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zalisa [80]
4 years ago
10

View Available Hint(s) Check all that apply. Hydrogen bonding occurs when a hydrogen atom is covalently bonded to an NN, OO, or

FF atom. The CH4CH4 molecule exhibits hydrogen bonding. A hydrogen atom acquires a partial positive charge when it is covalently bonded to an FF atom. A hydrogen bond is possible with only certain hydrogen-containing compounds. A hydrogen bond is equivalent to a covalent bond.
Chemistry
1 answer:
Sauron [17]4 years ago
7 0

Answer:

Hydrogen bonding occurs when a hydrogen atom is covalently bonded to an NN, OO, or FF atom.

A hydrogen atom acquires a partial positive charge when it is covalently bonded to an FF atom.

A hydrogen bond is possible with only certain hydrogen-containing compounds.

Explanation:

A hydrogen bond does not occur in all hydrogen containing compounds. Hydrogen bonds only occur in those compounds where hydrogen is bonded to a highly electronegative element such as fluorine, oxygen or nitrogen.

In a hydrogen bonded specie, hydrogen acquires a partial positive charge and the electronegative element acquires a partial negative charge which extends throughout the molecule.

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Calcule el volumen de disolución de LiOH a 3.5 M, necesario para neutralizar una disolución de 25 ml de H2SO3 cuya densidad es d
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Respuesta:

0.11 L

Explicación:

Paso 1: Escribir la ecuación balanceada

2 LiOH + H₂SO₃ ⇒ Li₂SO₃ + H₂O

Paso 2: Calcular la masa de solución de H₂SO₃

25 mL de solución de H₂SO₃ tiene una denisdad de 1.03 g/mL.

25 mL × 1.03 g/mL = 28 g

Paso 3: Calcular la masa de H₂SO₃ en 26 g de Solución de H₂SO₃

La riqueza de H₂SO₃ es 60%, es decir, cada 100 g de solución hay 60 g de H₂SO₃.

26 g Sol × 60 g H₂SO₃/100 g Sol = 16 g H₂SO₃

Paso 4: Calcular los moles correspondientes a 16 g de H₂SO₃

La masa molar de H₂SO₃ es 82.07 g/mol.

16 g × 1 mol/82.07 g = 0.19 mol

Paso 5: Calcular los moles de LiOH que reaccionan con 0.19 moles de H₂SO₃

La relación molar de LiOH a H₂SO₃ es 2:1. Los moles de LiOH que reaccionan son 2/1 × 0.19 mol = 0.38 mol.

Paso 6: Calcular el volumen de solución de LiOH

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0.38 mol × 1 L/3.5 mol = 0.11 L

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By determining the primary means of melting that would occur at divergent boundaries and the appropriate labels to the respective Convergent plate boundary.

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As the plate delves deeper into the convergent boundary, pressure and temperature rise along with it. When the pore water is released, this contributes to partial melting.

As a result, when the situation is at the melting curve, a rise in temperature will result in a large partial melt because of the presence of water. As seen in the wet peridotite melting curves in the aforementioned curves.

Only the peridotite's temperature changes during intraplate melting, not the pressure. Therefore, as the temperature rises, it will begin to melt.

Decompression causes melting at diverging boundaries. Temperature plays no part in this. As the plate delves deeper into the convergent boundary, pressure and temperature rise along with it.

When the pore water is released, this contributes to partial melting. As a result, when the situation is at the melting curve, a rise in temperature will result in a large partial melt because of the presence of water. As seen in the wet peridotite melting curves in the aforementioned curves.

Only the peridotite's temperature changes during intraplate melting, not the pressure. Therefore, as the temperature rises, it will begin to melt Decompression causes melting at diverging boundaries. Temperature plays no part in this.

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Answer:

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The physical area or space where the electron may be calculated to be present, as predicted by the specific mathematical shape of the orbital, is referred to as an atomic orbital.

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