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raketka [301]
3 years ago
10

Regarding water molecules, adhesion is best described as * the attraction water molecules have to other water molecules the attr

action water molecules have to ionic substances the attraction water molecules have to polar substances the attraction water molecules have to other surfaces
Chemistry
1 answer:
Helen [10]3 years ago
3 0

Answer:

the attraction water molecules have to other surfaces

Explanation:

Adhesion is defined as  the attractive forces between unlike substances, e.g water moving up a capillary tube.

Adhesion is the tendency of dissimilar particles or surfaces to cling to one another(Wikipedia).

So, what we mean by adhesion in this context, is the attraction water molecules to other surfaces.

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r = -kAB

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\frac{-1}{[100-2x]}= -kt -\frac{1}{100}\\\\or \\\\ \frac{1}{(100-2x)} = kt + \frac{1}{100}

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Insert the above value x into \frac{1}{(100-2x)} equation = kt + \frac{1}{100} to get k.  

\to  \frac{1}{(100-2\times \frac{10}{3})}  = k \times (7) + \frac{1}{100} \\\\ \to  \frac{1}{(100- 2 \times 3.33)} =   \frac{700k + 1}{100} \\\\ \to  \frac{1}{(100-6.66)} = \frac{700k + 1}{100}\\\\ \to \frac{1}{93.34} = \frac{700k + 1}{100} \\\\

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\to \frac{1}{(100-2x)} = kt +\frac{1}{100} \\\\\to \frac{1}{(100-2x)} = -0.0091t + \frac{1}{100}\\\\\to \frac{1}{(100-2x)} =  \frac{1 -0.91t}{100}\\\\\to \frac{1}{2(50-x)} =  \frac{1 -0.91t}{100}\\\\\to \frac{1}{(50-x)} =  \frac{1 -0.91t}{50}\\\\\to 50= (1-0.91t)(50-x)\\\\\to 50 = 50-45.5t-x-0.91tx\\\\\to x+0.91xt= -45.5t\\\\\to x(1+0.91t)= -45.5t\\\\\to x=\frac{-45.5t}{1+0.91t}

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y =3 \times \frac{-45.5t}{1+0.91t}

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