Answer:
See explanation
Explanation:
The boiling point of a substance is affected by the nature of bonding in the molecule as well as the nature of intermolecular forces between molecules of the substance.
2-methylpropane has only pure covalent and nonpolar C-C and C-H bonds. As a result of this, the molecule is nonpolar and the only intermolecular forces present are weak dispersion forces. Therefore, 2-methylpropane has a very low boiling point.
As for 2-iodo-2-methylpropane, there is a polar C-I bond. This now implies that the intermolecular forces present are both dispersion forces and dipole interaction. As a result of the presence of stronger dipole interaction between 2-iodo-2-methylpropane molecules, the compound has a higher boiling point than 2-methylpropane.
B. is ur answer I bieleve
Answer:
The equilibrium will shift to the left to favor the reactants.
Explanation:
Remember that the reaction quotient (Qc) is derived from initial concentrations of reactants and products. Since Qc is greater than Kc, this means that initial concentrations are heavily impacted by a high product concentration ([HI]). Therefore, the reverse reaction will occur and actually create more reactants again ([H2] and [I2]). Thus, the answer is that the equilibrium will shift to the left side to favor the reactants.
Answer:
E°(Ag⁺/Fe°) = 0.836 volt
Explanation:
3Ag⁺ + 3e⁻ => Ag°; E° = +0.800 volt
Fe° => Fe⁺³ + 3e⁻ ; E° = -0.036 volt
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Fe°(s) + 3Ag⁺(aq) => Fe⁺³(aq) + 3Ag°(s) ...
E°(Ag⁺/Fe°) = E°(Ag⁺) - E°(Fe°) = 0.800v - ( -0.036v) = 0.836 volt
