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3 years ago

2.714 mole of helium has a mass of ______ grams. (Enter just the number for this one.)

Chemistry

1 answer:

kifflom [539]3 years ago

6 0

Hello!

To find the mass of helium, we need to multiply the total moles by the mass of helium. We are given 2.714 moles of helium, and the mass of helium is about 4.00 grams. Now, we multiply the two values together to get the grams.

2.714 moles x 4.00 grams = 10.856 grams

According to the number of significant figures, 2.714 moles of helium has a mass of 10.9 grams (exact value: 10.856 grams).

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Answer :

(1) The frequency of photon is, $3\times 10^{10}Hz$

(2) The energy of a single photon of this radiation is $1.988\times 10^{-23}J/photon$

(3) The energy of an Avogadro's number of photons of this radiation is, 11.97 J/mol

Explanation : Given,

Wavelength of photon = $1.0cm=0.01m$ (1 m = 100 cm)

(1) Now we have to calculate the frequency of photon.

Formula used :

$\nu=\frac{c}{\lambda}$

where,

$\nu$ = frequency of photon

$\lambda$ = wavelength of photon

c = speed of light = $3\times 10^8m/s$

Now put all the given values in the above formula, we get:

$\nu=\frac{3\times 10^8m/s}{0.01m}$

$\nu=3\times 10^{10}s^{-1}=3\times 10^{10}Hz$ $(1Hz=1s^{-1})$

The frequency of photon is, $3\times 10^{10}Hz$

(2) Now we have to calculate the energy of photon.

Formula used :

$E=h\times \nu$

where,

$\nu$ = frequency of photon

h = Planck's constant = $6.626\times 10^{-34}Js$

Now put all the given values in the above formula, we get:

$E=(6.626\times 10^{-34}Js)\times (3\times 10^{10}s^{-1})$

$E=1.988\times 10^{-23}J/photon$

The energy of a single photon of this radiation is $1.988\times 10^{-23}J/photon$

(3) Now we have to calculate the energy in J/mol.

$E=1.988\times 10^{-23}J/photon$

$E=(1.988\times 10^{-23}J/photon)\times (6.022\times 10^{23}photon/mol)$

$E=11.97J/mol$

The energy of an Avogadro's number of photons of this radiation is, 11.97 J/mol

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