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Nady [450]
3 years ago
13

a certain skin lotion is a fine mixture of water and various oils. this lotion is cloudy and cannot be separated into oil and wa

ter by filtration. moreover, it’s compon do not separate when left standing. what type of mixture is it?
Chemistry
1 answer:
PolarNik [594]3 years ago
3 0

Answer: The skin lotion is a colloidal mixture of water and various oils.

Explanation:

Mixture: Blend of of two or more components which are chemically unreactive to give any product.

Mixture are defined as  of three types:

Solutions : Mixture in which one component is completely distributed in an other component resulting in clear mixture. They cannot be separated into its components on standing or by filtration.

Suspensions: Mixture in which one component is not completely or uniformly distributed in another component.They separate out into layers when kept undisturbed.

Colloids: Mixture in which one component is uniformly distributed in another component but the mixture formed is not clear. The color of mixture appears turbid or cloudy. They cannot get separated by filtration. They also cannot get separated into its components when left standing.

A skin lotion is a fine mixture of water and oils. it is a cloudy mixture and cannot be separated with help of filtration. Also its components also not separates out on standing. All these properties shown by skin lotion is of colloidal mixture.


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Answer:

(a) r = 6.26 * 10⁻⁷cm

(b) r₂ = 6.05 * 10⁻⁷cm

Explanation:

Using the sedimentation coefficient formula;

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s = M ( 1 - Vρ) / N*6πnr

making r sbjct of formula, r =  M (1 - Vρ) / N*6πnrs

Note: S = 10⁻¹³ sec, 1 KDalton = 1 *10³ g/mol, I cP = 0.01 g/cm/s

r = {(3.1 * 10⁵ g/mol)(1 - (0.732 cm³/g)(1 g/cm³)} / { (6.02 * 10²³)(6π)(0.01 g/cm/s)(11.7 * 10⁻¹³ sec)

r = 6.26 * 10⁻⁷cm

b. Using the formula r₂/r₁ = s₁/s₂

s₂ = 0.035 + 1s₁ = 1.035s₁

making r₂ subject of formula; r₂ = (s₁ * r₁) / s₂ = (s₁ * r₁) / 1.035s₁

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a) The concentration in ppm (mg/L) is 5.3 downstream the release point.

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The first step is to convert Million Gallons per Day (MGD) to Liters per day (L/d). In that sense, it is possible to calculate with data given previously in the problem.  

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We have one flow of wastewater released into a stream.  

First flow is F1 =5 MGD with a concentration of C1 =10.0 mg/L.

Second flow is F2 =10 MGD with a concentration of C2 =3.0 mg/L.  

After both of them are mixed, the final concentration will be between 3.0 and 10.0 mg/L. To calculate the final concentration, we can calculate the mass of pollutant in total, adding first and Second flow pollutant, and dividing in total flow. Total flow is the sum of first and second flow. It is shown in the following expression:  

C_f = \frac{F1*C1 +F2*C2}{F1 +F2}

Replacing every value in L/d and mg/L

C_f = \frac{18927059 L/d*10.0 mg/L +37854118 L/d*10.0 mg/L}{18927059 L/d +37854118 L/d}\\C_f = \frac{302832944 mg/d}{56781177 L/d} \\C_f = 5.3 mg/L

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