Phloem contains specialized cells that move sugars and other nutrients both
up and down in plants. Which object is the best model of one of these cells?
A. Water balloon

Answers

Answer 1

Answer:Drinking straw

Explanation: my quiz said so

Answer 2

The object that represents the best model of one of the phloem cells is the drinking straw.

What is Phloem?

Phloem may be defined as a type of vascular tissue in plants that significantly conducts the passage of sugars and other metabolic products from the leaves to the roots. This mediates the migration of sugar which is synthesized in the leaves during photosynthesis.

Apart from this, the phloem also regulates the movement of nutrients from the roots to the upper part of the body. This function of phloem cells shares homology with the object known as a drinking straw that mediates the transport of any fluid in both directions.

Therefore, a drinking straw is an object that represents the best model of one of the phloem cells.

To learn more about Phloem cells, refer to the link:

https://brainly.com/question/3138900

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Your question seems incomplete. The most probable complete question is as follows:

Can openerWater balloonDrinking StrawKitchen strainer

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Explanation:

Consider an electrochemical cell based on the spontaneous reaction 2AgCl(s) + Zn(s) → 2Ag(s) + 2Cl– + Zn2+. If the zinc ion concentration is kept constant at 1 M, and the chlorine ion concentration is decreased from 1 M to 0.001 M, the cell voltage should:

Answers

Answer:

there is an increase by 0.18 V in the cell voltage.

Explanation:

The given equation of the reaction can be well written as

[tex]2AgCl_{(s)} + Zn _{(s)} \to 2Ag_{(s)} + 2 Cl^- _{(aq)}+ Zn^{2+}_{(aq)}[/tex]

By application of Nernst Equation ; we have the expression

[tex]E_{cell} = E^0- \dfrac{0,059}{n}log (\dfrac{[product]}{[reactant]})[/tex]

here in the above equation;

n = number of electrons transferred in the equation of the reaction

n = 2

Also;

[tex]E^0 = E_{cathode} - E_{anode}[/tex]

[tex]E^0 = E_{Ag^+/Ag} - E_{Zn^+/Zn}[/tex]

[tex]E^0 = +(0.80 \ V) - (-0..76 \ V)[/tex]

[tex]E^0 = (0.80 \ V +0..76 \ V)[/tex]

[tex]E^0 = 1.56 \ V[/tex]

If the zinc ion concentration is kept constant at 1 M; we have:

[tex]E_{cell} = E^0- \dfrac{0.059}{n}log (\dfrac{[product]}{[reactant]})[/tex]

[tex]E_{cell} = 1.56 - \dfrac{0.059}{2}log ({[Zn^{2+} ]}{[Cl^{2-}]})[/tex]

[tex]E_{cell} = 1.56 - \dfrac{0.059}{2}log (1)[/tex]

Since log(1) = 0

Therefore;

[tex]E_{cell} = 1.56\ V[/tex]

When the chlorine ion concentration is decreased from 1 M to 0.001 M; we have;

[tex]E_{cell} = E^0- \dfrac{0.059}{n}log (\dfrac{[product]}{[reactant]})[/tex]

[tex]E_{cell} = 1.56 - \dfrac{0.059}{2}log ({[Zn^{2+} ]}{[Cl^{2-}]})[/tex]

[tex]E_{cell} = 1.56 - \dfrac{0.059}{2}log ({[1*0.001^2}]})[/tex]

[tex]E_{cell} = 1.56 - 0.0295 \ * \ log ({[1*10^{-6}}]})[/tex]

[tex]E_{cell} = + 1.737 \ V[/tex]

The change in voltage = [tex]E_{cell} - E^0[/tex]

=( 1.737 - 1.56 )V

= 0.177 V

≅ 0.18 V

Thus; from the following observation; there is an increase by 0.18 V in the cell voltage.

The voltage of the cell increased by 0.18 V.

The equation of the reaction is; 2AgCl(s) + Zn(s) → 2Ag(s) + 2Cl– + Zn2+

We know that;

E°cell = 1.36 - (-0.76) = 2.12 V

If the cells are both at 1M concentration the Ecell = E°cell = 2.12 V

When the concentration of Cl- decreased from 1 M to 0.001 M

Ecell =  E°cell - 0.0592/n log Q

Substituting values;

Ecell = 2.12 V -  0.0592/2 log (1 × (0.001)^2)

Ecell = 2.298 V

Increase in voltage = 2.298 V - 2.12 V = 0.18 V

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