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3 years ago

9

What factors affect water quality?In a typical drinking-water treatment process, ____________________ is when air is forced thro

ugh the water to reduce unpleasant odors and taste.

1 answer:

ki77a [65]3 years ago

3 0

Aeration is the answer to your question

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Chemical changes<br> involve the breaking and making of chemical bonds.<br><br> The answer is always

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Chemical reactions involve breaking chemical bonds between reactant molecules (particles) and forming new bonds between atoms in product particles (molecules). The number of atoms before and after the chemical change is the same but the number of molecules will change.

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3 years ago

start out as one kind of rock, but large amounts of pressure and heat change them into a different kind.

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Metamorphic

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3 years ago

Esters are generally pleasant smelling compounds<br>True<br>False

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True

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Esters are generally pleasantly smelling compounds. In fact, the fragrance industry uses esters to produce perfume, as well as uses esters as an ingredient to produce synthetic flavours and cosmetics, all of which have unique and pleasant smells.

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3 years ago

If the temperature of a liquid-vapor system at equilibrium increases, the new equilibrium condition will

nikdorinn [45]

If the temperature of a liquid-vapor system at equilibrium increases, it will shift towards the vapor phase, assuming that the pressure remains equal. The concentration of vapor will also increase relative to the concentration of liquid in the system. Thus, the new equilibrium condition will have more vapor than liquid.

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3 years ago

Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.

Ad libitum [116K]

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

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

T = $30^{o}C$ = (30 + 273) K = 303 K

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.

7 0

3 years ago