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AURORKA [14]
3 years ago
6

How to calculate the number of atoms, molecules, or ions

Chemistry
1 answer:
Leni [432]3 years ago
8 0
For atoms use a periodic table. The atomic number is the amount of atoms. For molecules and ions look up things revolved around the avogrados principle
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A container has a mixture of NO2 gas and N2O4 gas in equilibrium. The chemical reaction between the two gases is described by th
kondaur [170]

Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄

Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.

For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:

Kp = \frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }

where:

P(N₂O₄) and P(NO₂) are the partial pressure of each gas.

Calculating constant:

Kp = \frac{38.8}{61.2^{2} }

Kp = 0.0104

After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.

P(N₂O₄) + P(NO₂) = 200

P(N₂O₄) = 200 - P(NO₂)

Kp = \frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }

0.0104 = \frac{200 - P(NO_{2})  }{[P(NO_{2} )]^{2}}

0.0104[P(NO_{2} )]^{2} + P(NO_{2} ) - 200 = 0

Resolving the second degree equation:

P(NO_{2} ) = \frac{-1+\sqrt{9.32} }{0.0208}

P(NO_{2} ) = 98.7

Find partial pressure of N₂O₄:

P(N₂O₄) = 200 - P(NO₂)

P(N₂O₄) = 200 - 98.7

P(N₂O₄) = 101.3

The partial pressures are P(NO_{2} ) = 98.7 MPa and P(N₂O₄) = 101.3 MPa

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3 years ago
Los elementos están ordenados en la tabla periódica de acuerdo a​
Mice21 [21]

Answer: WHat

Explanation:

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2 years ago
What is the quantity of atoms in KMnO4 ?
tankabanditka [31]

Answer:

Potassium permanganate has a molar mass of 158.04 g/mol. This figure is obtained by adding the individual molar masses of <em><u>four oxygen atoms</u></em>, <em><u>one manganese atom</u></em> and <em><u>one potassium atom</u></em>

Explanation:

6 0
2 years ago
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Please help me out!!!
patriot [66]

Answer:

Use a scientific calculator

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2 years ago
I need anyone to answer me.. ​
artcher [175]
A. 2,3
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