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sdas [7]
3 years ago
5

Describe the relationship between predator and prey in a balanced ecosystem.

Chemistry
1 answer:
Annette [7]3 years ago
4 0
<h2><u>Relationship between predator and prey in a balanced ecosystem.</u></h2>

Predators grow along with their victims. Over time, prey animals are now developing and avoiding themselves to get eaten by their predators. Such tactics and modifications can take many forms that make their work easier, including disguise, mimicry, defense mechanisms, flexibility, distance, habits and even tool use.

Though fact an equilibrium appears to occur within an ecosystem between predators and prey, there are several factors which affect it, including the birth and death rates of predators and preys.

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Based on the law of conservation of mass, which of the following statements is correct? Matter is neither created nor destroyed.
Dmitrij [34]

The first statement (Matter is neither created nor destroyed) is correct.

The second statement would violate the law of conservation of mass (I will refer to this as LCM), as it would mean matter can "flow" into the universe, but not out, meaning the total matter will never be less than it was before.

The third statement violates LCM because it means matter is created during a reaction, which is not true.

The last statement violates LCM because it means matter is lost during a reaction, which is not true.

7 0
3 years ago
Very good electrical conductivity
bonufazy [111]
Huh? I need more info to answer this
7 0
3 years ago
The Ksp for BaCO3 is 5.1x10^-9. how many grams of BaCO3 will dissolve in 1000ml of water
Amanda [17]

Answer:

= 0.014 g of BaCO3

Explanation:

Let x = mol/L of BaCO3 that dissolve.  

This will give;

x mol/L Ba2+ and x mol/L CO32-  

But;

Ksp = 5.1x10^-9.

Therefore;

Ksp = 5.1 x 10^-9 = (x)(x)  

Thus;

x = molar solubility

  = √ (5.1 x 10^-9)

  = 7.1 x 10^-5 M  

Therefore;

Mass BaCO3 = 7.1 x 10^-5 M x 1 L x 197.34 g/mol

                      = 0.014 g

3 0
3 years ago
On the weak base/strong acid titration curve, label
aleksklad [387]

Answer:

i think the answer will be b. the point where the ph corresponds to a solution of the conjugate acid (bh+) in water;

Explanation:

7 0
3 years ago
When 1.025 g of naphthalene (c10h8) burns in a bomb calorimeter, the temperature rises from 24.25°c to 32.33°c. find δerxn for t
Goryan [66]

Answer : The \Delta E for the combustion of naphthalene is 5161.25KJ/mole

Solution : Given,

Mass of naphthalene = 1.025 g

Initial temperature = 24.25^oC

Final temperature = 32.33^oC

Specific heat capacity of calorimeter = 5.11KJ/^oC

Molar mass of naphthalene = 128 g/mole

First, we have to calculate the heat absorbed, q

Formula used :

q=c\times \Delta T=c\times (T_{final}-T_{initial})

Now put all the given values in this formula, we get

q=(5.11KJ/^oC)\times (32.33^oCT-24.25^oC)=41.29KJ

Now we have to calculate the moles of naphthalene.

Moles of C_{10}H_{8} = \frac{\text{ Mass of }C_{10}H_{8}}{\text{ Molar mass of }C_{10}H_{8}}=\frac{1.025g}{128g/mole}=0.0080moles

Now we have to calculate the \Delta E for combustion of naphthalene.

\Delta E=\frac{q}{n}

where,

q = heat absorbed

n = number of moles

Now put all the values in this formula, we get

\Delta E=\frac{41.29KJ}{0.0080moles}=5161.25KJ/mole

Therefore, the \Delta E for the combustion of naphthalene is 5161.25KJ/mole

6 0
3 years ago
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