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vesna_86 [32]
3 years ago
14

You have two sealed jars of water at the same temperature. in the first jar there is a large amount of water. in the second jar

there is a small amount of water. using 3 -4 sentences explain how the vapor pressure of water in the first jar compares with the vapor pressure of water in the second jar.
Chemistry
1 answer:
n200080 [17]3 years ago
4 0
The vapor pressure of the jar with a large amount of water and the jar with a small amount of water is the same. Vapor pressure is an intrinsic property. This means that this property is not dependent on the amount of the substance. Vapor pressure is dependent only on temperature and since the temperature is the same for both jars, their vapor pressures are also the same.
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stira [4]

Answer:

The answer is 5.7 minutes

Explanation:

A first-order reaction follow the law of Ln [A] = -k.t + Ln [A]_{0}. Where <em>[A]</em> is the concentration of the reactant at any <em>t</em> time of the reaction, [A]_{0} is the concentration of the reactant at the beginning of the reaction and <em>k</em> is the rate constant.

Dropping the concentration of the reactant to 6.25% means the concentration of A at the end of the reaction has to be [A]=\frac{6.25}{100}.[A]_{0}. And the rate constant (<em>k</em>) is 8.10×10−3 s−1

Replacing the equation of the law:

Ln \frac{6.25}{100}.[A]_{0} = -8.10x10^{-3}s^{-1}.t + Ln[A]_{0}

Clearing the equation:

Ln [A]_{0}.\frac{6.25}{100} - Ln [A]_{0} = -8.10x10^{-3}s^{-1}.t

<em>Considering the property of logarithms: </em>Ln A - Ln B = Ln \frac{A}{B}

Using the property:

Ln \frac{[A]_{0}}{[A]_{0}}.\frac{6.25}{100} = -8.10x10^{-3}s^{-1}.t

Clearing <em>t </em>and solving:

t = \frac{Ln \frac{6.25}{100} }{-8.10x10^{-3}s^{-1} } = 342.3s

The answer is in the unit of seconds, but every minute contains 60 seconds, converting the units:

342.3x\frac{1min}{60s} = 5.7min

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A block of ice has edge lengths of 8.00 cm each and a mass of 476
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