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Elena-2011 [213]
4 years ago
9

Calculate the amount of heat released when 27.0 g H2O is cooled from a liquid at 314 K to a solid at 263 K. The melting point of

H2O is 273 K. Other useful information about water is listed below.
Cp,liquid = 4.184 J/(g•K)

Cp,solid = 2.093 J/(g•K)

ΔHfusion = 40.7 kJ/mol
Chemistry
2 answers:
Liula [17]4 years ago
7 0

Answer:

Cool liquid from 314 K to 273 K, freeze liquid at 273 K, and cool solid to 263 K.

Explanation:

answer on edg

lubasha [3.4K]4 years ago
5 0
Cool liquid from 314 K to 273 K, freeze liquid at 273 K, and cool solid to 263 K.
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Which molecule has polar bonding and is nonpolar? A. H2O B. BF3 C. NH3 D. NCl3 E. CH2Cl2
Marina CMI [18]

Answer:

B. BF₃

Explanation:

All the molecules have polar bonds, but a molecule will be nonpolar if the molecule has the symmetry that makes the bond dipoles cancel.

To make the decision, we must

  1. Draw the Lewis structure
  2. Assign the VSEPR electron geometry
  3. Determine the molecular shape.
  4. Examine the symmetry of the molecule

===============

<em>A. Water </em>

Lewis structure = H-O-H (2 bonding pairs, 2 lone pairs)

Electron geometry = AX₂E₂ tetrahedral

Molecular geometry = bent

Symmetry (see Figure A): The two O-H bonds are polar, with their negative ends pointing towards the O. The horizontal components of the bond dipoles cancel, but the vertical components reinforce each other and give an upward pointing molecular dipole. This is a <em>polar molecule with polar bonds</em>.

===============

<em>B. Boron trifluoride </em>

Lewis structure = BF₃ (3 bonding pairs)

Electron geometry = AX₃, trigonal planar

Molecular geometry = trigonal planar

Symmetry (see Figure B): The three B-F bonds are polar, with their negative ends pointing towards the F. The horizontal components of the bond dipoles cancel, but the vertical components of the two downward -pointing dipoles reinforce each other and give a resultant that is equal and opposite to the upward dipole. Thus, the bond dipoles cancel. This is a nonpolar molecule with polar bonds.

===============

<em>C. Ammonia</em>

Lewis structure = :NH₃ (3 bonding pairs, 1 lone pairs)

Electron geometry = AX₃E, tetrahedral

Molecular geometry = trigonal pyramidal

Symmetry (see Figure C): The three N-H bonds are polar, with their negative ends pointing towards the N. The horizontal components of the bond dipoles cancel, but the vertical components reinforce each other and give an upward pointing molecular dipole. This is a <em>polar molecule with polar bonds</em>.

===============

<em>D. Nitrogen trichloride </em>

Lewis structure = :NCl₃ (3 bonding pairs, 1 lone pair)

Electron geometry = AX₃E, tetrahedral

Molecular geometry = trigonal pyramidal

Symmetry (see Figure D): The three N-Cl bonds are polar, with their negative ends pointing towards the Cl. The horizontal components of the bond dipoles cancel, but the vertical components reinforce each other and give a downward pointing molecular dipole. This is a <em>polar molecule with polar bonds</em>.

===============

<em>E. Dichloromethane </em>

Lewis structure = H₂CCl₂ (4 bonding pairs)

Electron geometry = AX₄, tetrahedral

Molecular geometry = tetrahedral

Symmetry (see Figure E): The two C-H bonds are nonpolar, but the two C-Cl bonds are polar with their negative ends pointing towards the Cl. The horizontal components of the bond dipoles cancel, but the vertical components reinforce each other and give a downward pointing molecular dipole. This is a <em>polar molecule with polar bonds</em>.

3 0
3 years ago
Calcular el valor de la presión osmótica que corresponde a una solución que contiene 2 moles de soluto en un litro de solución a
Brums [2.3K]

El valor de la presión osmótica que corresponde a una solución que contiene 2 moles de soluto en un litro de solución a una temperatura de 17°C es 47.56 atmósferas.

La presión osmótica se puede calcular usando la siguiente ecuación:

\pi = \frac{nRT}{V}   (1)

En donde:

n: es el número de moles = 2 moles

R: es la constante de los gases = 0.082 L*atm/(K*mol)

T: es la temperatura = 17 °C = 290 K

V: es el volumen = 1 L

Introduciendo lo valores anteriores en la ecuación (1), tenemos:

\pi = \frac{nRT}{V} = \frac{2 moles*0.082 L*atm/(K*mol)*290 K}{1 L} = 47.56 atm

Por lo tanto, la presión osmótica es 47.56 atmósferas.

Puedes encontrar más aca:

  • brainly.com/question/5041899?referrer=searchResults
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Espero que te sea de utilidad!

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What is the name of group 6a,7a,and 8a
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Answer:

Refering to Chemistry,

The Group names are:

Group 6A: Chalcogens

Group 7A: Halogens

Group 8A: Noble/Inert Gasses

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Answer:

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the movement of air parallel to earth’s surface is called wind

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