Chemistry, 25.10.2016 01:48 bfell92

Suppose you had a mixture of sand and small, hollow beads. how might you separate the mixture?

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Answer from: Quest

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Answer from: Quest

answer:

see below

step-by-step explanation:

1. colloid

b) microscopic insoluble particles that are suspended throughout another substance.

milk is a colloid. it consists of small fat globules suspended in water.

2. emulsion

d) two or more liquids that are not normally mixable.

mayonnaise is an emulsion. it consists of oil, egg yolks, vinegar, and seasonings. normally, the oil and vinegar (95 % water) do not mix. without the egg yolk , they would quickly separate. egg yolk contains the emulsifier, lecithin, which is attracted to both oil and water and stabilizes he mixture.

3. solution

a) a solute that is evenly distributed within the solvent.

sugar syrup is a solution of sugar in water.

4)homogeneous

c) a mixture that has its components distributed evenly within the substance.

a homogeneous mixture has only one phase. for example, the sugar syrup is homogeneous, because, even with a powerful microscope, you can't see a boundary between the sugar particles and the water.

Match the mixture with the substance// i really need on this guys (it’s a pic btw)

Answer from: Quest

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: 1 - α α α = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

$k_{c}$ = $\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)v }$

or, $k_{c}$ = $\frac{\alpha ^{2} }{(1-\alpha) }$ (at v = 1l)

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: (1 - α)/0.5 α/0.5 α/0.5 = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

$\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)0.5 }$

$\frac{\alpha ^{2} }{(1-\alpha) }$ = $\frac{\alpha ^{2} }{(1-\alpha) 0.5}$

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

Answer from: Quest

tin

metals

group 4a, period 5

atomic mass 118.71

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