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Sergeu [11.5K]
3 years ago
6

Models have been created that predict world sea levels will rise if the current global temperature increases continue. If these

models are correct, which of the following could be correctly inferred?
A. The amount of land area will decrease. 
B. Fossil fuels will rapidly become depleted.
C. Alternative energy sources will not meet demand. 
D. Deforestation continues to cause global warming.
Chemistry
1 answer:
lesya [120]3 years ago
5 0
To solve this question, think about which one of these options is a DIRECT effect of sea levels rising.

Choice A: If sea levels rise, they would start to cover up land that's at sea level, areas like beaches or coastal cities for example.

Choice B: Fossil fuels cause global warming. This is about the effects of global warming, not the causes. 

Choice C: Not relevant. It doesn't mention alternative energy at all in this.

Choice D: Again, effects, not causes (see Choice B explanation).

The correct answer is... CHOICE A
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Consider the reaction N2(g) + 2O2(g)2NO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr
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<u>Answer:</u> The value of \Delta S^o for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]

For the given chemical reaction:

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The equation for the entropy change of the above reaction is:

\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]

We are given:

\Delta S^o_{(NO_2(g))}=240.06J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(N_2)}=191.61J/K.mol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(2\times (240.06))]-[(1\times (191.61))+(2\times (205.14))]\\\\\Delta S^o_{rxn}=-121.77J/K

Entropy change of the surrounding = - (Entropy change of the system) = -(-121.77) J/K = 121.77 J/K

We are given:

Moles of nitrogen gas reacted = 1.90 moles

By Stoichiometry of the reaction:

When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K

So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = \frac{121.77}{1}\times 1.90=231.36 J/K

Hence, the value of \Delta S^o for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

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