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Nutka1998 [239]
2 years ago
7

How many molecules of CH4 are in 24.1 g of this compound?

Chemistry
1 answer:
jok3333 [9.3K]2 years ago
7 0

<u>1) find molar mass of CH₄</u>

=12.01+4(1.01)\\\\=16.05\ g/mol

<u>2) determine moles of CH₄</u>

=24.1g\ CH_{4}\ x\ \frac{1\ mol\ CH_{4}}{16.05g\ CH_{4}}\\\\=1.501557632\ mol\ CH_{4}

<u>3) multiply by avogadro's number</u>

=1.501557632\ mol\ CH_{4}\ x\ \frac{6.022\ x\ 10^{23}\ molecules\ CH_{4}}{1\ mol\ CH_{4}} \\\\=9.04238006\ x\ 10^{23}\\\\=9.04\ x\ 10^{23}\ molecules\ CH_{4}

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The initial pressure of a gas is 1.58 atm and occupies 1.76 L of space at constant temperature. This gas is compressed so that i
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Answer:

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3 years ago
How many moles at STP would take up 44.8 liters?
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Answer:

n =  2 mol

Explanation:

Given data:

Pressure = standard = 1 atm

Temperature = standard = 273.15 K

Volume = 44.8 L

Number of moles = ?

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

1 atm × 44.8 L = n × 0.0821 atm.L/ mol.K  × 273.15 K

44.8 atm.L = n × 22.43 atm.L/ mol

n = 44.8 atm.L /   22.43 atm.L/ mol

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4 0
3 years ago
Given the wavelength of the corresponding emission line, calculate the equivalent radiated energy from n = 4 to n = 2 in both jo
lions [1.4K]

Explanation:

It is given that,

Initial orbit of electrons, n_i=4

Final orbit of electrons, n_f=2

We need to find energy, wavelength and frequency of the wave.

When atom make transition from one orbit to another, the energy of wave is given by :

E=-13.6(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})

Putting all the values we get :

E=-13.6(\dfrac{1}{(4)^2}-\dfrac{1}{(2)^2})\\\\E=2.55\ eV

We know that : 1\ eV=1.6\times 10^{-19}\ J

So,

E=2.55\times 1.6\times 10^{-19}\\\\E=4.08\times 10^{-19}\ J

Energy of wave in terms of frequency is given by :

E=hf

f=\dfrac{E}{h}\\\\f=\dfrac{4.08\times 10^{-19}}{6.63\times 10^{-34}}\\\\f=6.14\times 10^{14}\ Hz

Also, c=f\lambda

\lambda is wavelength

So,

\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{6.14\times 10^{14}}\\\\\lambda=4.88\times 10^{-7}\ m\\\\\lambda=488\ nm

Hence, this is the required solution.

4 0
3 years ago
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