All categories

2 years ago

What are the non-living components of an ecosystem? The non-living components of an ecosystem are factors.

Chemistry

2 answers:

White raven [17]2 years ago

4 0

The answer is: abiotic factors.

Abiotic factors are non-living chemical and physical things of the environment.

On the contrary, biotic factors are all of the living organisms within an ecosystem.

Abiotic factors can include water, light, radiation, temperature, humidity, atmosphere, and soil.

For example, abiotic factors found in terrestrial ecosystems are sunlight, rain, wind, temperature, soil, pollution, nutrients, pH of soil and water, types of soil.

Nina [5.8K]2 years ago

4 0

Answer:

abiotic

Explanation:

got it right on plato!

You might be interested in

Which of the following best describes what happens when an ionic bond forms? A) two atoms come together and share electrons in o

Mars2501 [29]

I will try t do it and send it to you soon

5 0

2 years ago

Why is paper chromatography used for amino acids

vivado [14]

<em>Paper chromatography is especially useful in characterizing amino acids. The different amino acids move at differing rates on the paper because of differences in their R groups.</em>

5 0

1 year ago

What would happen if you used a warm washcloth if you had a fever.

aivan3 [116]

You will feel way better because in our class we learned that lukewarm baths are good for fevers and warm washcloths :)

3 0

2 years ago

A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.

irga5000 [103]

Answer:

1. pH = 1.23.

2. $H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

$H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+$

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the pKa is:

$pKa=-log(Ka)=-log(5.90x10^{-2})=1.23$

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

$pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23$

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

$n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol$

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

$H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Which is also shown in net ionic notation.

Best regards!

4 0

2 years ago

Zoom all the way in to Daphne Major. Describe its appearance in two sentences.

kicyunya [14]

The Daphne Major is located in the Colon Archipelago. It has a crater composed of volcanic rocks. These rocks are made up from volcanic ashes which hardened over time. The volcano is shaped like a cone which is surrounded by trees and houses several rare birds.

3 0

2 years ago

Read 2 more answers