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Pavlova-9 [17]
3 years ago
5

Explain how interactions can be both beneficial and harmful to the organism in a community

Chemistry
2 answers:
Marysya12 [62]3 years ago
7 0
Interactions to an organism can be both harmful and beneficial due to the organism's circumstances. For example by adding tadpoles to an eco system it results in the tadpoles eating all the food thus killing the other organisms. Whereas by adding plants to an ecosystem it provides shelter for the organisms and possibly food being beneficial.
Luda [366]3 years ago
4 0

Answer:

An ecological community consists of all the populations of all the different species that live together in a particular area.

Interactions between different species in a community are called interspecific interactions.

The main types of interspecific interactions include competition, predation, mutualism, commensalism, and parasitism.

Symbiosis

Symbiosis occur when two species live together in a long-term, intimate association. In everyday life, we sometimes use the term symbiosis to mean a relationship that benefits both parties

Parasitism

It involves two species have a close, lasting interaction that is beneficial to one, the parasite, and harmful to the other, the host.

Commensalism

In a commensalism, two species have a long-term interaction that is beneficial to one and has no positive or negative effect on the other. For instance, many of the bacteria that inhabit our bodies seem to have a commensal relationship with us. They benefit by getting shelter and nutrients and have no obvious helpful or harmful effect on us.

Mutualism

In a mutualism, two species have a long-term interaction that is beneficial to both of them. For example, some types of fungi form mutualistic associations with plant roots. The plant can photosynthesize, and it provides the fungus with fixed carbon in the form of sugars and other organic molecules

Predation

In predation, a member of one species—the predator—eats part or all of the living, or recently living, body of another organism—the prey. This interaction is beneficial for the predator, but harmful for the prey. Predation may involve two animal species, but it can also involve an animal or insect consuming part of a plant, a special case of predation known as herbivory

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Answer:

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2. What is the independent variable? Secret ingredient in the breath mint

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4. What should Mr. Krabs’ conclusion be? The breath mint with the secret ingredient appears to reduce the

amount of breath odor more than half the time, but it is not 100% effective.

5. Why do you think 10 people in group B reported fresher breath? This may be due to the placebo effect.

6 0
2 years ago
What is the mass in grams of 2.00 10 5 atoms of naturally occurring neon
lesantik [10]

Answer:

2.00X10^5 x 20gNe/6.02x10^23=6.46x10^-18 but books answer is 797.

Explanation:

3 0
2 years ago
What kind of change results in a new substance being produced
tatiyna
Chemical change creates a new substance
3 0
3 years ago
Read 2 more answers
Consider the reaction given below.
Drupady [299]

Answer:

  • <u>K =  0.167 s⁻¹</u>

Explanation:

<u>1) Rate law, at a given temperature:</u>

  • Since all the data are obtained at the same temperature, the equilibrium constant is the same.

  • Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:

        r = K [A]ᵃ [B]ᵇ

<u>2) Use the data from the table</u>

  • Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s

         Divide r₂ by r₁:     [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0

  • Use the first and second set of data to find the exponent a:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s

        Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]

                                  2ᵃ = 2 ⇒ a = 1

         

<u>3) Write the rate law</u>

  • r = K [A]¹ [B]⁰ = K[A]

This means, that the rate is independent of reactant B and is of first order respect reactant A.

<u>4) Use any set of data to find K</u>

With the first set of data

  • r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹

Result: the rate constant is K =  0.167 s⁻¹

6 0
3 years ago
the combustion of a sample butane, C4H10 (lighter fluid) produced 2.46 grams of water. how many moles of water formed
enyata [817]
2C_4H_{10}+13O_2 ⇒ 8CO_2 + 10H_2O

n= \frac{m}{M} =  \frac{m}{2M(H)+M(O)}= \frac{2,46}{2*1,0+16,0}  = 0,14mol

So 0,14mol are formed.


6 0
3 years ago
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