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3 years ago

Which of these would be affected

Chemistry

1 answer:

Makovka662 [10]3 years ago

7 0

Answer:

J. the growth of trees

Explanation:

The growth of trees would be greatly impaired if the energy from the sun were stopped.

Plants generally use the energy from the sun to obtain their nourishment. This same process makes food available in the ecosystem.

Plants are producers of food.
The process of photosynthesis allows plants to manufacture food by using sunlight to drive the reaction between carbon dioxide and water to produce glucose and oxygen gas.
The glucose is used by plant for nourishment.

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A piece of high-density Styrofoam measuring 24.0 cm by 36.0 cm by 5.0 cm floats when placed in a tub of water. When a 1.5 kg boo

zheka24 [161]

Explanation:

The given data is as follows.

Width of Styrofoam = 24.0 cm

Length of Styrofoam = 36.0 cm

Height of Styrofoam = 5.0 cm

Therefore, volume of the Styrofoam will be calculated as follows.

Volume = length × width × height

= (36.0 × 24.0 × 5.0) $cm^{3}$

= 4320 $cm^{3}$

or, = $4.32 \times 10^{3} cm^{3}$

As Styrofoam partially sinks at 3.0 cm and total height of Styrofoam is 5.0 cm. Hence, height of Styrofoam above the water is (5.0 - 3 cm) = 2 cm.

So, volume of water displaced is as follows.

24.0 cm × 36.0 cm × 2.0 cm

= $1.73 \times 10^{3} cm^{3}$

Hence, mass of displaced water is as follows.

mass = density × volume

= $1.00 g/cm^{3} \times 1.73 \times 10^{3} cm^{3}$

= $1.73 \times 10^{3} g$

Since, book is placed on the Styrofoam. Therefore, mass of water displaced is also equal to the following.

Mass of water displaced = mass of book + mass of Styrofoam

$1.73 \times 10^{3} g$ = 1500 g + mass of Styrofoam

(1730 - 1500) g = mass of Styrofoam

mass of Styrofoam = 230 g

Therefore, calculate the density of Styrofoam as follows.

Density = $\frac{mass}{volume}$

= $\frac{230}{4.32 \times 10^{3} cm^{3}}$

= $53.24 \times 10^{-3} g cm^{-3}$

Thus, we can conclude that the density of Styrofoam is $53.24 \times 10^{-3} g cm^{-3}$ .

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Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant

antoniya [11.8K]

Answer: The value of $E^{o}$ for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

$AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$

Its solubility product will be as follows.

$K_{sp} = [Ag^{+}][Cl^{-}]$

Cell reaction for this equation is as follows.

$Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)$

Reduction half-reaction: $Ag^{+} + 1e^{-} \rightarrow Ag(s)$ , $E^{o}_{Ag^{+}/Ag} = 0.799 V$

Oxidation half-reaction: $Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}$ , $E^{o}_{AgCl/Ag}$ = ?

Cell reaction: $Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)$

So, for this cell reaction the number of moles of electrons transferred are n = 1.

Solubility product, $K_{sp} = [Ag^{+}][Cl^{-}]$

= $1.77 \times 10^{-10}$

Therefore, according to the Nernst equation

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

At equilibrium, $E_{cell}$ = 0.00 V

Putting the given values into the above formula as follows.

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

$0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}$

$E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}$

= $0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}$

= 0.577 V

Hence, we will calculate the standard cell potential as follows.

$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}$

$0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}$

$0.577 V = 0.799 V - E^{o}_{AgCl/Ag}$

$E^{o}_{AgCl/Ag}$ = 0.222 V

Thus, we can conclude that value of $E^{o}$ for the half-cell reaction is 0.222 V.

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