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nata0808 [166]
2 years ago
6

How would you describe the size of atoms and molecules in your own words ?

Chemistry
2 answers:
topjm [15]2 years ago
8 0

Answer:

tomic size is the distance from the nucleus to the valence shell where the valence electrons are located. Atomic size is difficult to measure because it has no definite boundary. The electrons surrounding the nucleus exist in an electron cloud.

Explanation:

Mazyrski [523]2 years ago
8 0

Answer:

Atomic size is the distance from the nucleus to the valence shell where the valence electrons are located. The separation that occurs because electrons have the same charge. The number of protons in the nucleus. The core electrons in an atom interfere with the attraction of the nucleus for the outermost electrons. and Molecular size is a measure of the area a molecule occupies in three-dimensional space. The amount of space any mass takes up in three-dimensional space is known specifically as its volume.

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NaN₃ is the chemical formula for Sodium Azide
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2 years ago
uranium atoms each have 92 protons. how many protons does each u238 atom have? a. 92 b. 146 c. 238 d. 330
Brilliant_brown [7]
Number of protons as stated in the question = 92.

a. 92
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3 years ago
What is the difference between the lithosphere and the crust?
Dmitrij [34]

Answer:

The answer is B "The lithosphere is characterized by its physical state while the crust is characterized by its composition (mostly oxygen, aluminum, and silicon)

8 0
3 years ago
When a soda can is dropped, it should not be immediately opened. Why?
Kipish [7]
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7 0
3 years ago
Se hace reaccionar 4,00 g de aluminio y 42,00 g de bromo, según la reacción: Al(s)+Br2(l)⟶AlBr3(s) Calcular las moles de AlBr3(s
NeX [460]

Answer:

0.145 moles de AlBr3.

Explanation:

¡Hola!

En este caso, al considerar la reacción química dada:

Al(s)+Br2(l)⟶AlBr3(s)

Es claro que primero debemos balancearla como se muestra a continuación:

2Al(s)+3Br2(l)⟶2AlBr3(s)

Así, calculamos las moles del producto AlBr3 por medio de las masas de ambos reactivos, con el fin de decidir el resultado correcto:

n_{AlBr_3}^{por\ Al}=4.00gAl*\frac{1molAl}{27gAl} *\frac{2molAlBr_3}{2molAl}=0.145mol AlBr_3\\\\n_{AlBr_3}^{por\ Br_2}=42.00gr*\frac{1molr}{160g Br_2} *\frac{2molAlBr_3}{3molBr_2}=0.175mol AlBr_3

Así, inferimos que el valor correcto es 0.145 moles de AlBr3, dado que viene del reactivo límite que es el aluminio.

¡Saludos!

3 0
2 years ago
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