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RoseWind [281]
3 years ago
9

When a soda can is dropped, it should not be immediately opened. Why?

Chemistry
1 answer:
Kipish [7]3 years ago
7 0
<span>The pressure inside a coke bottle is really high. This helps keep the soda carbonated. That is, the additional pressure at the surface of the liquid inside the bottle forces the bubbles to stay dissolved within the soda. </span><span>When the coke is opened, there is suddenly a great pressure differential. The initial loud hiss that is heard is this pressure differential equalizing itself. All of the additional pressure found within the bottle pushes gas out of the bottle until the pressure inside the bottle is the same as the pressure outside the bottle. </span><span>However, once this occurs, the pressure inside the bottle is much lower and the gas bubbles that had previously been dissolved into the soda have nothing holding them in the liquid anymore so they start rising out of the liquid. As they reach the surface, they pop and force small explosions of soda. These explosions are the source of the popping and hissing that continues while the soda is opened to the outside air. Of course, after a while, the soda will become "flat" when the only gas left dissolved in the liquid will be the gas that is held back by the relatively weak atmospheric pressure.</span>
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Explanation:

The given data is as follows.

    Concentration = 0.1 mol/dm^{3}

                             = 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

                             = 6.022 \times 10^{25} ions/m^{3}

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Formula for electric double layer thickness (\lambda_{D}) is as follows.

            \lambda_{D} = \frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}

where, n^{o} = concentration = 6.022 \times 10^{25} ions/m^{3}

Hence, putting the given values into the above equation as follows.

                 \lambda_{D} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}                    

                          = \sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}  

                         = 9.669 \times 10^{-10} m

or,                     = 9.7 A^{o}

                          = 1 nm (approx)

Also, it is known that \lambda_{D} = \sqrt \frac{1}{n^{o}}

Hence, we can conclude that addition of 0.1 mol/dm^{3} of KCl in 0.1 mol/dm^{3} of NaBr "\lambda_{D}" will decrease but not significantly.

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