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igor_vitrenko [27]
3 years ago
15

Giant sequoia trees (Sequoiadendron giganteum) are among the largest and longest-living organisms on Earth. These trees grow in

weight by up to 1000 pounds of wood per year. As these trees are growing, they:
A. All of the answer choices are correct.
B. are probably a net sink of CO2.
C. incorporate CO2 into their biomass (wood, roots, and leaves).
D. generate some CO2 through cellular respiration.
Chemistry
1 answer:
nevsk [136]3 years ago
3 0

Answer:

A. All of the answer choices are correct.

Explanation:

Giant sequoia trees

<u>They are long - living and the largest living organism on the planet Earth .</u>

They are majorly found in the western slopes of the Sierra Mountains present in the California .

These tress can grow to an average height of 50 to 85 meters , with diameter measuring 6 to 8 meters  .

  • They are also responsible for the generation of carbon dioxide via cellular respiration .
  • It is the net sink of the carbon dioxide .
  • It incorporates the carbon dioxide into the biomass .
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A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a
Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = m C .ΔT

m is mass of the solution = 151.8 mL * \frac{1 g}{1 mL} = 151.8 g

C is the specific heat of solution = 4.18 \frac{J}{g. ^{0}C}

ΔT is the temperature change = 31.50^{0}C - 21.45^{0}C = 10.05^{0}C

q = 151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J

Moles of NaOH = 101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol NaOH

Moles of H_{2}SO_{4} = 50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}

Enthalpy of the reaction = \frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol

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Answer:

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4 0
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3 years ago
What is the volume of a weather balloon that has 39 grams of helium with a density 0.017 g/mL?
olchik [2.2K]

Answer:

<h3>The answer is 235.29 mL</h3>

Explanation:

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volume =  \frac{mass}{density}  \\

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