How does the DNA of a daughter cell compare to the DNA of the parent cell in asexual reproduction?

What are the roles of producers, primary consumers, secondary consumers, tertiary consumers for the trophic levels in the energy

Nesterboy [21]

This question is more for Biology than Chemistry, but the role of producers is to make energy (food) to be consumed. In a pyramid diagram, the producers would be at the bottom. Now going up the pyramid, the primary conumers are the first to consume producers and obtain energy from them. As you go up the pyramid, the secondary consumers will consume the primary consumers as a way to obtain energy, and the same goes for tertiary consumers towards secondaries.

As you go up the energy pyramid, you will notice a trend that there is less energy being obtained from each consumer. In other words, the producers will ALWAYS have more energy than the tertiary consumers.

I hope this answers your question.

3 0

3 years ago

What is the concentration of the acid in this titration 1.2 m 2.4 m 1.95 m 0.98 m 1.98 m

tresset_1 [31]

Answer:

1.95

Explanation:

Just did the test and it was correct!

5 0

4 years ago

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A gas that exerts a pressure of ? bar in a container with a volume of 0.5650 L will exert a pressure of 715.1 bar when transferr

uranmaximum [27]

Robert Boyle, the 17th century British chemist, first noticed that the volume of a given amount of gas is inversely proportional to its pressure when kept at a constant temperature. When working with ideal gases we use PV = nRT, but remember n, R, and T are all constant. Therefore we have:

PV(before) = PV(after)

P(0.5650) = (715.1)(1.204)

8 0

3 years ago

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Bromine reacts with nitric oxide to form nitrosyl bromide as shown in this reaction: br2(g) + 2 no(g) → 2 nobr(g) a possible mec

Svetach [21]

The overall reaction is given by:

$Br_{2}(g) + 2 NO(g) \rightarrow 2 NOBr(g)$

The fast step reaction is given as:

$NO(g) + Br_{2}(g) \rightleftharpoons NOBr_{2}(g) (fast; k_{eq}= \frac{k_{1}}{k_{-1}})$

The slow step reaction is given as:

$NOBr_{2}(g) + NO(g) \rightarrow 2 NOBr(g)$ (slow step $k_{2}$ )

Now, the expression for the rate of reaction of fast reaction is:

$r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}]$

The expression for the rate of reaction of slow reaction is:

$r_{2}=k_{2}[NOBr_{2}] [NO]$

Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as $[NOBr_{2}]$ takes place in this reaction.