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gavmur [86]
3 years ago
15

A jar is filled with water, and the lid is closed tightly. The jar is kept at constant temperature. What happens at the surface

of the water?
A.
Water molecules evaporate and condense at the same rate.
B.
Water molecules evaporate and condense, but evaporation occurs more quickly.
C.
Water molecules evaporate only and do not condense.
D.
Water molecules condense only and do not evaporate.
Chemistry
1 answer:
Katarina [22]3 years ago
5 0
<h2>Hello!</h2>

The correct answer is A: Water molecules evaporate and condense at the same rate.

<h2>Why?</h2>

Evaporation is defined as the physical change from liquid to gas, and Condensation is the physical change from gas to liquid.

At any given temperature, these two processes occur at once, in a dynamic equilibrium.

When the lid is closed, evaporation occurs faster than condensation, and pressure increases. Then, when pressure reaches a critical value, condensation starts to occur faster than evaporation, until an equilibrium is reached, and the pressure of the water molecules in the gas phase is maximum for that temperature. The pressure at that point is called Vapor Pressure.

Have a nice day!

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration of BC_2 that should used originally is C_Z_o = 0.4492M

Explanation:

     From the question we are told that

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          The  time taken for sixth of 0.5 M of reactant to react t = 9 hr

           The time available is t_a = 3.5 hr

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Generally the elementary rate  law is mathematically as

                    -r_Z = kC_Z^2 = - \frac{d C_Z}{dt}

Where k is the rate constant ,  C_Z is the concentration of Z

From the elementary rate law we see that the reaction is second order (This because the concentration of the reactant is raised to power 2 )

 For second  order reaction

            \frac{1}{C_Z}  - \frac{1}{C_Z_o}  = kt

Where C_Z_o is the initial concentration of Z which a value of   C_Z_o = 0.5M

       From the question we are told that it take  9 hours  for the concentration of  the reactant to become

                 C_Z =  C_Z_o - \frac{1}{6}  C_Z_o

                  C_Z = 0.5  - \frac{0.5}{6}

                       = 0.4167 M

So      

                     \frac{1}{0.4167}  - \frac{1}{0.50}  =  9 k

                          0.400 = 9 k

                =>    k = 0.044\  L/ mol \cdot hr^{-1}

  For   C_Z = 0.42M

                \frac{1}{0.42} - \frac{1}{C_Z_o}  = 3.5 * 0.044

                2.38 -  0.154  =    \frac{1}{C_Z_o}

                           2.226  =    \frac{1}{C_Z_o}

                            C_Z_o = \frac{1}{2.226}

                             C_Z_o = 0.4492M

                       

           

             

         

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