Answer: The correct answer is "B" two bonding domains(or bonding pairs) or two non bonding domains(or lone pairs)
Explanation:
Molecular geometry/structure is a three dimensional shape of a molecule. It is basically an arrangement of atoms in a molecule.It is determined by the central atom, its surrounding atoms and electron pairs.According to VSEPR theory, there are 5 basic shapes of a molecule: linear, trigonal planar, tetrahedral, trigonal bipyramidal and octahedral.
A)Four bonding domains and zero non bonding domains: shape is tetrahedral and bond angle is 109.5°
B)Two bonding domains and two non bonding domains(lone pairs): shape is bent and bond angle is 104.5°
C)Three bonding domains and one non bonding domain: shape is trigonal pyramidal and bond angle is 107°
D)Two bonding domain and zero non bonding domain: shape is linear and bond angle is 107°
E)Two bonding domain and one non bonding domain: bent shape and bond angle is 120°
F)Three bonding domains and zero nonbonding domain: shape is trigonal planar and bond angle is 120°
Hence Two bonding domains and two non bonding domains have the smallest bond angle.
Answer:
The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol
Explanation:
Step 1: Data given
The combustion reaction of octane produces 5104.1 kJ per mol octane
Step 2: The balanced equation
C8H18(g) + 12.5 O2 ⟶ 8CO2 (g) + 9 H2O (g) ∆H°rxn = -5104.1 kJ/mol
Step 3:
∆H°rxn = ∆H°f of products minus the ∆H° of reactants
∆H°rxn = ∆H°f products - [∆H°f reactants]
-5104.1 kJ/mol = (8*∆H°fCO2 + 9*∆H°fH20) - (∆H°fC8H18 + 12.5∆H°fO2)
∆H°f C8H18 = ∆H°f 8CO2 + ∆H°f 9H2O+ 5104.1 kJ/mol
∆H°f C8H18 = 8 * (-393.5 kJ)/mol + 9 * (-241.8 kJ/mol)] + 5104.1 kJ
/mol
∆H°f C8H18 = -220.1 kJ/mol
The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol
Answer:
I just did it home slice the first one's Ag+ and Zn2+ and the second one is A
Explanation:
I just did the assignment
