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Usimov [2.4K]
3 years ago
10

The growing of plants, burning of wood, and decomposition of living matter demonstrate how carbon can be in the _______ at the s

ame time. a. atmosphere and biosphere b. biosphere and geosphere c. geosphere and atmosphere d. atmosphere, biosphere, and geosphere
Chemistry
2 answers:
JulijaS [17]3 years ago
8 0

Answer:

d. atmosphere, biosphere, and geosphere

Explanation:

Carbon is one of the most common elements on the planet, with large reservoirs in the biosphere, atmosphere, geosphere and hydrosphere in the form of carbohydrates, proteins, nucleic acids, carbon dioxide and calcium carbonate. Events such as plant cultivation, wood burning and decomposition of living matter prove that carbon is in the above mentioned environments simultaneously as these events are the result of adding to the carbon removal from these environments.

With its strong influence on the interaction between molecules, the carbon cycle is a fundamental process for maintaining life on earth. Responsible for photosynthesis and keeping the planet warm through the greenhouse effect, without it we would not be here.

Fittoniya [83]3 years ago
3 0
I think the answer is in the ground at the same time            

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For the reaction:
Rom4ik [11]

Answer:

Half-life at 310 K is 6.54 × 10³ s.

Explanation:

Let's consider the following reaction:

2 N₂O₅(g) → 4 NO₂(g) + O₂(g)

The rate law is:

(Δ[O₂]/Δt) = k . [N₂O₅]

Since [N₂O₅] is raised to the power of 1, the reaction order is 1.

For a first-order reaction:

t_{1/2}=\frac{ln2}{k}

where,

t_{1/2} is the half-life

k is the rate constant

At 300 K,

(t_{1/2})_{1}=\frac{ln2}{k_{1}}\\k_{1}=\frac{ln2}{(t_{1/2})_{1}} =\frac{ln2}{2.50 \times 10^{4} s} =2.77 \times 10^{-5} s^{-1}

We can use two-point Arrhenius equation to solve for k₂ at 310 K

ln\frac{k_{2}}{k_{1}} =\frac{-Ea}{R} (\frac{1}{T_{2}} -\frac{1}{T_{1}} )\\ln\frac{k_{2}}{k_{1}} =\frac{-103.3kJ/mol}{8.314 \times 10^{-3} kJ/mol.K} .(\frac{1}{310K}-\frac{1}{300K}  )\\ln\frac{k_{2}}{k_{1}}=1.34\\k_{2}=1.06 \times 10^{-4} s^{-1}

At 310 K,

(t_{1/2})_{2}=\frac{ln2}{k_{2}}=\frac{ln2}{1.06 \times 10^{-4} s^{-1}   } =6.54 \times 10^{3} s

4 0
3 years ago
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iren [92.7K]

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