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Ulleksa [173]
2 years ago
11

The thermostat in a refrigerator filled with cans of soft drinks malfunctions and the temperature of the refrigerator drops belo

w zero celcius. The contents of the cans of diet soft drinks freeze, rupturing many of the cans and causing an awful mess. However, none of the cans containing regular non diet soft drinks rupture. Select the statement or statements below that best describe this behavior.A. As the temperature drops the solubility of the dissolved carbon dioxide gas decreases in the diet soft drinks. The pressure caused by this released gas builds up and finally ruptures the can.B.Water expands on freezing. Since water is the principle ingredient in soft drinks when the soft drink freezes it will rupture the can.C. There is more water in diet soft drink so they will freeze point of the solution sufficiently so that the solution does not freeze.D. Diet soft drink are inherently messier than non diet soft drinks.
Chemistry
2 answers:
butalik [34]2 years ago
5 0

Answer:

B.

Explanation:

Water expands on freezing and there is more water in the diet drinks than in the other drinks.

AleksandrR [38]2 years ago
4 0

Answer:

The sugar in the non-diet soft drinks decreases the freezing point of the solution sufficiently so that the solution does not freeze.

Explanation:

Hello,

None of the options is the correct one as long as this is a case in which the presence-absence of a solute modifies the freezing point. It is seen that regular non-diet soft drinks have a higher concentration of sugar which decreases the normal freezing point, that is why they do not freeze in this situation, however, the absence of sugar normalizes the freezing point, causing the dietetic cans of diet soft drinks to rupture due to the freezing of the drink. Therefore the answer is the sugar in the non-diet soft drinks decreases the freezing point of the solution sufficiently so that the solution does not freeze.

Best regards.

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P\cdot V = n\cdot R\cdot T,

where

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The question is asking for the final volume V of the gas. Rearrange the ideal gas equation for volume:

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V_1 = V_0 \cdot \dfrac{T_1}{T_2} = 5.57\times 10^{4}\;\text{L}\times \dfrac{258.65\;\textbf{K}}{305.15\;\textbf{K}} = 4.72122\times 10^{4}\;\text{L}.

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The volume of the gas is proportional to the reciprocal of its absolute temperature \dfrac{1}{T} if both n and T stays constant. In other words,

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(3 sig. fig. as in the question.).

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