Chemistry, 03.08.2020 14:01 GiaTeyy6536

Define the term “Tyndall effect”. Which of the following would show Tyndall effect:- (Starch solution, sodium chloride solution, alcohol and water mixture)

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Answer from: gwendallinesikes

Tyndall Effect is the phenomenon of scattering of light by particles in a colloid or in a very fine suspension.

Starch solution would show Tyndall effect.

N/B: Heterogeneous mixtures show Tyndall effect.

Answer from: alexahrnandez4678

The Tyndall effect is light scattering by particles in a colloid or in a very fine suspension. Also known as Willis–Tyndall scattering, it is similar to Rayleigh scattering, in that the intensity of the scattered light is inversely proportional to the fourth power of the wavelength, so blue light is scattered much more strongly than red light. An example in everyday life is the blue colour sometimes seen in the smoke emitted by motorcycles, in particular two-stroke machines where the burnt engine oil provides these particles.

Under the Tyndall effect, the longer wavelengths are more transmitted while the shorter wavelengths are more diffusely reflected via scattering. The Tyndall effect is seen when light-scattering particulate matter is dispersed in an otherwise light-transmitting medium, when the diameter of an individual particle is the range of roughly between 40 and 900 nm, i.e. somewhat below or near the wavelengths of visible light (400–750 nm).

It is particularly applicable to colloidal mixtures and fine suspensions; for example, the Tyndall effect is used in nephelometers to determine the size and density of particles in aerosols and other colloidal matter (see ultramicroscope and turbidimeter).

It is named after the 19th-century physicist John Tyndall.

Explanation:

Answer from: Quest

Forgot the question hahaha

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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