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Nimfa-mama [501]
3 years ago
14

Potato plants with a Jellyfish gene will glow when they need to be watered.

Chemistry
2 answers:
Sliva [168]3 years ago
6 0

Answer:

Hello!!! Princess Sakura here ^^

Explanation:

True the chemist are working to develop more feature on to this plant as well.

maksim [4K]3 years ago
5 0

This statement is true, potato plants with jellyfish genes do in fact glow when they need to be watered!

May i have brainliest? I need 1 more!

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Because it represents 2 atoms of carbon not 2 molecules of carbon
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Only one of the listed choices are correct here:
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ascorbic acid is a diprotic acid (Ka= 8.0x10^-5 and Ka2= 1.6x10^-12). What is the pH of a 0.260 M solution of ascorbic acid
zlopas [31]

The pH of a 0.260 M solution of ascorbic acid is 0.585. Details about pH can be found below.

<h3>How to calculate pH?</h3>

The pH of a solution can be calculated using the following expression:

pH = - log {H+}

According to this question, ascorbic acid is a diprotic acid and posseses a concentration of 0.260M. The pH can be calculated as follows;

pH = - log {0.260}

pH = 0.585

Therefore, the pH of a 0.260 M solution of ascorbic acid is 0.585.

Learn more about pH at: brainly.com/question/15289741

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7 0
2 years ago
What is one effect of convection currents in the mantle?
Pie

Answer:

D

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due yo high energy from the ocean to the earth

4 0
2 years ago
Consider the reaction given below.
Drupady [299]

Answer:

  • <u>K =  0.167 s⁻¹</u>

Explanation:

<u>1) Rate law, at a given temperature:</u>

  • Since all the data are obtained at the same temperature, the equilibrium constant is the same.

  • Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:

        r = K [A]ᵃ [B]ᵇ

<u>2) Use the data from the table</u>

  • Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s

         Divide r₂ by r₁:     [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0

  • Use the first and second set of data to find the exponent a:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s

        Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]

                                  2ᵃ = 2 ⇒ a = 1

         

<u>3) Write the rate law</u>

  • r = K [A]¹ [B]⁰ = K[A]

This means, that the rate is independent of reactant B and is of first order respect reactant A.

<u>4) Use any set of data to find K</u>

With the first set of data

  • r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹

Result: the rate constant is K =  0.167 s⁻¹

6 0
4 years ago
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