the answer is A......
it is supported by practical evidence and examples. this is the answer because he tried and tested many different ways to see what would happen so he is happy with the conclusion that what he tested is what he gets.
Electrical energy is the energy of electrons
The theoretical yield of NaBr given that 2.36 moles of FeBr₃ reacts is 7.08 moles
<h3>Balanced equation </h3>
2FeBr₃ + 3Na₂S → Fе₂S₃ + 6NaBr
From the balanced equation above,
2 moles FeBr₃ reacted to produce 6 moles of NaBr
<h3>How to determine the theoretical yield of NaBr</h3>
From the balanced equation above,
2 moles FeBr₃ reacted to produce 6 moles of NaBr
Therefore,
2.36 moles FeBr₃ will react to produce = (2.36 × 6) / 2 = 7.08 moles of NaBr
Therefore,
Thus, the theoretical yield of NaBr is 7.08 moles
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Answer:
It has to do with increasing the entropy of the universe.
Explanation:
The modern definition of entropy is that it is the extent to which a system is able to disperse its energy. Energy (such as heat!) likes to spread itself out, so that as many states as possible are occupied with the least amount of energy.
Nonane (b) has the highest melting point.
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A caveat: I'm assuming that we're dealing with the straight-chain isomers of these alkanes (specifically pentane and nonane). The straight-chain isomer of pentane (<em>n</em>-pentane, CH3-[CH2]3-CH3) has a melting point of -129.8 °C; the straight-chain isomer of nonane (<em>n-</em>nonane, CH3-[CH2]7-CH3) has a melting point of -53.5 °C. The pattern holds as you go down (or up): The more carbon atoms, the higher the melting point. So, in decreasing order of melting points here, you'd have the following: nonane > pentane > butane > ethane.
However, one structural isomer of pentane, neopentane, has a melting point of -16.4 °C, which is <em>higher </em>that the melting point of <em>n</em>-nonane despite neopentane having the same molecular formula as its straight-chain isomer. Of course, you're not to blame for coming up with this question; this is just some extra info to keep in mind.