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Ganezh [65]
3 years ago
5

An ideal gas is kept in a 10​-liter ​[L] container at a pressure of 2.5 atmospheres​ [atm] and a temperature of 310 kelvin​ [K].

If the gas is compressed until its pressure is raised to 5 atmospheres​ [atm] while holding the temperature​ constant, what is the new volume in units of liters​ [L]?
Chemistry
1 answer:
Hatshy [7]3 years ago
8 0

Answer:

5 L.

Explanation:

From the question given above, the following data were obtained:

Initial volume (V1) = 10 L

Initial pressure (P1) = 2.5 atm

Final pressure (P2) = 5 atm

Final volume (V2) =.?

Since the temperature is constant, we shall apply the Boyle's law equation to determine the new volume of the gas. This can be obtained as follow:

P1V1 = P2V2

2.5 × 10 = 5 × V2

25 = 5 × V2

Divide both side by 5

V2 =25/5

V2 = 5 L

Thus, the new volume of the gas is 5 L

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Explanation:

The number of electron pairs is 4 that means the hybridization will be sp^3 but as there are three bonding domains and one nonbonding domain, thus electronic geometry is tetrahedral and the molecular geometry will be trigonal pyramidal.

Linear electron geometry is possible when number of electron pairs is 2 and the hybridization will be sp^2.

Trigonal planar geometry is possible when number of electron pairs is 3 and the hybridization will be sp^3.

Trigonal bipyramidal geometry is possible when number of electron pairs is 5 and the hybridization will be sp^3d.

Octahedral geometry is possible when number of electron pairs is 6 and the hybridization will be sp^3d^2.

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B) Quelle est la masse de tétraoxyde de trifer (Fe3O4) produite si 3,60 moles de trioxyde de
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Answer:

626,4 g de Fe₃O₄

Explanation:

Nous commencerons par écrire l'équation équilibrée de la réaction entre le fer (Fe) et le trioxyde d'aluminium. Ceci est donné ci-dessous:

9Fe + 4Al₂O₃ -> 3Fe₃O₄ + 8Al

De l'équation équilibrée ci-dessus,

4 moles d'Al₂O₃ ont réagi pour produire 3 moles de Fe₃O₄.

Par conséquent, 3,6 moles d'Al₂O₃ réagiront pour produire = (3,6 × 3) / 4 = 2,7 moles de Fe₃O₄

Ainsi, 2,7 moles de Fe₃O₄ sont produites à partir de la réaction.

Enfin, nous déterminerons la masse massique de Fe₃O₄ produite par la réaction. Ceci peut être obtenu comme suit:

Mole de Fe₃O₄ = 2,7 moles

Masse molaire de Fe₃O₄ = (3 × 56) + (4 × 16)

= 168 + 64

= 232 g / mol

Masse de Fe₃O₄ =?

Mole = masse / masse molaire

2,7 = Masse de Fe₃O₄ / 232

Croiser multiplier

Masse de Fe₃O₄ = 2,7 × 232

Masse de Fe₃O₄ = 626,4 g

Par conséquent, 626,4 g de Fe₃O₄ sont produits à partir de la réaction

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