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3 years ago

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The mass of a sulfur atom is 32.0 u, and the mass of a fluorine atom is 19.0 u. What is the mass of a mole of sulfur hexafluorid

e (UF.)

1 answer:

givi [52]3 years ago

5 0

<u>Answer:</u> The mass of sulfur hexafluoride is 146 u.

<u>Explanation:</u>

Mass of a mole of a substance is defined as the molar mass of that substance.

It is defined as the sum of every component each multiplied by the times they appear in that substance.

We are given:

Mass of sulfur atom = 32.0 u

Mass of fluorine atom = 19.0 u

The chemical formula for sulfur hexafluoride is $SF_6$

So, the molar mass of $SF_6=[32+(6\times 19)=146u$

Hence, the mass of sulfur hexafluoride is 146 u.

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Read 2 more answers

(a) How many fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit

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Answer:

a

The number of fringe is z = 3 fringes

b

The ratio is $I = 0.2545I_o$

Explanation:

a

From the question we are told that

The wavelength is $\lambda = 600 nm$

The distance between the slit is $d = 0.117mm = 0.117 *10^{-3} m$

The width of the slit is $a = 35.7 \mu m = 35.7 *10^{-6}m$

let z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is and this mathematically represented as

$z = \frac{d}{a}$

Substituting values

$z = \frac{0.117*10^{-3}}{35.7 *10^{-6}}$

z = 3 fringes

b

From the question we are told that the order of the bright fringe is n = 3

Generally the intensity of a pattern is mathematically represented as

$I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]$

Where $I_o$ is the intensity of the central fringe

And Generally $sin \theta = \frac{n \lambda }{d}$

$I = I_o co^2 [ \frac{\pi (\frac{n \lambda}{d} )}{\lambda} ] [\frac{\frac{sin (\pi a (\frac{n \lambda}{d} ))}{\lambda} }{\frac{\pi a (\frac{n \lambda}{d} )}{\lambda} } ]$

$I = I_o cos^2 (n \pi)[\frac{\frac{sin(\pi a (\frac{n \lambda}{d} ))}{\lambda} )}{ \frac{ \pi a (\frac{n \lambda }{d} )}{\lambda} } ]$

$I = I_o cos^2 (3 \pi) [\frac{sin (\frac{3 \pi }{6} )}{\frac{3 \pi}{6} } ]$

$I = I_o (1)(0.2545)$

$I = 0.2545I_o$

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