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svetlana [45]
3 years ago
7

Which molecule will undergo only London dispersion forces when interacting with other molecules of the same kind? a.HCl b.C2H5OH

c.NH3 d.CH4
Chemistry
1 answer:
ohaa [14]3 years ago
5 0

Answer:

CH4

Explanation:

London dispersion forces are a type of force acting between atoms and molecules that are normally electrically symmetric; that is, the electrons are symmetrically distributed with respect to the nucleus. They are part of the van der Waals forces(Wikipedia).  It is a temporary attractive force that results when the electrons in two adjacent atoms occupy positions that make the atoms form temporary dipoles. This force is sometimes called an induced dipole-induced dipole attraction(Chem Purdue).

If we consider HCl, C2H5OH  and NH3, we will discover that they all have polar bonds. Consequently, dipole - dipole interaction as well as hydrogen bonds exist between their molecules.

CH4 is nonpolar. Being a nonpolar molecule, the only possible intermolecular interaction forces between its molecules is the London dispersion forces.

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Which of the following statements is not true for an exothermic reaction? Question options: The products have a higher heat cont
dsp73

Answer:

The products have a higher heat content than the reactants.

Explanation:

The statement above is not true for an exothermic reaction because in an exothermic reaction heat is released to the surroundings. This simply means that the total energy of the products is less than that of the reactants.

5 0
3 years ago
Calculate the energy required to heat 406.0mg of cyclohexane from 33.5°C to 38.9°C . Assume the specific heat capacity of cycloh
N76 [4]

Answer:

Q = 4.056 J

Explanation:

  • Q = m<em>C</em>ΔT

∴ m = 406.0 mg = 0.406 g

∴ <em>C </em>= 1.85 J/g.K

∴ T1 = 33.5°C ≅ 306.5 K

∴ T2 = 38.9°C = 311.9 K

⇒ ΔT = 311.9 - 306.5 = 5.4 K

⇒ Q = (0.406 g)(1.85 J/gK)(5.4 K)

⇒ Q = 4.056 J

6 0
3 years ago
C (graphite) is used as a lubricant, whereas C (diamond) is used as an abrasive. Why is this?
zysi [14]

Answer:

Carbon atoms in graphite and diamond are arranged in different ways. Hence, the two allotropes of carbon have different physical properties.

Explanation:

Both graphite and diamond are both made of only carbon atoms. However, their physical properties differ from each other. Hence, they are called allotropes. Think about how these carbon atoms are arranged in each of the allotropes.

<h3>Graphite</h3>

In graphite, each carbon atom is bonded to three other carbon atoms. These carbon atoms will be located in the same plane. A chunk of graphite can contain many of these planes.

Each carbon atom has four valence electrons. Three of these electrons will be used in the bonds. The other electron will be delocalized. These electrons would flow between the sheets of carbon atoms. That keeps the sheets separate and allow them to slide on top of each other.

<h3>Diamond</h3>

In diamond, each carbon atom is bonded to four other carbon atoms. These carbon atoms will form a tetrahedral network.  

In graphite, there's a significant separation between two adjacent sheets of carbon atoms. The force between the two sheets is rather weak. When a piece of graphite is between two objects that move over one another, the layers in the graphite would also slide over one another. Since the attraction between two adjacent sheets isn't very strong, there wouldn't be much resistance. Hence the graphite acts as a lubricant.  

In contrast, most of the carbon atoms in a piece of diamond would be connected to each other. Unlike the sheets in graphite, in a diamond there are almost no moving parts. Also, the forces between neighboring carbon atoms are very strong. When an external force acts on a chunk of diamond, the carbon atoms would barely move. Hence, the structure appears to be very rigid. That gives diamond its abrasive properties.

4 0
3 years ago
Cuando se quema 1 mol de metano –o sea, 16 g–, se desprenden 802
Anvisha [2.4K]

Answer:

1 gramo de metano aporta 50.125 kilojoules.

1 gramo de metano aporta 48.246 kilojoules.

Explanation:

La cantidad de energía liberada por la combustión de una unidad de masa del hidrocarburo (Q), en kilojoules por mol, es igual a la cantidad de energía liberada por mol de compuesto (\bar {Q}), en kilojoules por mol, dividido por su masa molar (M), en gramos por mol:

Q = \frac{\bar Q}{M} (1)

A continuación, analizamos cada caso:

Metano

Q = \frac{802\,\frac{kJ}{mol} }{16\,\frac{g}{mol} }

Q = 50.125\,\frac{kJ}{g}

1 gramo de metano aporta 50.125 kilojoules.

Octano

Q = \frac{5500\,\frac{kJ}{mol} }{114\,\frac{g}{mol} }

Q = 48.246\,\frac{kJ}{mol}

1 gramo de metano aporta 48.246 kilojoules.

3 0
2 years ago
For the reaction: h2(g) + cl2(g) → 2hcl(g), how many moles hcl will be produced from 10.0 g of h2? the reaction occurs in the pr
motikmotik
Hope this would help you

3 0
3 years ago
Read 2 more answers
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