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miss Akunina [59]
3 years ago
9

Which one of the following exhibits dipole-dipole attraction between molecules? Group of answer choices SF6 HCl O2 CF4 C10H22

Chemistry
1 answer:
Cerrena [4.2K]3 years ago
6 0

Answer:

The correct answer is HCl

Explanation:

There is a dipole-dipole force in hydrochloric acid (HCl), this type of attraction occurs in polar covalent molecules. In this case, the positive end of the molecule (H) attracts the negative (Cl).

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What is the photoelectric effect?
Gennadij [26K]
I believe A is the is the right answer
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3 years ago
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Label the molecular shape around each of the central atoms in the amino acid glycine. hint
svetlana [45]

The Structure of Glycine is attached below and each central atom is encircled with different colors.

Molecular Shape around Nitrogen Atom (Orange):

As shown, Nitrogen is making three single bonds with two hydrogen atoms and one carbon atom hence, it has three bonded pair electrons and a single lone pair of electron. Therefore, according to VSEPR theory it has a tetrahedral electronic geometry but due to repulsion created by lone pair of electrons its molecular geometry becomes Trigonal Pyramidal.

Molecular Shape around Carbon Atom (Green):

As shown, Carbon is making four single bonds with two hydrogen atoms and one nitrogen atom one with carbon atom of carbonyl group hence, it has four bonded pair electrons. Therefore, according to VSEPR theory it has Tetrahedral geometry.

Molecular Shape around Carbon Atom (Blue):

As shown, Carbon is making two single bonds with oxygen and carbon atoms and a double bond with oxygen. Hence, it has a Trigonal Planar geometry.

Molecular Shape around Oxygen Atom (Red):

As shown, Oxygen is making two single bonds with one carbon atoms and one hydrogen atom hence, it has two bonded pair electrons and two lone pair of electrons. Therefore, according to VSEPR theory it has a tetrahedral electronic geometry but due to repulsion created by lone pair of electrons its molecular geometry becomes Bent.


6 0
3 years ago
Read 2 more answers
How many joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point?
Korolek [52]

Answer: 1123000 Joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point

Explanation:

Latent heat of vaporization is the amount of heat required to convert 1 mole of liquid to gas at atmospheric pressure.

Amount of heat required to vaporize 1 mole of lead =  177.7 kJ

Molar mass of lead = 207.2 g

Mass of lead given = 1.31 kg = 1310 g       (1kg=1000g)

Heat required to vaporize 207.2 of lead = 177.7 kJ

Thus Heat required to vaporize 1310 g of lead =\frac{177.7}{207.2}\times 1310=1123kJ=1123000J

Thus 1123000 Joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point

7 0
3 years ago
Where is the friction that allows the skater to complete the move that is occurring in this image? a girl ice skating © Wildcow
adelina 88 [10]

Answer:

C. The point of contact between the skates and the ice

Explanation:

That is the only point of friction with the ice that provides the ability to complete the move.

4 0
4 years ago
When sodium metal is added to water, an orange flame is observed on the metal surface. Based on this observation, what can best
pickupchik [31]
All you can conclude is that something must be burning with an orange flame.

Actually, the "something" that must be burning is the hydrogen that is produced when the sodium reacts with the water:

2Na + 2H₂O → 2NaOH + H₂ + heat

So much heat is produced that the hydrogen catches fire and some of the sodium evaporates into the flame.

The electrons in the sodium atoms get "excited" in the flame. When they drop back to a lower energy level, they emit energy in the form of an orange-yellow light.
3 0
4 years ago
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