What is the most important reason that caffeine (C8H10N4O2) has a lower melting point than copper (II) chloride (CuCl2)?A. The intermolecular forces holding the caffeine molecules together are weaker than the ionic bonds in CuCl2.
B. The heavy caffeine molecules are more likely to fall apart than the lighter CuCl2 molecules.
C. The metallic bonds in copper chloride are stronger than the ionic bonds in caffeine.
D. Caffeine is a network solid and has a low melting point like other network solids.

Answers

Answer 1

Answer:

Option-A " The intermolecular forces holding the caffeine molecules together are weaker than the ionic bonds in CuCl₂ ".

Explanation:

There are two types of interactions among the atoms and molecules. One are known as intramolecular forces while the other are known as intermolecular forces.

Examples of Intramolecular forces are ionic bonds and covalent bonds e.t.c. while examples of intermolecular forces are hydrogen bond interactions, dipole-dipole interactions e.t.c.

Remember that intramolecular forces ar far more greater in strength than the intermolecular forces. Hence, in given statement the interactions between Caffeine molecules are intermolecular forces while, that between Cu and Cl ions in CuCl₂ are intramolecular forces.

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A child drags a 0.398 kg toy dogacross flat ground at constant speed,
with a 4.63 N force at a 63.0° angle.
What is the force of friction on the
toy?
(Hint: You do NOT need to know the
coefficient of friction.)

Answers

The force of friction on the toy is 2.10 N and it acts opposite to the direction of its motion.

What is friction?

A force called friction prevents one solid object from rolling or sliding over another. Although frictional forces, such the traction required to walk without slipping, may be advantageous, they can provide a significant amount of resistance to motion.

Given parameters:

mass of the toy dog: M = 0.398 kg.

Applied force: F = 4.63 N.

Angle of application of force: θ = 63.0°

Hence, horizontal component of applied force = F cosθ

= 4.63 × cos63.0° N

= 2.10 N.

As the child drags a 0.398 kg toy dog across flat ground at constant speed, this component of applied force is nullified by force of friction.

Hence, the force of friction on the toy is 2.10 N and it acts opposite to the direction of its motion.

Learn more about friction here:

brainly.com/question/13000653

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