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3 years ago

Identify the following as abiotic or biotic. A paramecium swimming in a lake

Chemistry

1 answer:

sertanlavr [38]3 years ago

4 0

<h3>Answer:</h3>

Biotic

<h3>Explanation:</h3>

We are being tested on whether we can differentiate between abiotic and biotic factors in an ecosystem.

We need to know that factors that affect living organisms in an ecosystem are classified into abiotic factors and biotic factors.

Then, what is the difference between abiotic and bioitic factors?

Abiotic factors are non-living factors that affect living organisms in an ecosystem while biotic factors are living factors that affect organisms in an ecosystem.

What are examples of Biotic and abiotic factors.

Examples of abiotic factors include wind, temperature, humidity, climate,. etc.
Examples of abiotic factors include parasitism, mutualism, commensalism, etc.

How do we arrive at the answer to our question?

A Paramecium swimming in a lake; this is a biotic factor.
Because paramecium is a living organism which relates with other organisms in the lake ecosystem in various way.

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3 years ago

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What was the main reason why Alfred Wegener's theory about the movement of the continents was dismissed in 1912?

eduard

Answer:

His map of the ocean floor did not support his theory

Explanation:

Alfred Wegner was a great scientist
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It was dismissed after viewing his ocean floor

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3 years ago

What is the balance equation for copper metal and aqueous zinc nitrate

AnnZ [28]

Answer:

No reaction

Explanation:

Single-displacement reaction is the chemical reaction in which one metal which has higher negative redox potential replaces the another from it's salt solution.

Thus, Cu cannot replaces zine from its salt as it is less reactive than zince.

However, zinc can replace copper from its salt as:

$Zn_{(s)}+Cu(NO_3)_2_{(aq)}\rightarrow Cu_{(s)}+Zn(NO_3)_2_{(aq)}$

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3 years ago

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kap26 [50]

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3 years ago

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The following mechanism has been suggested for the reaction between nitrogen monoxide and oxygen: NO(g) + NO(g) → N2O2(g) (fast)

Karo-lina-s [1.5K]

Answer:

b. Second order in NO and first order in O₂.

Explanation:

A. The mechanism

$\rm 2NO\xrightarrow[k_{-1}]{k_{1}}N_{2}O_{2} \, (fast)\\\rm N_{2}O_{2} + O_{2}\xrightarrow{k_{2}} 2NO_{2} \, (slow)$

B. The rate expressions

$-\dfrac{\text{d[NO]} }{\text{d}t} = k_{1}[\text{NO]}^{2} - k_{-1} [\text{N}_{2}\text{O}_{2}]^{2}\\\\\rm -\dfrac{\text{d[N$_{2}$O$_{2}$]}}{\text{d}t} = -\dfrac{\text{d[O$_{2}$]}}{\text{d}t} = k_{2}[ N_{2}O_{2}][O_{2}] - k_{1} [NO]^{2}\\\\\dfrac{\text{d[NO$_{2}$]}}{\text{d}t}= k_{2}[ N_{2}O_{2}][O_{2}]$

The last expression is the rate law for the slow step. However, it contains the intermediate N₂O₂, so it can't be the final answer.

C. Assume the first step is an equilibrium

If the first step is an equilibrium, the rates of the forward and reverse reactions are equal. The equilibrium is only slightly perturbed by the slow leaking away of N₂O₂ to form product.

$\rm k_{1}[NO]^{2} = k_{-1} [N_{2}O_{2}]\\\\\rm [N_{2}O_{2}] = \dfrac{k_{1}}{k_{-1}}[NO]^{2}$

D. Substitute this concentration into the rate law

$\rm \dfrac{\text{d[NO$_{2}$]}}{\text{d}t}= \dfrac{k_{2}k_{1}}{k_{-1}}[NO]^{2} [O_{2}] = k[NO]^{2} [O_{2}]$

The reaction is second order in NO and first order in O₂.

8 0

3 years ago