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HACTEHA [7]
3 years ago
13

When a 235 92U 92235U nucleus is bombarded by neutrons (10n01n) it undergoes a fission reaction, resulting in the formation of t

wo new nuclei and neutrons. The following equation is an example of one such fission process: 235 92U+10n→AZBa+9436Kr+310n 92235U+01n→ZABa+3694Kr+301n Enter the isotope symbol for the barium (BaBa) nucleus in this reaction. Express your answer as an isotope.
Chemistry
1 answer:
r-ruslan [8.4K]3 years ago
6 0

Answer:

139/56 Ba

Explanation:

In writing nuclear equations, there must be a balance of the mass and charge on both sides of the reaction equation. That is, the sum of masses on the lefthand side of the nuclear reaction equation must equal the sum of masses on the righthand side of the nuclear reaction equation. Similarly, the charges on the lefthand side of the nuclear reaction equation must equal the sum of the charges on the right hand side of the nuclear reaction equation.

Hence, the isotope of barium formed in the equation mentioned in the question is 139/56 Ba.

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Determina el grado de pureza de un marmol (CaCO3), si al descomponerse 125 g del mismo se desprenden 20 litros de dióxido de car
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Answer:

67.8%

Explanation:

La reacción de descomposición del CaCO₃ es:

CaCO₃ → CO₂ + CaO

<em>Donde 1 mol de CaCO₃ al descomponerse produce 1 mol de CO₂ y 1 mol de CaO.</em>

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PV = nRT.

<em>Donde P es presión (1atm), V es volumen (20L), n son moles de gas, R es la constante de los gases (0.082atmL/molK) y T es temperatura absoluta (15 + 273.15 = 288.15K). </em>Reemplazando los valores en la ecuación:

PV / RT = n

1atmₓ20L / 0.082atmL/molKₓ288.15K = 0.846 moles

Como 1 mol de CO₂ es producido desde 1 mol de CaCO₃, las moles iniciales de CaCO₃ son 0.846moles.

La masa molar de CaCO₃ es 100.087g/mol. Así, la masa de 0.846moles de CaCO₃ es:

0.846moles ₓ (100.087g / mol) = <em>84.7g de CaCO₃</em>

Así, la pureza del marmol es:

(84.7g de CaCO₃ / 125g) ₓ 100<em> = </em>

<h3>67.8%</h3>
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3 years ago
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