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vitfil [10]
3 years ago
14

The byproduct of photosynthesis is _____. nitrogen water oxygen carbon dioxide

Geography
2 answers:
professor190 [17]3 years ago
6 0

Answer:

An extremely important byproduct of photosynthesis is oxygen, on which most organisms depend.

Explanation:

Photosynthesis occurs in green plants, seaweeds, algae, and certain bacteria. These organisms are veritable sugar factories, producing millions of new glucose molecules per second.

Tatiana [17]3 years ago
5 0

Answer:

Option (3)

Explanation:

The photosynthesis refers to the process by which the green plants can prepare their own food in the presence of sunlight, water, and carbon dioxide, and in turn, it liberates oxygen. These plants are called autotrophs or producers.

This green plants are the major producers of oxygen and helps in the reduction of carbon dioxide from the atmosphere.

The equation that shows the process of photosynthesis is given below-

6CO₂ + 6H₂0 + (sunlight/energy) → C₆H₁₂O₆ + 6O₂

Thus, the byproduct of the photosynthesis process is Oxygen (O₂)

Hence, the correct answer is option (3).

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5 0
3 years ago
Read 2 more answers
Calculate the amount, in grams, of an original 300-gram sample of radioactive isotope Potassium-40 remaining after
Aleksandr [31]

Answer:

The current amount of the Potassium-40 sample is approximately 37.521 grams.

Explanation:

The amount of the sample of the radioactive isotope decays exponentially in time, the amount of mass of the sample as a function of time (m (t)), in grams, is described below:

m(t) = m_{o}\cdot e^{-\frac{t}{\tau} } (1)

Where:

m_{o} - Initial mass, in grams.

t - Time, in years.

\tau - Time constant, in years.

The time constant can be found from half life (t_{1/2}), in years, described in statement:

\tau = \frac{t_{1/2}}{\ln 2} (2)

If we know that m_{o} = 300\,g, t = 3.9\times 10^{9}\,yr and t_{1/2} = 1.3\times 10^{9}\,yr, then the current amount of the sample is:

\tau = \frac{1.3\times 10^{9}\,yr}{\ln 2}

\tau \approx 1.876\times 10^{9}\,yr

m = (300\,g)\cdot e^{-\frac{3.9\times 10^{9}\,yr}{1.876\times 10^{9}\,yr} }

m\approx  37.521\,g

The current amount of the Potassium-40 sample is approximately 37.521 grams.

4 0
3 years ago
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