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3 years ago

Must all species reproduce in order for the species to survive?

Physics

2 answers:

Airida [17]3 years ago

5 0

ANSWER:

the correct answer is D.Yes; without reproduction, once the living organisms died, there would be no offspring to replace them.

TIP: bc if a species dies and it doesnt reproduce that species will go extinct, and remember every specie plays a role in life (but they are some animals that dont and are just a threat to the world). so reproducing is the only way to keep their generation going!

~batmans wife dun dun dun...aka ~serenitybella

fomenos3 years ago

4 0

The correct answer is :

D. Yes; without reproduction, once the living organisms died, there would be no offspring to replace them.

If every species didn't reproduce they'd eventually all start dying and then there would be none left. (reproduce or reproducing basically means when a species gets pregnant and has more species. :D)

Hope this helps,

Davinia.

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Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

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$2Ag_2O(s)\rightarrow 4Ag(s)+O_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{product}$

$\Delta H^o=[n_{Ag}\times \Delta H_f^0_{(Ag)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta H_f^0_{(Ag_2O)}]$

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$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

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

$\Delta H^o=[4mole\times (0kJ/mol)+1mole\times (0kJ/mol)}]-[2mole\times (-31.1kJ/mol)]$

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conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{product}$

$\Delta S^o=[n_{Ag}\times \Delta S_f^0_{(Ag)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta S_f^0_{(Ag_2O)}]$

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$\Delta S^o$ = entropy of reaction = ?

n = number of moles

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

$\Delta S^o=[4mole\times (42.55J/K.mole)+1mole\times (205.07J/K.mole)}]-[2mole\times (121.3J/K.mole)]$

$\Delta S^o=132.67J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

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

At room temperature, the temperature is 298 K.

$\Delta G^o=(62200J)-(298K\times 132.67J/K)$

$\Delta G^o=22664.34J=22.66kJ$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

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