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eduard
3 years ago
8

What is isotope the percentage of the solid components of soil​

Chemistry
1 answer:
zubka84 [21]3 years ago
4 0

Answer:

an isotope consists of two or more forms of the same elements that contains equal number of protons but different number of neutrons in their nuclei but differ in relative atomic mass but not in chemical properties

The percentage of soil components of the soil is: inorganic components

-soil particles

-mineral elements = 45%

-water or moisture= 25%

- air= 25%

organic components

-humus or decayed organic matter= 5%

-living organisms

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What is density? How do you find it?
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2 years ago
What is the similarity between radioactive iodine and stable iodine
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7 0
3 years ago
An unknown compound contains 38.7 % calcium, 19.9 % phosphorus and 41.2 % oxygen. what is the empirical formula of this compound
Softa [21]

Answer: Ca3(PO4)2

Explanation:

  • We assume that the sample is 100.0 g.
  • The sample contains 38.7 g of Ca, 19.9 g of P, and 41.2 g of O as the proportions show.
  • Then we can calculate the number of moles of each component (n = m/atomic mass).
  • The number of moles of Ca = 38.7/40.078 = 0.96 mol.
  • The number of moles of P = 19.9/30.97 = 0.64 mol.
  • The number of moles of O = 41.2/15.99 = 2.75 mol.
  • Now, we can get the molar ratios of different components (Ca, P, and O) in the sample by dividing the number of moles of each component by the lower number of moles, that we should divide the number of moles by (0.64).
  • Ca: P: O = (0.96/0.64) : (0.64/0.64) : (2.75/0.64) = 1.5 : 1 : 4.
  • To avoid the fraction of the ratios, we can multiply all ratios by 2.0.
  • Now, the ratio of Ca : P : O will be 3 : 2 : 8.
  • That main the empirical formula of the compound is Ca3P2O8 which can be expressed as calcium phosphate (Ca3(PO4)2).
8 0
3 years ago
"A sphere of radius 0.50 m, temperature 27oC, and emissivity 0.85 is located in an environment of temperature 77oC. What is the
Aleksandr-060686 [28]

Explanation:

It is known that formula for area of a sphere is as follows.

                     A = 4 \pi r^{2}

                        = 4 \times 3.14 \times (0.50 m)^{2}

                        = 3.14 m^{2}

    T_{a} = (27 + 273.15) K = 300.15 K

          T = (77 + 273.15) K = 350.15 K

Formula to calculate the net charge is as follows.

             Q = esA(T^{4} - T^{4}_{a})

where,    e = emissivity = 0.85

               s = stefan-boltzmann constant = 5.6703 \times 10^{-8} Wm^{-2} K^{-4}

                A = surface area

Hence, putting the given values into the above formula as follows.

                 Q = esA(T^{4} - T^{4}_{a})

                     = 0.85 \times 5.6703 \times 10^{-8} Wm^{-2} K^{-4} \times 3.14 \times ((350.15)^{4} - (300.15)^{4})

                     = 1046.63 W

Therefore, we can conclude that the net flow of energy transferred to the environment in 1 second is 1046.63 W.

8 0
3 years ago
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