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3 years ago

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A mixture of three noble gases has a total pressure of 1.25 atm. The individual pressures exerted by neon and argon are 0.22 atm

and 0.35 atm, respectively. What is the partial pressure of the third gas, helium?
Group of answer choices

0.57 atm

0.68 atm

1.38 atm

1.82 atm

1 answer:

Lerok [7]3 years ago

3 0

Answer:

0.68 atm

Explanation:

am i right?

tell me in the comment if im right

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If 0.25 moles of KBr is dissolved in 0.5 liters of

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Answer:

[KBr] = 454.5 m

Explanation:

m is a sort of concentration that indicates the moles of solute which are contianed in 1kg of solvent.

In this case, the moles of solute are 0.25 moles.

Let's determine the mass of solvent in kg.

Density of heavy water, solvent, is 1.1 g/L and our volume is 0.5L.

1.1 g = mass of solvent / 0.5L, according to density.

mass of solvent = 0.5L . 1.1g/L = 0.55 g

We convert the mass to kg → 0.55 g . 1kg /1000g = 5.5×10⁻⁴ kg

m = mol/kg → 0.25 mol /5.5×10⁻⁴ kg = 454.5 m

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When the products of a reaction are hotter than the reactants: The reaction is exothermic. The reaction is endothermic. The reac

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<span>Exothermic reaction evolves energy due to which products get hot...</span>

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What is the electron configuration of an element with atomic number 20? A. 1s2 2s2 2p6 3s2 3p5 B. 1s2 2s2 2p6 3s2 3p6 C. 1s2 2s2

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D. The number of electrons equals the atomic number for a neutral element. Each number after the letter refers to the number of electrons in that shell. So for D, 2+2+6+2+6+2 = 20 electrons, which is equal to the atomic number.

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1. Unas de las formas de producir nitrógeno gaseoso (N2) es mediante la oxidación de metilamina (CH3NH2), tal como se muestra en

Maslowich

Answer:

a) 4CH₃NH₂ + 9O₂ ⇄ 4CO₂ + 10H₂O + 2N₂

b) m = 5,043 g

c) % = 69,4 %

Explanation:

a) La ecuación balanceada es la siguiente:

4CH₃NH₂ + 9O₂ ⇄ 4CO₂ + 10H₂O + 2N₂

En el balanceo, se tiene en la relación estequiométrica que 4 moles de metilamina reacciona con 9 moles de oxígeno para producir 4 moles de dióxido de carbono, 10 moles de agua y 2 moles de nitrógeno.

b) Para determinar la masa de nitrógeno se debe calcular primero el reactivo limitante:

$n_{O_{2}} = \frac{m}{M} = \frac{25,6 g}{31,99 g/mol} = 0,800 moles$

$n_{CH_{3}NH_{2}} = \frac{4}{9}*0,800 moles = 0,356 moles$

De la ecuación anterior se tiene que la cantidad de moles de metilamina necesaria para reaccionar con 0,800 moles de oxígeno es 0,356 moles, y la cantidad de moles iniciales de metilamina es 0,5 moles, por lo tanto el reactivo limitante es el oxígeno.

Ahora, podemos calcular la masa de nitrógeno producida:

$n_{N_{2}} = \frac{2}{9}*n_{O_{2}} = \frac{2}{9}*0,8 moles = 0,18 moles$

$m_{N_{2}} = n_{N_{2}}*M = 0,18 moles*28,014 g/mol = 5,043 g$

Por lo tanto, se pueden producir 5,043 g de nitrógeno.

c) El redimiento de la reacción se puede calcular usando la siguiente fórmula:

$\% = \frac{R_{r}}{R_{T}}*100$

<u>Donde</u>:

$R_{r}$ : es el rendimiento real

$R_{T}$ : es el rendimiento teórico

$\% = \frac{3,5}{5,043}*100 = 69,4$

Entonces, el procentaje de rendimiento de la reacción es 69,4%.

Espero que te sea de utilidad!

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Answer:

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