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MakcuM [25]
3 years ago
13

The product of nuclear ___ is several nuclei

Chemistry
1 answer:
garri49 [273]3 years ago
3 0

I think Its "fission"

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Four students measure the length of a paper clip whose actual length is 3.2 cm. whose measurements are precise but not accurate?
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A. Samuel

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although he is not right he gets the same measurement several times

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Which event would be impossible to explain by using John Dalton’s model of the atom?
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The theoretical yield of zinc oxide in a reaction is 486 g. What is the percent
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the correct answer is c

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5 0
2 years ago
Sabiendo que a 20 °C, 0,275 moles de una sustancia desconocida cuya densidad
dimulka [17.4K]

Answer:

a) M=154.224g/mol

b) moleculas=4.33x10^{23}moleculas

c) m=2.56x10^{-22}g

Explanation:

Hola.

Para resolver los numerales a), b) y c), debemos seguir los siguientes procedimientos:

a) En este caso, primeo calculamos la masa que 45.9 cm³ ocupan con una densidad 0.924 g/cm³ como se muestra a continuación:

\rho=\frac{m}{V}\\ \\m=\rho V =0.924\frac{g}{cm^3}*45.9cm^3=42.4g

Seguidamente, sabiendo que estos gramos ocupan 0.275 moles, podemos facilmente calcular la masa molar como la relación entre ellas:

M=\frac{m}{n}=\frac{42.4g}{0.275mol}\\  \\M=154.224g/mol

b) En este caso, como se tienen 120 cm³ calculamos la nueva masa:

m=0.924\frac{g}{cm^3} *120cm^3=110.88g

Luego, con la masa molar, las moles:

n=110.88g*\frac{1mol}{154.224g} =0.719mol

Finalmente, con el número de Avogadro, las moléculas presentes:

moleculas=0.719mol*\frac{6.022x10^{23}moleculas}{1mol} \\\\moleculas=4.33x10^{23}moleculas

c) Aquí, consideramos que hay una molécula de la sustancia, por lo que usando el número de Avogadro, podemos calcular las moles que una molécula tiene:

n=1molecula*\frac{1mol}{6.022x10^{23}moleculas}=1.66x10^{-24}mol

Luego, usando la masa molar, calculamos la masa de una molécula:

m=1.66x10^{-24}mol*\frac{154.224g}{1mol} \\\\m=2.56x10^{-22}g

¡Saludos!

5 0
3 years ago
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