Which of these objects converts light to chemical energy? four objects

Chemistry

2 answers:

Papessa [141]2 years ago

3 0

It would be a ehtanol plant

bazaltina [42]2 years ago

3 0

Answer:

Corn

Explanation:

Being a plant and having chlorophyll in its structures, corn is able to capture sunlight and converting it into chemical energy, through photosynthesis.

Photosynthesis is a process carried out by plants to produce their own food. The plant draws carbon dioxide from the air and energy from the sun. Through this process, the plant produces its own food consisting essentially of glucose.

Briefly, we can say that photosynthesis uses carbon dioxide, water and sunlight (in the form of heat) to produce organic compounds (glucose) and oxygen.

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What are they waiting? or what are their waiting?<br><br>please correct me if I'm wrong

kiruha [24]

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3 0

2 years ago

What is the name of the acid made from hydrogen and NO3 1-?

gizmo_the_mogwai [7]

Answer:

nitric acid - HNO3

Explanation:

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7 0

2 years ago

The concentrations of the products at equilibrium are [pcl3] = 0.180 m and [cl2] = 0.250 m. what is the concentration of the rea

sineoko [7]

We have Kc = 4.2 x 10^-2 (given but missing in the question)
and When the balanced equation for this reaction is:
PCl5(g) ↔ PCl3(g) + Cl2(g)
so, according to the Kc formula:
Kc = the concentration of products / the concentration of the reactants
so, to get the concentration of the reactants in equilibrium, the concentration of the products / the concentration of the reactants should equal the Kc value which is given in the question (missing in your question).
So by substitution in Kc formula:
Kc = [PCl3]*[Cl2] / [PCl5]
4.2 x 10^-2 = 0.18 * 0.25 /[PCl5]
∴[PCl5] = 0.18*0.25 / 4.2x10^-2 = 1.07
So the concentration of the reactants in equilibrim = 1.07

4 0

3 years ago

What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )

yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution : Given,

Concentration (c) = 0.150 M

Acid dissociation constant = $k_a=4.5\times 10^{-4}$