Oxygen is produced by what part of photosynthesis? HURRY

2 answers:

5 0

Answer:The oxygen during photosynthesis comes from split water molecules. During photosynthesis, the plant absorbs water and carbon dioxide. After the absorption, the water molecules are disassembled and converted into sugar and oxygen.

andriy [413]3 years ago

3 0

Answer:

The chloroplast is involved in both stages of photosynthesis. The light reactions take place in the thylakoid. There, water (H2O) is oxidized, and oxygen (O2) is released. All photosynthetic eukaryotic cells contain chloroplasts that use the radiant energy of sunlight to convert carbon dioxide and water into carbohydrates. As a byproduct of photosynthesis, oxygen gas is also released into the atmosphere through tiny openings in the leaves called stomata.

Explanation:

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Suppose you are working in a chemistry lab and trying to make a solution of a substance called Sodium Chloride (which is a solid

Alenkasestr [34]

Answer:

a weighing balance, a measuring cylinder, a spatula, a beaker/flask, and a stirrer

Explanation:

The lab apparatus that would be needed to prepare a solution of sodium chloride would be <em>a weighing balance, a measuring cylinder, a spatula, a beaker/flask, and a stirrer.</em>

The weighing balance would be used to weigh out the required amount of sodium chloride. The beaker or flask would be placed on the weighing balance and its weight zeroed. The spatula would then be sued to take out the sodium chloride from its container into the beaker till the required amount is reached. The measuring cylinder would then be used to measure out the required volume of water which would be added to the salt in the beaker. The stirrer would then be used to stir the mixture in order for the salt to dissolve.

7 0

2 years ago

Part III.The two reactions involved in quantitatively determining the amount of iodate in solution are: IO3-(aq) 5 I-(aq) 6 H (a

slega [8]

Answer:

$\large \boxed{\math{\dfrac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}$

Explanation:

The I₂ is the common substance in the two equations.

(1) IO₃⁻ + 5I⁻ + 6H⁺ ⟶ 3I₂ + 3H₂O

{2) I₂ + 2S₂O₃²⁻ ⟶ 2I⁻ + S₄O₆²⁻

From Equation (1), the molar ratio of iodate to iodine is

$\dfrac{\text{I}_{2}}{\text{IO}_{3}^{-}} = \dfrac{3}{1}$

From Equation (2), the molar ratio of iodine to thiosulfate is

$\dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{I}_{2}} = \dfrac{2}{1}$

Combining the two ratios, we get

$\text{Stoichiometric factor} = \dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{IO}_{3}^{-}} = \dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{I}_{2}} \times \dfrac{\text{I}_{2}}{\text{IO}_{3}^{-}} = \dfrac{2}{1} \times \dfrac{3}{1} = \mathbf{\dfrac{6}{1}}\\\\\text{The stoichiometric factor is $\large \boxed{\mathbf{\dfrac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}$}$