Concept explainers
(a)
Interpretation:
A 35.0 mL solution of
Concept Introduction:
Solubility product constant:
The equilibrium constant of a more soluble ionic compound in water at the higher solubility is known as solubility product constant.
The equilibrium constant of more soluble ionic compound is given by
Ion product:
The multiplication of concentrations of ions which are raised to the power of number of respective ions present in the molecular formula. This gives the ion product.
Precipitation occurs when solution is supersaturated,
Precipitation will not occurs when solution is saturated and in equilibrium,
Precipitation will not occurs when solution is unsaturated,
(b)
Interpretation:
A 35.0 mL solution of
Concept Introduction:
Solubility product constant:
The equilibrium constant of a more soluble ionic compound in water at the higher solubility is known as solubility product constant.
The equilibrium constant of more soluble ionic compound is given by
Ion product:
The multiplication of concentrations of ions which are raised to the power of number of respective ions present in the molecular formula. This gives the ion product.
Precipitation occurs when solution is supersaturated,
Precipitation will not occurs when solution is saturated and in equilibrium,
Precipitation will not occurs when solution is unsaturated,
(c)
Interpretation:
A 35.0 mL solution of
Concept Introduction:
Solubility product constant:
The equilibrium constant of a more soluble ionic compound in water at the higher solubility is known as solubility product constant.
The equilibrium constant of more soluble ionic compound is given by
Ion product:
The multiplication of concentrations of ions which are raised to the power of number of respective ions present in the molecular formula. This gives the ion product.
Precipitation occurs when solution is supersaturated,
Precipitation will not occurs when solution is saturated and in equilibrium,
Precipitation will not occurs when solution is unsaturated,
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Chapter 19 Solutions
CHEMISTRY MOLECULAR NATURE CONNECT ACCES
- true or false The equilibrium constant for this reaction is 0.20. N2O4(g) ⇔ 2NO2(g) Based on the above, the equilibrium constant for the following reaction is 0.4. 2N2O4(g) ⇔ 4NO2(g)arrow_forwardtrue or false Using the following equilibrium, if heat is added the equilibrium will shift toward the reactants. N2(g) + 3H2(g) ⇔ 2NH3(g) + heatarrow_forwardTrue or False Using the following equilibrium, if heat is added the equilibrium will shift toward the products. N2O4(g) + heat ⇔ 2NO2(g)arrow_forward
- true or false Using the following equilibrium, if solid carbon is added the equilibrium will shift toward the products. C(s) + CO2(g) ⇔ 2CO(g)arrow_forwardProvide the complete mechanism for the reaction below. You must include appropriate arrows,intermediates, and formal charges. Please also provide a reason to explain why the 1,4-adduct is preferred over the 1,3-adduct.arrow_forwardWhich of the following pairs are resonance structures of one another? I. III. || III IV + II. :0: n P !༠ IV. EN: Narrow_forward
- Predict the major organic product(s) and byproducts (either organic or inorganic) for thefollowing reactions.arrow_forwardA 8.25 g sample of aluminum at 55°C released 2500 J of heat. The specific heat of aluminum is 0.900 J/g°C. The density of aluminum is 2.70 g/mL. Calculate the final temperature of the aluminum sample in °C.arrow_forwardPredict the major organic product(s) and byproducts (either organic or inorganic) for thefollowing reactions.arrow_forward
- Predict the major organic product(s) and byproducts (either organic or inorganic) for thefollowing reaction.arrow_forwardplease helparrow_forwardExperiment 1 Data Table 1: Conservation of Mass - Initial Mass Data Table 1 Data Table 2 Data Table 3 Data Table 4 Panel 1 Photo 1 Data Table 5 Reaction Mass of test tube and 5.0% HC₂H₂O2 (g) # (A) (B) Mass of NaHCO, (g) Mass of balloon and NaHCO, (g) (C) 0.10 1 0829 14.38g 0.20 2 0.929 14.29g 0.35 1.00g 3 14.25g 0.50 1.14g 14.29 Experiment 1 Data Table 2: Moles of HC2H3O2 Reaction Volume of Mass of Moles of HC₂H₂O₂ 5.0% Vinegar (g) (ML) 5.0 0.25 0042 mol 2 5.0 0.25 0042 mol 3 5.0 0.25 0042 mol 5.0 0.25 0042 mol Experiment 1 Data Table 3: Moles of NaHCO3 Reaction Mass of NaHCO (g) 10g 20g 35g 50g Experiment 1 Data Table 4: Theoretical Yield of CO₂ Reaction # 1 2 3 Experiment 1 Total mass before reaction (g) (D=A+C) 15.29 15.21g 15.25g 15.349 Exercise 1 Data Table 1 Data Table 2 Data Table 3 Data Table 4 Panel 1 Photo 1 Data Table 5 Exercise 1- Data Table 1 Data Table 2 DataTable 3 Data Table 4 Panel 1 Photo 1 Data Table 5 Exercise 1- Moles of NaHCO 0012 mol 0025 mol 0044 mol 0062 mol…arrow_forward
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