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AfilCa [17]
4 years ago
7

How matter cycles through our world

Chemistry
1 answer:
Dovator [93]4 years ago
7 0

Answer:

There are multiple cycles in our world and here are some examples:

The nitrogen cycle. Nitrogen circulates between air, the soil and living things.

(Nitrogen fixation (N2 to NH3/ NH4+ or NO3-)

Nitrification (NH3 to NO3-)

Assimilation (Incorporation of NH3 and NO3- into biological tissues)

Ammonification (organic nitrogen compounds to NH3)

Denitrification(NO3- to N2)

The carbon cycle. Carbon dioxide circulates between the air, soil, and living things.

(Photosynthesis, Decomposition, Respiration and Combustion.)

Photosynthesis. This process followed by respiration recycles oxygen.

During photosynthesis in green plants, light energy is captured and used to convert water, carbon dioxide, and minerals into oxygen and energy-rich organic compounds.

The water cycle.

( Step 1: Evaporation.

Step 2: Condensation.

Step 3: Sublimation.

Step 4: Precipitation.

Step 5: Transpiration.

Step 6: Runoff.

Step 7: Infiltration. )

Explanation:

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Answer:

120 extracted from the gas

Explanation:

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3 years ago
Please help.
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Explanation: An element is represented in the form of _{A}^{Z}\textrm{X} where,

X = Symbol of the element

A = Atomic Mass of the element X

Z = Atomic Number of element X

Hence, For the element Justwondoricium,

Symbol = Jw

Atomic number = 120

Atomic Mass = 224

Now, Atomic number = Number of electrons = Number of Protons

Number of electrons = 120

Number of Protons = 120

and Atomic Mass = Number of neutrons + Number of protons

Number of neutrons can be calculated as we know atomic mass and number of neutrons, Putting the numbers we get

Number of neutrons = 224 - 120 = 104

The nearest noble gas to the element having atomic number 120 is Oganesson (Og), which has an atomic number of 118, so the next two electrons will be filled in the 8s orbital.

Electronic Configuration of Jw is [Og]8s^2

This electronic configuration lets us know about the location of the element in periodic table.

As the electron is entering the 8th shell, it belongs to the 8th period and as the last electron enters the s-orbital, it belongs to the S-block of the periodic table.

When the s-electrons are 1, it belongs to Group 1.

When the s-electrons are 2, it belongs to Group 2.

As this element has 2 s-electrons, it belongs to the Group 2.

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Answer:

[OH-] = 6.17 *10^-10

Explanation:

Step 1: Data given

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Step 2: Calculate [OH-]

pOH = -log [OH-] = 9.21

[OH-] = 10^-9.21

[OH-] = 6.17 *10^-10

Step 3: Check if it's correct

pOH + pH = 14

[H+]*[OH-] = 10^-14

pH = 14 - 9.21 = 4.79

[H+] = 10^-4.79

[H+] = 1.62 *10^-5

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CaCO3(s)+2H*(aq) →Ca2+ (aq)+H2001+CO262)
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