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Pie
3 years ago
10

explain why particles of colloidal solution donot settle down when left undisturbed , while in the case of suspensions they do .

?​
Chemistry
1 answer:
Leni [432]3 years ago
7 0

Answer:

the partial of colloidal are smaller and they are not heavy whereas the partial of suspension are larger but they are heavy and less movement thus they settle down due to gravity

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What value of ℓℓ is not allowed for an electron in an n = 3 shell?
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Every electron shell, designated by "n", has a specific number of electron subshells. The shell number has values equivalent to natural numbers (1, 2, 3, 4...), while the subshells are s, p, d and f.

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4 years ago
Calculate the specific volume of Helium using the compressibility factor. 500 kPa, 60 degrees Celsius
IrinaK [193]

Explanation:

Formula for compressibility factor is as follows.

                     z = \frac{P \times V_{m}}{R \times T}

where,     z = compressibility factor for helium = 1.0005

               P = pressure

          V_{m} = molar volume

                R = gas constant = 8.31 J/mol.K

                T = temperature

So, calculate the molar volume as follows.

                V_{m} = \frac{z \times R \times T}{P}

                             = \frac{1.0005 \times 8.314 \times 10^{-3} m^{3}.kPa/mol K \times (60 + 273)K}{500 kPa}

                             = 0.0056 m^{3}/mol

As molar mass of helium is 4 g/mol. Hence, calculate specific volume of helium as follows.

                    V_{sp} = \frac{V_{m}}{M_{w}}

                           = \frac{0.0056 m^{3}/mol}{4 g/mol}

                           = 0.00139 m^{3}/g

                           = 0.00139 m^{3}/g \times \frac{1 g}{10^{-3}kg}

                                 = 1.39 m^{3}/kg

Thus, we can conclude that the specific volume of Helium in given conditions is 1.39 m^{3}/kg.

7 0
3 years ago
Aluminum reacts with chlorine gas to form aluminum chloride
VikaD [51]

Answer: 48.95g

Explanation:

no. of moles of Cl2 = 39/(2*35.5) = 0.55 mol

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hence, aluminium is in excess so we'll do calculation using no. of moles of Cl2 as it will be the only reactant to be used up completely. So,

no of moles of AlCl3 = 2/3 * (0.55) = 0.367 mol

hence amount of AlCl3 = 0.367 * (27+3*35.5) = 48.95g

8 0
3 years ago
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