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xenn [34]
3 years ago
8

Identical wire loops are dipped into Liquid X and Liquid Y, so that a film of liquid forms across the loops (like the bubble sol

ution on a child's bubble blowing wand). The width of each loop is increased slowly and the forces FX and FY needed to make the loops 5% wider are measured.
a. FX will be greater than F Y
b. FX will be less than FY
c. FX will be equal to FY
d. It's impossible to predict whether F X or FY will be greater without more information.
Chemistry
1 answer:
muminat3 years ago
6 0

Answer:

a.

Explanation:

Assuming that Liquid X is considered to possess a greater viscosity as well as higher surface tension than liquid Y. Then, liquid X will tend to harbour more pressure inside the liquid.

In addition to that, the greater the surface tension, the greater the force required to expand the liquid's surface area.

This in turn makes the force required to make the loop 5% wider to be greater in FX rather than FY.

Thus, option a is the correct answer.

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Answer:

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The dissociation of calcium carbonate has an equilibrium constant of Kp= 1.20 at 800°C. CaCO3(s) ⇋ CaO(s) + CO2(g)
Fed [463]

Explanation:

(a)   Formula that shows relation between K_{c} and K_{p} is as follows.

                 K_c = K_p \times (RT)^{-\Delta n}

Here, \Delta n = 1

Putting the given values into the above formula as follows.

        K_c = K_p \times (RT)^{-\Delta n}

                  = 1.20 \times (RT)^{-1}

                  = \frac{1.20}{0.0820 \times 1073}

                  = 0.01316

(b) As the given reaction equation is as follows.

               CaCO_{3}(s) \rightleftharpoons CaO(s) + CO_{2}(g)

As there is only one gas so ,

                p[CO_{2}] = K_{p} = 1.20

Therefore, pressure of CO_{2} in the container is 1.20.

(c)   Now, expression for K_{c} for the given reaction equation is as follows.  

             K_{c} = \frac{[CaO][CO_{2}]}{[CaCO_{3}]}

                        = \frac{x \times x}{(a - x)}

                        = \frac{x^{2}}{(a - x)}[/tex]

where,    a = initial conc. of CaCO_{3}

                  = \frac{22.5}{100} \times 9.56

                  = 0.023 M

          0.0131 = \frac{x^{2}}{0.023 - x}

                  x = 0.017

Therefore, calculate the percentage of calcium carbonate remained as follows.

       % of CaCO_{3} remained = (\frac{0.017}{0.023}) \times 100

                                  = 75.46%

Thus, the percentage of calcium carbonate remained is 75.46%.

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