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Talja [164]
3 years ago
5

Most of the energy from a lower trophic level is converted to

Chemistry
1 answer:
frutty [35]3 years ago
5 0
The correct answer to the question above is heat. Most of the energy from a lower trophic level is converted into heat. When an organism from a higher trophic level consumed an organism from a lower trophic level, it is mostly heat that is being converted to.
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What is 256ml to liters
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That would be 0.26 liters

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If the elastic modulus of cobalt (co) is 200 gpa, and the elastic modulus of tungsten carbide (wc) is 700 gpa, calculate the upp
Dmitry_Shevchenko [17]
To determine the upper bond
Ec(u) = EmVm + EpVp
Em is the elastic modulus of cobalt.
E₁ is the elastic modulus of the particulate
Vm is the volume fraction of cobalt
Vp is the volume fraction of particulate
substitute
Ec(u) = 200 (Vm) + 700 (Vp)
To determine the lower bound
Ec (l) = EmEp/VmEp+ VpEm
Substitute
Ec (l) = 200(700)/Vm(700) + Vp (200)
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7 0
3 years ago
How to do them 11-15 most difficult ones for me
BigorU [14]
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6 0
3 years ago
Newtons first law second law and third law <br><br> help
Romashka [77]

Answer:

In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

Explanation:

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7 0
3 years ago
One mole of an ideal gas doubles its volume in a reversible isothermal expansion. (a) What is the change in entropy of the gas?
Andre45 [30]

Explanation:

The given data is as follows.

       n = 1 mol,     V_{f} = 2V_{i}

       Q = 1500 J,      R = 8.314 J/mol k

(a)    \Delta S = \frac{dQ}{dT}

And, according to the first law of thermodynamics

                \Delta E_{int} = Q - W

And, in an isothermal process the change in internal energy of the gas is zero.

Hence,    0 = Q - W

or,             W = Q

Expression for work done in an isothermal process is as follows.

                   W = nRT ln \frac{V_{f}}{V_{i}}

As W = Q, Hence expression for Q will also be given as follows.

            Q = nRT ln \frac{V_{f}}{V_{i}}

Now,  

        \Delta S = \frac{nRT ln \frac{V_{f}}{V_{i}}}{T}

        [/tex]\Delta S = nR ln \frac{V_{f}}{V_{i}}[/tex]

                      = nR ln \frac{2V_{i}}{V_{i}}

                       = nR ln 2

                        = 1 \times 8.314 \times 0.693

                        = 5.76 J/K

Therefore, change in entropy is 5.76 J/K.

(b)    As,  Q = nRT ln \frac{V_{f}}{V_{i}}

                   = nRT ln \frac{2V_{i}}{V_{i}}

                   = nRT ln 2

           T = \frac{Q}{nR ln 2}

              = \frac{1500}{1 \times 8.314 ln 2}

              = 260.4 K

Therefore, temperature of the gas is 260.4 K.

7 0
2 years ago
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