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2 years ago

Why did plants perform cellular respiration?

2 answers:

7 0

Plants perform cellular respiration Because it allows glucose to break down into something called ATP and use it to release energy from glucose.

harina [27]2 years ago

4 0

Answer:process of cellular respiration allows plants to break down glucose into ATP. ... Although plants use photosynthesis to produce glucose, they use cellular respiration to release energy from the glucose.

Explanation:

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Calculate the concentrations of h2so3, hso−3, so2−3, h3o+ and oh− in 0.025 m h2so3.

sammy [17]

We will use this two reaction equation:

H2SO3 + H2O ↔ H3O+ + HSO3- Ka1 = 1.3 x 10^-2

HSO3- + H2O ↔ H3O+ + SO3 2- Ka2= 6.3 x 10^-8

we will use the ICE table for the first equation:

H2SO3 + H2O ↔ H3O+ + HSO3-

initial 0.025 0 0

change -X +X +X

Equ (0.025-X) X X

Ka1 = [H3O+] [HSO3-] / [H2SO3]

1.3 x 10^-2 = X^2 / (0.025 - X) by solving for X

∴ X = 0.0127

when [H3O+] = X

∴[H3O+] = 0.0127 M

and when [HSO3-] = X

∴[HSO3-] = 0.0127 M

and when [H2SO3] = 0.025 - X

∴[H2SO3] = 0.025 - 0.0127

= 0.0123 M

when Kw = [OH-][H3O+]

and Kw = 1.1 x 10^-14 / 0.0127

∴[OH-] = 1.1 x 10^-14 / 0.0127

= 8.66 x 10^-13 M

- by using the ICE table for the second equation:

HSO3- + H2O ↔ H3O+ + SO3 2-

initial 0.0127 0.0127 0

change -X +X +X

Equ (0.0127-X) (0.0127+X) X

when Ka2 = [SO32-] [H3O+] / [HSO3-]

by substitution:

6.3 x 10^-8 = X(0.0127+X) / (0.0127-X)

as the Ka2 is so small so we can assume that (0.01271 + X) & (0.01271-X) = 0.01271 and neglect X

6.3 x 10^-8 = 0.0127X /0.0127

∴X = 6.3 x 10^-8

when [SO3 2-] = X

∴[SO32-] = 6.3 x 10^-8

3 0

3 years ago

Which of the following is true for the reaction below? CO2(g) + 393.5 kJ C(s) + O2(g)

AleksandrR [38]

It is endothermic and the ∆H = +393.5 kJ.

6 0

2 years ago

Read 2 more answers

Which of the following keeps the planets in their orbits in the Solar System?

posledela

Answer:

B- gravity

Explanation:

Gravity is a force in which everything in the universe is trying to pull everyother thing towards themsef. Here, the sun's gravity keeps the planet in orbit.

I hope im right! !!

6 0

2 years ago

A sample of an unknown gas with a mass of 8.21 g has a volume of 4.8064 L when the temperature is 200oC and the pressure is 1.81

Pie

Answer:

The answer to your question is M = 36.49 g

Explanation:

Data

mass = 8.21 g

volume = 4.8064 L

Temperature = 200°C

Pressure = 1.816 atm

M = ?

Process

1.- Convert temperature to °K

°K = 273 + 200

°K = 473

2.- Calculate the number of moles

n = (PV)/RT

n = (1.816)(4.8064)/(0.082)(473)

n = 0.225

3.- Calculate the molar mass

M --------------- 1 mol

8.21 g ---------- 0.225 moles

M = (1 x 8.21)/0.225

M = 36.49 g

4 0

3 years ago

In Part III, the phenolphthalein indicator is used to monitor the equilibrium shifts of the ammonia/ammonium ion system. The phe

AnnZ [28]

The pink color in the solution fades. Some of the colored indicator ion converts to the colorless indicator molecule.

<h3>Explanation</h3>

What's the initial color of the solution?

$\text{NH}_4\text{Cl}$ is a salt soluble in water. $\text{NH}_4\text{Cl}$ dissociates into ions completely when dissolved.

$\text{NH}_4\text{Cl} \; (aq)\to {\text{NH}_4}^{+} \; (aq) +{\text{Cl}}^{-} \; (aq)$ .

The first test tube used to contain $\text{NH}_4\text{OH}$ . $\text{NH}_4\text{OH}$ is a weak base that dissociates partially in water.

$\text{NH}_4\text{OH} \; (aq) \rightleftharpoons {\text{NH}_4}^{+} \;(aq)+ {\text{OH}}^{-} \; (aq)$ .

There's also an equilibrium between $\text{OH}^{-}$ and ${\text{H}_3\text{O}}^{+}$ ions.

${\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l)$ .

$\text{OH}^{-}$ ions from $\text{NH}_4\text{OH}$ will shift the equilibrium between $\text{OH}^{-}$ and ${\text{H}_3\text{O}}^{+}$ to the right and reduce the amount of ${\text{H}_3\text{O}}^{+}$ in the solution.

The indicator equilibrium will shift to the right to produce more ${\text{H}_3\text{O}}^{+}$ ions along with the colored indicator ions. The solution will show a pink color.

What's the color of the solution after adding NH₄Cl?

Adding $\text{NH}_4\text{Cl}$ will add to the concentration of ${\text{NH}_4}^{+}$ ions in the solution. Some of the ${\text{NH}_4}^{+}$ ions will combine with $\text{OH}^{-}$ ions to produce $\text{NH}_4\text{OH}$ .

The equilibrium between $\text{OH}^{-}$ and ${\text{H}_3\text{O}}^{+}$ ions will shift to the left to produce more of both ions.

${\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l)$

The indicator equilibrium will shift to the left as the concentration of ${\text{H}_3\text{O}}^{+}$ increases. There will be less colored ions and more colorless molecules in the test tube. The pink color will fade.

7 0

3 years ago