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2 years ago

The descriptions below explain two ways that water is used by plants on a sunny day.

Chemistry

1 answer:

lubasha [3.4K]2 years ago

3 0

In a sunny day, the plant makes use of water through photosynthesis and transpiration.

<h3>What is photosynthesis?</h3>

Photosynthesis is defined as the process by which green plants, which has chlorophyll, makes use of water and carbondioxide in the presence of sunlight as energy to form oxygen and glucose.

Transpiration is the movement is of water out of the plant through its pores located on the leaves, roots and stem. This process is increased in sunny seasons.

Therefore, In a sunny day, the plant makes use of water through photosynthesis and transpiration.

Learn more about photosynthesis here:

brainly.com/question/19160081

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Which of the following is not a concern regarding the use of corn as biofuel

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The color of corn is different than the color of gasoline or coal.

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(Help!) What is the mass of hydrogen atoms that is measured at 72 u?

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Answer:

=71.439911 u

Explanation:

We know that 1 mass of hydrogen atom = 1.00784 u

If it's measured at 72u: 72/1.00784=71.439911 u

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a compound is found to contain 31.1% sulfur and 68.9% chlorine determine the empirical formula for the sulfur and chlorine sampl

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Answer:

Empirical formula = S₁Cl₂ = SCl₂

Explanation:

The empirical formula of a compound is the simplest whole number ratio of each type of atom in a compound.

This compound contains 31.1% sulfur and 68.9% chlorine.

That is 100g of the compound contains 31.1 g of sulfur and 68.9 g of chlorine.

Convert the mass of each element to moles using the molar mass from the periodic table.

Moles of Sulfur = $\frac{31.1}{32.065 } \\$

= 0.9699

Moles of Chlorine= $\frac{68.9}{35.453 } \\$

= 1.9434

Divide each mole value by the smallest number of moles calculated.

Units of sulfur = $\frac{0.9699}{0.9699} \\$

= 1

Units of Chlorine = $\frac{1.9434}{0.9699} \\$

= 2

Empirical formula = S₁Cl₂ = SCl₂

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3 years ago

Which class of molecules functions as chemical signals? hormones water carbohydrates proteins

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hormones - the first choice

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Practice Problem: True Stress and Strain A cylindrical specimen of a metal alloy 49.7 mm long and 9.72 mm in diameter is stresse

amm1812

Answer:

The true stress required = 379 MPa

Explanation:

True Stress is the ratio of the internal resistive force to the instantaneous cross-sectional area of the specimen. True Strain is the natural log to the extended length after which load applied to the original length. The cold working stress – strain curve relation is as follows,

σ(t) = K (ε(t))ⁿ, σ(t) is the true stress, ε(t) is the true strain, K is the strength coefficient and n is the strain hardening exponent

True strain is given by

Epsilon t =㏑ (l/l₀)

Substitute㏑(l/l₀) for ε(t)

σ(t) = K(㏑(l/l₀))ⁿ

Given values l₀ = 49.7mm, l =51.7mm , n =0.2 , σ(t) =379Mpa

379 x 10⁶ = K (㏑(51.7/49.7))^0.2

K = 379 x 10⁶/(㏑(51.7/49.7))^0.2

K = 723.48 MPa

Knowing the constant value would be same as the same material is being used in the second test, we can find out the true stress using the above formula replacing the value of the constant.

σ(t) = K(㏑(l/l₀))ⁿ

l₀ = 49.7mm, l = 51.7mm, n = 0.2, K = 723.48Mpa

σ(t) = 723.48 x 106 x (㏑(51.7/49.7))^0.2

σ(t) = 379 MPa

The true stress necessary to plastically elongate the specimen is 379 MPa.

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