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antiseptic1488 [7]
4 years ago
6

PLEASSEE HELP I WILL GIVE 20 POINTS no wrong answers pllzz

Biology
2 answers:
finlep [7]4 years ago
7 0

The given temperature of T=50 ∘C and the given pressure of p=92.5 mmHg approximately correspond to equilibrium conditions for liquid water and steam. (Strictly speaking, the corresponding saturation pressure for a given temperature of T=50.000 ∘C is p=92.647 mmHg, which could be rounded to p=92.6 mmHg and not p=92.5 mmHg; in the opposite direction, however, the corresponding saturation temperature for a given pressure of p=92.500 mmHg is T=49.968 ∘C, which could be rounded to T=50 ∘C.) The question does not explain whether these values describe the initial or final state; however, since the values correspond to equilibrium conditions, we may assume that these values apply to the final state when equilibrium is established (e.g. this can be achieved by keeping the closed system at a constant temperature of T=50 ∘C). Strictly speaking, the question doesn’t even explain whether the added water initially is liquid or steam; however, this is not relevant for the final state. If the water is added as liquid, a part of it evaporates until the equilibrium state is reached in the closed container; if the water is introduced as steam, a part of it condenses until the equilibrium state is reached. Therefore, when the defined equilibrium is established, the container contains certain amounts of liquid water and steam irrespective of the initial conditions. The question does not mention any air in the container. For simplicity’s sake, we may assume that the container has been evacuated before the experiment and contains only liquid water and vapour. The available volume of the container is reduced by the volume of the liquid water. However, since the density of liquid water at the given temperature and pressure is about ρ=988 kg m−3, the given mass of m=1.00 g corresponds to a maximum volume of Vliquid=1.01 ml=0.00101 l, assuming that no water has evaporated. Considering that the total volume is given as V=5.00 l (note the number of significant digits), the difference caused by the liquid water is not significant and may be neglected. Note that the use of the non-SI unit “conventional millimetre of mercury” (unit symbol: mmHg) is deprecated; the use of SI units is to be preferred.

Inessa05 [86]4 years ago
5 0

A. 13.5 grams is the answer.

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