Answer:
7. .........................
Ionic bond
Explanation:
Ca - Cl₂ is an ionic bond because it involves a true metal and a non-metal.
An ionic bond is an inter-atomic bond between two chemical species.
- This bond forms between two atom with a large electronegative difference.
- Often times, this corresponds to a metal and a non-metal.
- The metal is the less electronegative specie and the non-metal is more electronegative
- The metal such as Ca loses two electrons to become positively charge ion.
- The non-metal gains the electrons. Here two atoms of Cl are turned to ions.
- Electrostatic attraction between the metal and non-metal forms the ionic bonding.
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Question:
Part D) An object at 20∘C absorbs 25.0 J of heat. What is the change in entropy ΔS of the object? Express your answer numerically in joules per kelvin.
Part E) An object at 500 K dissipates 25.0 kJ of heat into the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.
Part F) An object at 400 K absorbs 25.0 kJ of heat from the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.
Part G) Two objects form a closed system. One object, which is at 400 K, absorbs 25.0 kJ of heat from the other object,which is at 500 K. What is the net change in entropy ΔSsys of the system? Assume that the temperatures of the objects do not change appreciably in the process. Express your answer numerically in joules per kelvin.
Answer:
D) 85 J/K
E) - 50 J/K
F) 62.5 J/K
G) 12.5 J/K
Explanation:
Let's make use of the entropy equation: ΔS =
Part D)
Given:
T = 20°C = 20 +273 = 293K
Q = 25.0 kJ
Entropy change will be:
ΔS =
= 85 J/K
Part E)
Given:
T = 500K
Q = -25.0 kJ
Entropy change will be:
ΔS =
= - 50 J/K
Part F)
Given:
T = 400K
Q = 25.0 kJ
Entropy change will be:
ΔS =
= 62.5 J/K
Part G:
Given:
T1 = 400K
T2 = 500K
Q = 25.0 kJ
The net entropy change will be:
ΔS =
= 12.5 J/K
Answer: Water, alcohols, aldehydes, and ketones.
Explanation:
Answer:
<span>Dipole-Dipole Forces are common </span><span>to all polar molecules but not non polar molecules.
Explanation:
An Asymmetrical molecule having a region of high electron density (partial negative) and lower density (partial positive) interacts with its neighbor molecules through Dipole-Dipole Interactions. The partial positive part of one molecule interacts with the partial negative part of another.
Example:
Acetone having a partial positive carbon (of carbonyl group) and partial negative oxygen interacts through Dipole-Dipole forces. Hence, acetone does not involves Hydrogen Bond interactions.</span>