(a) Interpretation: The value of change in enthalpy for the reaction between NaOH and HCl needs to be determined. Concept introduction: For a reaction as follows: A ⇄ B + C The change in enthalpy can be calculated as follows: Δ H r e a c t i o n = Δ H B + Δ H C − Δ H A Or, change in enthalpy of reaction is equal to difference in the sum of enthalpy of formation of product and sum of enthalpy of formation of reactants.
(a) Interpretation: The value of change in enthalpy for the reaction between NaOH and HCl needs to be determined. Concept introduction: For a reaction as follows: A ⇄ B + C The change in enthalpy can be calculated as follows: Δ H r e a c t i o n = Δ H B + Δ H C − Δ H A Or, change in enthalpy of reaction is equal to difference in the sum of enthalpy of formation of product and sum of enthalpy of formation of reactants.
Solution Summary: The author explains the value of change in enthalpy for the reaction between NaOH and HCl.
The value of change in enthalpy for the reaction between NaOH and HCl needs to be determined.
Concept introduction:
For a reaction as follows:
A⇄B+C
The change in enthalpy can be calculated as follows:
ΔHreaction=ΔHB+ΔHC−ΔHA
Or, change in enthalpy of reaction is equal to difference in the sum of enthalpy of formation of product and sum of enthalpy of formation of reactants.
Interpretation Introduction
(b)
Interpretation:
The value of change in enthalpy for the reaction between NaOH and HF needs to be determined.
Concept introduction:
For a reaction as follows:
A⇄B+C
The change in enthalpy can be calculated as follows:
ΔHreaction=ΔHB+ΔHC−ΔHA
Or, change in enthalpy of reaction is equal to difference in the sum of enthalpy of formation of product and sum of enthalpy of formation of reactants.
A solution containing 100.0 mL of 0.155 M EDTA buffered to pH 10.00 was titrated with 100.0 mL of 0.0152 M Hg(ClO4)2 in a cell:
calomel electrode (saturated)//titration solution/Hg(l)
Given the formation constant of Hg(EDTA)2-, logKf= 21.5, and alphaY4-=0.30, find out the cell voltage E.
Hg2+(aq) + 2e- = Hg(l) E0= 0.852 V
E' (calomel electrode, saturated KCl) = 0.241 V
From the following reduction potentials
I2 (s) + 2e- = 2I- (aq) E0= 0.535 V
I2 (aq) + 2e- = 2I- (aq) E0= 0.620 V
I3- (aq) + 2e- = 3I- (aq) E0= 0.535 V
a) Calculate the equilibrium constant for I2 (aq) + I- (aq) = I3- (aq).
b) Calculate the equilibrium constant for I2 (s) + I- (aq) = I3- (aq).
c) Calculate the solubility of I2 (s) in water.
2. (3 pts) Consider the unit cell for the spinel compound,
CrFe204. How many total particles are in the unit cell?
Also, show how the number of particles and their
positions are consistent with the CrFe204
stoichiometry - this may or may not be reflected by the
particle colors in the diagram. (HINT: In the diagram,
the blue particle is in an interior position while each red
particle is either in a corner or face position.)
Chapter 13 Solutions
Student Solutions Manual For Masterton/hurley's Chemistry: Principles And Reactions, 8th
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