = 1.669 a

Now put (c = 1.669 a) and (a = 2R) in this formula, we get:

V=6R^2\times (1.669a)\sqrt{3}

V=6R^2\times (1.669\times 2R)\sqrt{3}

V=(1.669)\times (12\sqrt{3})R^3

Now put all the given values in this formula, we get:

1.06\times 10^{-22}cm^3=(1.669)\times (12\sqrt{3})R^3

R=1.45\times 10^{-8}cm

Therefore, the atomic radius for Ti is, 1.45\times 10^{-8}cm

3 0
3 years ago
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