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Nuetrik [128]
3 years ago
9

Describe how organisms in the tropical rainforest both depend on and compete for biotic and abiotic factors. In your answer, ide

ntify which factors are biotic and which are abiotic.
Chemistry
1 answer:
nadezda [96]3 years ago
7 0

What are biotic and abiotic factors?

Biotic and abiotic are the two essential factors responsible for shaping the ecosystem. The biotic factors refer to all the living beings present in an ecosystem, and the abiotic factors refer to all the non-living components like physical conditions (temperature, pH, humidity, salinity, sunlight, etc.) and chemical agents (different gases and mineral nutrients present in the air, water, soil, etc.) in an ecosystem. Therefore, both the abiotic and biotic resources affect the survival and reproduction process.

Answer:

Abiotic factors in the tropical rainforest include humidity, soil composition, temperature, and sunlight. A limiting factor in the ecosystem is that canopy layers block sunlight from reaching the forest floor, causing shorter plants to not be able to grow. Biotic factors in the tropical rainforest include orchids, lilies, heliconia, and bromeliads. Tropical rainforests can have various fungi, shrubs, herbs, woody vines, lichens, and mosses. The trees making up the rainforest canopy include the tonka bean wood, teak, rubber, and several species of evergreens and palm trees. This way the organisms in the tropical rainforest both depend on biotic and abiotic factors.

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A 0.20 M solution of a weak acid has a pH of 5.40. What is the Ka for the acid?
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The Ka for this weak acid is 7,92 * 10^-11

Explanation:

First of all, let's think the equation

HA + H2O <------> H3O+  +   A-

When we add water to a weak acid, it dissociates in an equilibrium to generate the corresponding anion and the hydronium cation  (the acid form of water)

How do you calculate Ka??  Ka is the acid equilibrium constant.

( [H3O+]  . [A-] ) / [HA] where all the concentrations are in equilibrium.

We don't have the concentration in equilibrium but we have the initial concentration. So...

 HA + H2O <------> H3O+  +   A-

initial- 0.2 M             I don't have H3O+, either A-

reaction - an specific amount reacted  (X)

in equilibrium (0,2 - X)  <----->  X   +  X

And now, how's the formula for Ka

( [H3O+]  . [A-] ) / [HA] = Ka

(X . X) / (0.2-X)

X^2 / (0.2-X) = Ka

Look, that we don't have X as the [H3O+] but we know the pH, so we can know the [H3O+] indeed.

10^-pH = [H3O+]

10^-5,40 = 3,98 * 10^-6

Let's go back to Ka

( [H3O+]  . [A-] ) / [HA] = Ka

(3,98 * 10^-6)^2 / (0.2 - 3,98 * 10^-6) = Ka

(3,98 * 10^-6 is an small number, soooo small that we can approximate to 0)

If we have in order 10^-6, 10^-5 we can consider that.

So now, we have

(3,98 * 10^-6)^2 / (0.2) = Ka

1,58 * 10^-11 / (0.2) = Ka = 7,92* 10^-11

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