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3 years ago

11

Would the topsoil most likely be thicker in a lush, tropical area, or in a dry area? Why?

1 answer:

Marina CMI [18]3 years ago

6 0

The top soil would most likely be thicker in lush tropical area.

Explanation:

The top soil presents just above the earth's surface in the upper position of soil profile. It's thickness is 2 to 3 inches. Tropical area receives highest rainfall and direct sunlight which promotes the growth of lush vegetation that tightly holds the soil.

Dry region is devoid of vegetation. And it is free zone for air passing. With high speed wind top soil blow out from the upper surface and make surface bare. The thickness of soil in tropical area is maximum and pure.

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<u>Answer:</u> The percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 77.5% and 22.5% respectively.

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Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope be 'x'. So, fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope will be '1 - x'

<u>For $_{17}^{35}\textrm{Cl}$ isotope:</u>

Mass of $_{17}^{35}\textrm{Cl}$ isotope = 35 amu

Fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope = x

<u>For $_{17}^{37}\textrm{Cl}$ isotope:</u>

Mass of $_{17}^{37}\textrm{Cl}$ isotope = 37 amu

Fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope = 1 - x

Average atomic mass of chlorine = 35.45 amu

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$35.45=[(35\times x)+(37\times (1-x))]\\\\x=0.775$

Percentage abundance of $_{17}^{35}\textrm{Cl}$ isotope = $0.775\times 100=77.5\%$

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Hence, the percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 77.5% and 22.5% respectively.

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