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2 years ago

6

You can obtain a rough estimate of the size of a molecule with the following simple experiment: Let a droplet of oil spread out

on a fairly large but smooth water surface. The resulting "oil slick" that forms on the surface of the water will be approximately one molecule thick. Given an oil droplet with a mass of 9.00 × 10−7 kg and a density of 918 kg/m3 that spreads out to form a circle with a radius of 41.8 cm on the water surface, what is the approximate diameter of an oil molecule?

1 answer:

mina [271]2 years ago

6 0

Answer:

The diameter of the oil molecule is $4.4674\times 10^{-8} cm$ .

Explanation:

Mass of the oil drop = $m=9.00\times 10^{-7} kg$

Density of the oil drop = $d=918 kg/m^3$

Volume of the oil drop: v

$d=\frac{m}{v}$

$v=\frac{m}{d}=\frac{9.00\times 10^{-7} kg}{918 kg/m^3}$

Thickness of the oil drop is 1 molecule thick.So, let the thickness of the drop or diameter of the molecule be x.

Radius of the oil drop on the water surface,r = 41.8 cm = 0.418 m

1 cm = 0.01 m

Surface of the sphere is given as: a = $4\pi r^2$

$a=4\times 3.14\times (0.418 m)^2=2.1945 m^2$

Volume of the oil drop = v = Area × thickness

$\frac{9.00\times 10^{-7} kg}{918 kg/m^3}=2.1945 m^2\times x$

$x= 4.4674\times 10^{-10} m= 4.4674\times 10^{-8} cm$

The thickness of the oil drop is $4.4674\times 10^{-8} cm$ and so is the diameter of the molecule.

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2 years ago

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sertanlavr [38]

Answer:

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We have to find out the molar mass of each pesticide to calculate the moles, and then the milimoles

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3.51×10⁻⁶ . 1000 = 3.51×10⁻³ mmoles of C₆H₆Cl₆

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2 years ago

Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produces

Alchen [17]

The question is incomplete, here is the complete question:

Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produces only nitrogen and water vapor, and in the liquid form it is easily transported. An industrial chemist studying this reaction fills a 5.0 L flask with 2.2 atm of ammonia gas and 2.4 atm of oxygen gas at 44.0°C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of nitrogen gas to be 0.99 atm.

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<u>Answer:</u> The pressure equilibrium constant for the reaction is 32908.46

<u>Explanation:</u>

We are given

Initial partial pressure of ammonia = 2.2 atm

Initial partial pressure of oxygen gas = 2.4 atm

Equilibrium partial pressure of nitrogen gas = 0.99 atm

The chemical equation for the reaction of ammonia and oxygen gas follows:

$4NH_3(g)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(g)$

<u>Initial:</u> 2.2 2.4

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Evaluating the value of 'x':

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Equilibrium partial pressure of water vapor = 6x = (6 × 0.495) = 1.98 atm

The expression of $K_p$ for above equation follows:

$K_p=\frac{(p_{N_2})^2\times (p_{H_2O})^6}{(p_{NH_3})^4\times (p_{O_2})^3}$

Putting values in above equation, we get:

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Hence, the pressure equilibrium constant for the reaction is 32908.46

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2 years ago

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Jlenok [28]

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2 years ago

Read 2 more answers