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2 years ago

Which organisms in soil helps relieve element so they can be recycled

Chemistry

2 answers:

klio [65]2 years ago

8 0

Answer: Hewo there! your answer is Below

Soil bacteria perform recycling of soil organic matter through different processes, and as a result they produce and release into the soil inorganic molecules

Explanation:

Bacteria consume organic matter and other compounds and recycle them into substances that can be used by other organisms. Bacteria can live anywhere that has water. They are more numerous, can reproduce faster and can survive harsher conditions than any other organism on Earth.

Hope this helps you!!

Have a great day❤️❤️

-August-

natita [175]2 years ago

6 0

Answer:

Explanation:

Protozoa and Bacteria: Protozoa are very small one-celled creatures. They play an important role in helping to make the soil better by feeding on bacteria and releasing excess nitrogen, which is really good for plants. Bacteria are among the tiniest and most numerous organisms in the soil.

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WORTH 20 POINTS Confirmation bias is where a scientist may change the data to support a certain bias.

Orlov [11]

Answer:

True

Explanation:

Confirmation Bias is the tendency to look for information that supports, rather than rejects, one’s preconceptions, typically by interpreting evidence to confirm existing beliefs while rejecting or ignoring any conflicting data

8 0

2 years ago

In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.

Mamont248 [21]

Answer:

Approximately $1.30 \times 10^{-2}$ , assuming that this acid is monoprotic.

Explanation:

Assume that this acid is monoprotic. Let $\rm HA$ denote this acid.

$\rm HA \rightleftharpoons H^{+} + A^{-}$ .

Initial concentration of $\rm HA$ without any dissociation:

$[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}$ .

After $12.5\%$ of that was dissociated, the concentration of both $\rm H^{+}$ and $\rm A^{-}$ (conjugate base of this acid) would become:

$12.5\% \times 0.730\; \rm mol \cdot L^{-1} = 0.09125\; \rm mol \cdot L^{-1}$ .

Concentration of $\rm HA$ in the solution after dissociation:

$(1 - 12.5\%) \times 0.730\; \rm mol \cdot L^{-1} = 0.63875\; \rm mol\cdot L^{-1}$ .

Let $[{\rm HA}]$ , $[{\rm H}^{+}]$ , and $[{\rm A}^{-}]$ denote the concentration (in $\rm mol \cdot L^{-1}$ or $\rm M$ ) of the corresponding species at equilibrium. Calculate the acid dissociation constant $K_{\rm a}$ for $\rm HA$ , under the assumption that this acid is monoprotic:

$\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}$ .