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Maslowich
3 years ago
7

The energy available to primary consumers in an energy pyramid is 800 kilocalories. Approximately how much energy is used by the

primary consumers for their physical functions?
A. Less than 720 kilocalories
B. 720 kilocalories
C. 880 kilocalories
D. More than 880 kilocalories
Chemistry
2 answers:
ki77a [65]3 years ago
5 0
The answer is B, Let me know if you get it right!
torisob [31]3 years ago
3 0

Answer:

B. 720 kilocalories

Explanation:

The primary consumers use about 90% of available energy pyramid for their physical functions. In order to solve this exercise, we need to calculate the 90% of 800 kilocalories.

We can state that there is  a direct linear relationship between the kilocalories and the percentage. This means that for a higher percentage the amount of kilocalories will be greater.

We can write :

\frac{800kilocalories}{100}=\frac{x}{90}

Where the denominators are percentages.

Solving for x :

x=\frac{(800kilocalories).(90)}{100}

x=720kilocalories

The correct option is B. 720 kilocalories

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For the reaction C2H4(g) + H2O(g) --&gt; CH3CH2OH(g)
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Answer : The value of equilibrium constant for this reaction at 262.0 K is 3.35\times 10^{2}

Explanation :

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

where,

\Delta G^o = standard Gibbs free energy  = ?

\Delta H^o = standard enthalpy = -45.6 kJ = -45600 J

\Delta S^o = standard entropy = -125.7 J/K

T = temperature of reaction = 262.0 K

Now put all the given values in the above formula, we get:

\Delta G^o=(-45600J)-(262.0K\times -125.7J/K)

\Delta G^o=-12666.6J=-12.7kJ

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  = -12666.6 J

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T = temperature  = 262.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

-12666.6J=-(8.314J/K.mol)\times (262.0K)\times \ln k

k=3.35\times 10^{2}

Therefore, the value of equilibrium constant for this reaction at 262.0 K is 3.35\times 10^{2}

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