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Talja [164]
4 years ago
10

A colloidal liquid suspended in a gas is called an aerosol. true or false

Chemistry
2 answers:
iragen [17]4 years ago
6 0

Aerosol:

Aerosol is a colloidal dispersion of liquid or solid in gas.

Thus a colloidal liquid suspended in a gas is called an aero sol is true  

Aerosol is an example of colloidal dispersion and when a beam of light is fall on colloids then light is scattered means light is spread in all direction. It is a characteristic property of collides.  


ella [17]4 years ago
6 0
I think the correct answer is true. A colloidal liquid suspended in a gas is called an aerosol. It can be directly emitted into the atmosphere or they can be formed in the atmosphere by condensation. Hope this answers the question. Have a nice day.
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I need help with my chemistry homework. SHOW WORK
Veronika [31]

1) The gold will occupy a volume of 2.99 dm³.


\text{Volume = 17 200 g} \times \frac{\text{1 cm}^{3}}{\text{5.75 g}} = \text{2990 cm}^{3} = \textbf{2.99 dm}^{3}\\

2) The density of the olive oil is 0.850 g/cm³.

\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{\text{425 g}}{\text{500 cm}^{3}}= \textbf{0.850 g/cm}^{3}

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3 0
3 years ago
Part A - Calculate the molar solubility in water
Andrei [34K]
Part 1)

when the balanced equation for this reaction is:

and by using ICE table

                Mg(OH)2(s) ↔ Mg2+(aq) + 2OH-(aq)

initial                                     0                0

change                                +X               +2X

Equ                                        X                 2X

 When KSp expression = [Mg2+][OH-]

when we have KSp = 5.61 x 10^-11

and when we assumed [Mg2+] = X

and [ OH-] = (2X)^2

when we assume X is the value of molar solubility of Mg(OH)2
 
so, by substitution:

5.61 x 10^-11 = 4X^3

∴ X = 2.4 x 10^-4 M

∴ molar solubility of Mg(OH)2 = X = 2.4 x 10^-4 M 

Part 2) 

the molar solubility of Mg(OH)2 in 0.16 m NaOH we assumed it = X

by using the ICE table:

                Mg(OH)2(s) → Mg2+(aq)  +  2OH-

initial                                   0                  0.16m

change                               +X                +2X

equ                                        X                (0.16+2x)

when Ksp = [mg2+][OH-]^2

5.61 x 10^-11 = X * (0.16+2X)^2 by solving for X

∴ X = 1.3 x 10^-5 M

∴ the molar solubility = X = 1.3 x 10^-5 M          



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4 years ago
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Explanation:

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