Answer:
- <u>Decreasing the temperature of the system will shift the reaction rightward.</u>
Explanation:
The complete question is:
Given the equation representing a system at equilibrium:
- N₂(g) + 3H₂(g) ⇌ 2NH₃(g) + energy
what changes occur when the temperature of this system is decreased?
<h2>Solution</h2>
Modifying the temperature of a system in equilibrium changes the equilibrium constant and the equilibrium position (concentrations) of the system.
When the temperature is decreased, following LeChatelier's principle that the system will react in a way that seeks to counteract the disturbance, the reaction will shift toward the reaction that produces more heat energy to compensate the temperature decrease.
Thus, decreasing the temperature of the system will favor the forward reaction, more N₂(g) and H₂(g) will be consumed and more NH₃(g) and energy will be produced. Hence, the equilibrium will shift rightward.
Answer:
I think C but it could be D
Explanation:
I would say go for C
a) They are solid at ambient temperatures of 25 ° C and pressure of 1 atm.
b) Ionic compounds represent high temperature melting and boiling.
c) They are hard and brittle and then subjected to the impact, break easily, creating planar faces.
d) When dissolved in water, or pure liquid, carry electrical current due to the existence of ions that move freely and can be attracted by the electrodes, closing the electric circuit.
<span>e) His solvent is water. </span>
Answer:
Explanation:
(a) Hybridization of orbitals
See the Lewis structure of propyne in the first diagram below.
C1 is directly bonded to two other atoms (H and C2) so it is sp hybridized.
C2 is directly bonded to two other atoms (C1 and C3) so it is sp hybridized.
C3 is directly bonded to four other atoms (C2 and 3 H) so it is sp³ hybridized.
(b) C2-C3 Sigma Bond
See the atomic and molecular orbitals in the second picture below.
C3 is using its hybrid atomic orbitals to form sigma molecular bonds. The C2-C3 sigma bond is formed by the overlap of the C2 sp atomic orbital with the C3 sp³ atomic orbital to make a σ(sp-sp³) molecular orbital.
(c) Bond angles
C1 and C2 are sp hybridized. Since the angle between sp orbitals is 180°, all atoms directly attached the C1 and C2 must be in a straight line. The C-C-C bond angle is 180°.
