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MrMuchimi
2 years ago
3

Read the article Looking at Leaves and answer the question. What causes leaves to change color in the autumn? Is the change phys

ical or chemical? Explain your answer.
Chemistry
2 answers:
ElenaW [278]2 years ago
7 0

Answer:

Both chemical and physical change

Explanation:

Physically, the change in color is due to the addition and subtraction of pigments such as chlorophyll, carotenoids and anthocyanins

Chemically, the change results from the conversion of the pigments chemicals such as chlorophyll and anthocyanins product one form to another product.

From the article, it is noted that during the autumn, there is cooler weather and less sunshine due to shorter days, and hence the trees instead of relying on the sunshine for the production of plant food through photosynthesis, the trees absorbs the food stored in the leaf for storage and nutrition during the winter periods, in the process, chlorophyll, the substance that gives the leaves their green color, is broken down and the leaves changes color.

After the chlorophyll is broken down other chemicals present carotenoids and anthocyanins (formed by trapped sugar in leaves) provide pigments which gives the leave.

ivann1987 [24]2 years ago
7 0

Answer:

Leaf color comes from three pigments: chlorophyll, carotenoid, and anthocyanin. Chlorophyll is the pigment that makes leaves green and allows plants to make their own food using sunlight, water, and carbon dioxide. In the autumn, trees stop producing chlorophyll because there is less sunlight available. This allows carotenoid to show through, making the leaves yellow, orange, and brown. This is an example of a chemical change.

Explanation:

You might be interested in
list down some examples of solutions that we need to prepare/make in the form of unsaturated and saturated solution​
Vladimir79 [104]

Answer:

1)Carbonated water is saturated with carbon, hence it gives off carbon through bubbles.

2)Adding sugar to water until it no longer dissolves creates a saturated solution.

3)Continuing to dissolve salt in water until it will no longer dissolve creates a saturated solution.

An unsaturated tea and sugar solution would be one into which you could add more sugar and have the sugar still dissolve

5 0
2 years ago
A student titrated 25.0 cm3 portions of dilute sulfuric acid with a 0.105 mol/dm3 sodium hydroxide solution.The equation for the
Rzqust [24]

Answer:

This question is incomplete

Explanation:

This question is incomplete as the volume of the base that was used during the titration was not provided. However, the completed question is in the attachment below.

The formula to be used here is CₐVₐ/CbVb = nₐ/nb

where Cₐ is the concentration of the acid = unknown

Vₐ is the volume of the acid used = 25 cm³ (as seen in the question)

Cb is the concentration of the base = 0.105 mol/dm³ (as seen in the question)

Vb is the volume of the base = 22.13 cm³ (22.1 + 22.15 + 22.15/3)

nₐ is the number of moles of acid = 1 (from the chemical equation)

nb is the number of moles of base = 2 (from the chemical equation)

Note that the Vb was based on the concordant results (values within the range of 0.1 cm³ of each other on the table) of the student

Cₐ x 25/0.105 x 22.13 = 1/2

Cₐ x 25 x 2 = 0.105 x 22.13 x 1

Cₐ x 50 = 0.105 x 22.13

Cₐ = 0.105 x 22.13/50

Cₐ = 0.047  mol/dm³

The concentration of the sulfuric acid is 0.047  mol/dm³

Download docx
7 0
2 years ago
The town of Natrium, West Virginia, derives its name from the sodium produced in the electrolysis of molten sodium chloride (NaC
Alona [7]

Explanation:

The reaction equation will be as follows.

           Na^{+} + e^{-} \rightarrow Na(s)

Hence, moles of Na = moles of electron used

Therefore, calculate the number of moles of sodium as follows.

       No. of moles = \frac{mass}{\text{molar mass}}

                             = \frac{4500 g}{23 g/mol}    (as 1 kg = 1000 g)

                             = 195.65 mol

As,     Q = n \times F       where F = Faraday's constant

              = 195.65 mol \times 96500 C

              = 1.88 \times 10^{7} mol C

Relation between electrical energy and Q is as follows.

               E = Q \times V

Hence, putting the given values into the above formula and then calculate the value of electricity as follows.

              E = Q \times V

                 = 1.88 \times 10^{7} \times 5

                 = 9.4 \times 10^{7} J

As 1 J = 2.77 \times 10^{-7} kWh

Hence,      \frac{9.4 \times 10^{7}}{2.77 \times 10^{-7}} kWh

                = 3.39 kWh

Thus, we can conclude that 3.39 kilowatt-hours of electricity is required in the given situation.

7 0
2 years ago
Read 2 more answers
Gina wants to use models to better understand how the types of bonds in a molecule relate to the presence of geometric isomers.
vazorg [7]
The answer to this question is D! The ball and stick model! Hope this helps :)
8 0
2 years ago
Read 2 more answers
A small portion of a crystal lattice is sketched below. What is the name of the unit cell of this lattice? Your answer must be a
Nikolay [14]

This is an incomplete question, the given sketch is shown below.

Answer : The name of given unit cell is, FCC (face-centered cubic unit cell)

Explanation :

Unit cell : It is defined as the smallest 3-dimensional portion of a complete space lattice which when repeated over the and again in different directions produces the complete space lattice.

There are three types of unit cell.

  • SCC (simple-centered cubic unit cell)
  • BCC (body-centered cubic unit cell)
  • FCC (face-centered cubic unit cell)

In SCC, the atoms are arranged at the corners.

Z=\frac{1}{8}\times 8=1

The number of atoms of unit cell = Z = 1

In BCC, the atoms are arranged at the corners and the body center.

Z=\frac{1}{8}\times 8+1=2

The number of atoms of unit cell = Z = 2

The given unit cell is, FCC because the atoms are arranged at the corners and the center of the 6 faces.

Z=\frac{1}{8}\times 8+\frac{1}{2}\times 6=4

The number of atoms of unit cell = Z = 4

Thus, the name of given unit cell is, FCC (face-centered cubic unit cell)

7 0
2 years ago
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