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Gemiola [76]
3 years ago
14

Compare and contrast a cation and an anion.

Chemistry
2 answers:
lubasha [3.4K]3 years ago
5 0
Those with a negative charge are called anions<span> and those with a positive charge are called </span>cations. 
ipn [44]3 years ago
4 0
Anion, come from atoms, or molecules that have increased or decreased one or more valence electrons. Which gives them a positive or a negative charge. Cation, come from a negative charge. :)
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What would happen if you tried to cut a sheet of aluminum foil in half over and over again?
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Answer:

Its properties remain unchanged as it is undergoing physical changes.

Explanation:

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In this case, since aluminum foil is made up of aluminum leaves having a 0.2-mm thick it is widely known that is deformed when wrinkled due to its tiny average thick. Moreover, when cut half over and over again smaller pieces are obtained every time it is done. Nevertheless, even when its shape and size can be modified via physical actions, its properties remain unchanged because cutting is a physical change which is not affecting its chemical composition.

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Also called superposition. When overlapping waves produce a wave with an amplitude that is the sum of the individual waves. When overlapping waves produce a wave with an amplitude that is less than the sum of the individual waves.

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A quantity of 0.0250 mol of a gas initially at 0.050 L and 19.0°C undergoes a constant-temperature expansion against a constant
KiRa [710]

Answer:

V_2=2.995L\\\\W=248.5J

Explanation:

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In this case, for us to compute the final volume we apply the Boyle's law that analyzes the pressure-volume temperature as an inversely proportional relationship:

P_1V_1=P_2V_2

So we solve for V_2 by firstly computing the initial pressure:

P_1=\frac{nRT}{V_1}=\frac{0.025mol*0.082\frac{atm*L}{mol*K}*(19+273.15)K}{0.050L}  =11.98atm

V_2=\frac{P_1V_1}{P_2}=\frac{11.98atm*0.050L}{0.200atm}\\ \\V_2=2.995L

Finally, we can compute the work by using the following formula:

W=nRTln(\frac{V_2}{V_1} )=0.025mol*8.314\frac{J}{mol*K}*(19.0+273.15)K*ln(\frac{2.995L}{0.050L}) \\\\W=248.5J

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4 years ago
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