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pychu [463]
3 years ago
15

How is energy passed along in the food chain and food web PLEASE HELP QUICK

Chemistry
1 answer:
LenKa [72]3 years ago
5 0
<span>The energy (from food) an organism consumes is mostly burned for the purposes of movement and keeping warm. In terms of the trophic levels, this is energy "wasted" (heat radiated by an animal out to the surrounding environment) and not passed on to the next level as consumable food. For instance, imagine that the sun's rays (the ultimate source of energy in the "food chain") striking a portion of the ocean on the Earth's surface is equal to one million energy units. Much of that energy is absorbed by the ocean's water molecules, but the amount left over is intercepted by phytoplankton and turned into food by photosynthesis. At this first trophic level, the amount of energy the phytoplankton store as food is 20,000 units. At the next trophic level, the zooplankton consume the phytoplankton, but just 2,000 units of energy is passed on as food in the bodies of the tiny animals. Small fish (at the next trophic level) scoop up the little zooplankton, but their bodies only represent 10% of the previous level's energy units - therefore, the amount of energy left is now 200 units. By the time large fish at their trophic level eat the smaller fish, 20 energy units remain. Finally, at the point where humans (at the final trophic level) catch these fish, only 1 or 2 energy units are left. As this example shows, the approximate efficiency at which energy is transferred through the "food chain" from one level to the next highlights the great amount of energy lost in an ecosystem, and how a relatively small amount of energy is actually available to an organism in the form of food that it needs to survive and perpetuate.</span>
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Explanation:

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Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti
gayaneshka [121]

Answer : The value of K for this reaction is, 2.6\times 10^{15}

Explanation :

The given chemical reaction is:

CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)

Now we have to calculate value of (\Delta G^o).

\Delta G^o=G_f_{product}-G_f_{reactant}

\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]

where,

\Delta G^o = Gibbs free energy of reaction = ?

n = number of moles

\Delta G^0_{(HCH_3CO_2(g))} = -389.8 kJ/mol

\Delta G^0_{(CH_3OH(g))} = -161.96 kJ/mol

\Delta G^0_{(CO(g))} = -137.2 kJ/mol

Now put all the given values in this expression, we get:

\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]

\Delta G^o=-89.4kJ/mol

The relation between the equilibrium constant and standard Gibbs, free energy is:

\Delta G^o=-RT\times \ln K

where,

\Delta G^o = standard Gibbs, free energy  = -89.4 kJ/mol = -89400 J/mol

R = gas constant  = 8.314 J/L.atm

T = temperature  = 30.0^oC=273+30.0=303K

K = equilibrium constant = ?

Now put all the given values in this expression, we get:

-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K

K=2.6\times 10^{15}

Thus, the value of K for this reaction is, 2.6\times 10^{15}

4 0
4 years ago
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Vadim26 [7]

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7 0
3 years ago
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2 years ago
Answer and work for this problem
MArishka [77]
We can write the balanced equation for the synthesis reaction as 
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We also use the molar masses of hydrogen chloride gas HCl and chlorine gas CL2 to calculate for the mass of hydrogen gas H2:
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Therefore, we need 4.056 grams of hydrogen gas and 142.4 grams of chlorine gas to produce 146.4 grams of hydrogen chloride gas.
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