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Firlakuza [10]
4 years ago
10

A saturated solution of lead(II) nitrate was made at 50 °C using 100 g of water, and this was then allowed to cool to 30 °C. Pre

dict what would happen as the solution cooled?
Physics
2 answers:
olchik [2.2K]4 years ago
8 0
The solubility of the solution would decrease and some of the lead(II) nitrate would precipitate out 
Sunny_sXe [5.5K]4 years ago
4 0

Answer: The temperature of solution will decease along with the precipitation of solid lead nitrate in solution.

Explanation:

Solubility of the substance is directly effect by the temperature of the solvent.

This is because the kinetic energy of the solvent increase as temperature increase due to which solvent easily breaks apart the solute particles easily and faster which will result in solubility of more solute in solvent and vice versa.

So, when prepared saturated solution of lead nitrate at 50°C was allowed to cool down to 30°C one will observe that the precipitation of solid lead nitrate in solution. And this means that solubility of lead nitrate with decrease in temperature decreases as well.

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A Rankine oval is formed by combining a source-sink pair, each having a strength of 36 ft2/s and separated by a distance of 15 f
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Answer:

0.28 ft

Explanation:

We are given that

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Using the formula

\frac{l}{\frac{15}{2}}=(\frac{36}{\pi\times 12\times \frac{15}{2}}+1)^{\frac{1}{2}}

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l=0.14

Length of oval=2l=2(0.14)=0.28 ft

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The correct answer to your question and how to solve it is
 
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