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3 years ago

6

Which shows the correct order of increasing trophic level, from producer to tertiary consumer?

2 answers:

polet [3.4K]3 years ago

6 0

Answer:

Grass, deer, wolf, fungi.

Explanation:

The tropical level of an organism is defined as the position of organism occupied in the food chain. The food chain is starts from tropical level one which is primary producers (such as plants), then move to herbivores (such as deer) in level two, carnivores, (such as wolf) at level three, apex predators in level four (such as lions), and it ends in decomposers such as fungi.

Fungi play important role in ecosystem to help in breakdown of dead organic matter and return nutrients to the soil. Without fungi nutrients cannot cycle in an ecosystem, and causing the breakdown of entire food web.

Anna007 [38]3 years ago

4 0

Answer:

Grass, deer, wolf, fungi.

Explanation:

i just took the test

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Explanation:

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What is the volume of 12.0 grams of oxygen gas at STP<br><br> Atomic mass:O = 15.99 grams/moles

strojnjashka [21]

Answer:

16.82 L.

Explanation:

We can use the general law of ideal gas: PV = nRT.

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4 years ago

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3 years ago

The half-life of a first-order reaction is 13 min. If the initial concentration of reactant is 0.13 M, it takes ________ min for

Zigmanuir [339]

Answer:

Therefore it takes 8.0 mins for it to decrease to 0.085 M

Explanation:

First order reaction: The rate of reaction is proportional to the concentration of reactant of power one is called first order reaction.

A→ product

Let the concentration of A = [A]

$\textrm{rate of reaction}=-\frac{d[A]}{dt} =k[A]$

$k=\frac{2.303}{t} log\frac{[A_0]}{[A]}$

[A₀] = initial concentration

[A]= final concentration

t= time

k= rate constant

Half life: Half life is time to reduce the concentration of reactant of its half.

$t_{\frac{1}{2} }=\frac{0.693}{k}$

Here $t_{\frac{1}{2} }=0.13 min$

$k=\frac{0.693}{t_{\frac{1}{2}} }$

$\Rightarrow k=\frac{0.693}{13 }$

To find the time takes for it to decrease to 0.085 we use the below equation

$k=\frac{2.303}{t} log\frac{[A_0]}{[A]}$

$\Rightarrow t=\frac{2.303}{k} log\frac{[A_0]}{[A]}$

Here , $k=\frac{0.693}{13 }$ , [A₀] = 0.13 m and [ A] = 0.085 M

$t=\frac{2.303}{\frac{0.693}{13} } log(\frac{0.13}{0.085})$

$\Rightarrow t= 7.97\approx 8.0$

Therefore it takes 8.0 mins for it to decrease to 0.085 M

7 0

3 years ago