<span>
<span>
</span></span>Volume = 4/3 * PI * r^3
1.59 x 10^24 copper atoms = 2.64 moles of copper
Atomic Mass of copper = 63.55
2.64 * 63.55 = 167.77 grams of copper
Volume of Copper = Mass / Density
Volume of Copper = 167.77 grams / 8.96
Volume of Copper = <span>
<span>
18.72</span></span> cubic centimeters
r^3 = Volume / (4/3 * PI)
r^3 = 18.72 / 4.188
r^3 =
<span>
<span>
<span>
4.47
radius = </span></span></span><span><span><span>1.647</span> centimeters
</span></span>
Answer:
The boiling point of 1-chlorobutane is substantially lower than that of 1-butanol
Explanation:
Fractional distillation is a separation process based on difference in boiling point of two compounds.
1-chlorobutane is a polar aprotic molecule due to presence of polar C-Cl bond. Hence dipole-dipole intermolecular force exists in 1-chlorobutane as a major force.
1-butanol is a polar protic molecule. Hence dipole-dipole force along with hydrogen bonding exist in 1-butanol.
Therefore intermolecular force is stronger in 1-butanol as compared to 1-chlorobutane.
So, boiling point of 1-butanol is much higher than 1-chlorobutane.
Hence mixture of 1-chlorobutane and 1-butanol can be separated by fractional distillation based on difference in boiling point.
So, option (D) is correct.
Answer
If the temperature is increased , the number of collision per second increases.
Explanation
Temperature is proportional to the average kinetic energy of a sample of a gas according to the equation PV=n R T. An increased in temperature , increases the kinetic energy of the gas particles which in turn rises the velocity of the gas particles hitting the walls of the container. The more the number of particles the higher the collision rate and greater the pressure as long as the volume of container and the temperature are constant.
The given question is incomplete. The complete question is as follows.
A solution of chloroform (
) and acetone (
) exhibits a negative deviation from Raoult's law. This result implies that:
chloroform-chloroform interactions are stronger than chloroform-acetone interactions.
chloroform-chloroform interactions are weaker than chloroform-acetone interactions.
acetone-acetone interactions are stronger than chloroform-acetone interactions.
acetone-acetone interactions are weaker than chloroform-acetone interactions
chloroform-chloroform interactions are weaker than chloroform-acetone interactions AND acetone-acetone interactions are weaker than chloroform-acetone interactions.
Explanation:
It is known that when the forces between the particles of the mixture are stronger than the forces between the particles in the pure liquids then negative deviations from Raoult's law are observed.
Hence, when a solution of chloroform () and Acetone () will exhibit a negative deviation from Roult's law.
Then, chloroform-chloroform interactions should be weaker than chloroform-acetone interactions and acetone-acetone interactions must be weaker than Chloroform-Acetone interactions.
thus, we can conclude that for the given situation chloroform-chloroform interactions are weaker than chloroform-acetone interactions AND acetone-acetone interactions are weaker than chloroform-acetone interactions.
