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ololo11 [35]
3 years ago
9

How are the carbon, oxygen, and nitrogen cycles vital to sustaining life on Earth?. a.. Carbon, oxygen, and nitrogen are vital c

omponents of life on Earth.. b.. The carbon, oxygen, and nitrogen cycles allow vital elements to return to usable form by organisms.. c.. The carbon, oxygen, and nitrogen cycles are an important interface between biotic and abiotic factors.. d.. All of the above are reasons the carbon, oxygen, and nitrogen cycles are vital to life on Earth
Chemistry
2 answers:
kumpel [21]3 years ago
5 0

The correct answer is (D)


All the above are reasons the carbon , oxygen, and nitrogen cycles are vital to life on earth.


The explanation:


because :


1) Carbon, oxygen, and nitrogen are vital components of life on Earth.


2) The carbon, oxygen, and nitrogen cycles allow vital elements to return to usable form by organisms.


3)The carbon, oxygen, and nitrogen cycles are an important interface between biotic and abiotic factors


4) They are all biogeochemical cycles.


5) They all involve an interaction between living and nonliving elements.

6)They are all part of the Earth system.

Llana [10]3 years ago
4 0
The correct answer among the choices listed is option D. Carbon, oxygen, and nitrogen cycles vital to sustaining life on Earth because they are vital components of life, vital elements to return to usable form by organisms and an important interface between biotic and abiotic factors.
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Use the equation: 2Mg + O2→ 2MgO
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What is the pH of a solution has a hydrogen ion concentration of 6.3 x 10–10 M? Show work
NISA [10]

Answer:

pOH = 4.8

pH = 9.2

Explanation:

Given data:

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pH = -log [H⁺]

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An unknown compound with a molar mass of 223.94 g/mol consists of 32.18% c, 4.50% h, and 63.32% cl. find the molecular formula f
Dima020 [189]

The actual number of atoms of each element present in the molecule of the compound is represented by the formula known as molecular formula.

Molar mass of the unknown compound = 223.94 g/mol (given)

Mass of each element present in the unknown compound is determined as:

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\frac{32.18}{100}\times 223.94 = 72.06 g

  • Mass of hydrogen, H:

\frac{4.5}{100}\times 223.94 = 10.08 g

  • Mass of chlorine, Cl:

\frac{63.32}{100}\times 223.94 = 141.79 g

Now, the number of each element in the unknown compound is determined by the formula:

number of moles = \frac{given mass}{molar mass}

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number of moles = \frac{72.06}{12} = 6.005 mole\simeq 6 mole

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number of moles = \frac{10.08}{1} = 10.08 mole\simeq 10 mole

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number of moles = \frac{141.79}{35.5} = 3.99 mole\simeq 4 mole

Dividing each mole with the smallest number of mole, to determine the empirical formula:

C_{\frac{6}{4}}H_{\frac{10}{4}}Cl_{\frac{10}{4}}

C_{1.5}H_{2.5}Cl_{1}

Multiplying with 2 to convert the numbers in formula into a whole number:

So, the empirical formula is C_{3}H_{5}Cl_{2}.

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In order to determine the molecular formula:

n = \frac{molar mass}{empirical mass}

n = \frac{223.94}{112} = 1.99 \simeq 2

So, the molecular formula is:

2\times C_{3}H_{5}Cl_{2} =  C_{6}H_{10}Cl_{4}

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