Concept explainers
(a)
Interpretation:
The chemical equation for burning of lithium in oxygen has to be given.
(a)
Answer to Problem 8H.1E
The chemical equation for burning of lithium in oxygen is,
Explanation of Solution
Lithium burns in the presence of oxygen to form lithium oxide
Four moles of lithium are required in the reactant side and two moles of lithium oxide are required in the product side to make the equation a balanced one. The balanced chemical equation for burning of lithium in oxygen is,
(b)
Interpretation:
The chemical equation for sodium metal with water has to be given.
(b)
Answer to Problem 8H.1E
The chemical equation for sodium metal with water is,
Explanation of Solution
Sodium reacts with water to form sodium hydroxide with the evolution of hydrogen gas. The chemical equation is given as,
Two moles of sodium, two moles of water are required in the reactant side and two moles of sodium hydroxide are required in the product side to make the equation a balanced one. The balanced chemical equation for sodium metal with water is,
(c)
Interpretation:
The chemical equation for fluorine gas with water has to be given.
(c)
Answer to Problem 8H.1E
The chemical equation for fluorine gas with water is,
Explanation of Solution
Fluorine reacts with water to form hydrogen fluoride
Two moles of fluorine, two moles of water are required in the reactant side and four moles of hydrogen fluoride are required in the product side to make the equation a balanced one. The balanced chemical equation for fluorine gas with water is,
(d)
Interpretation:
The chemical equation for oxidation of water at the anode of an electrolytic cell has to be given.
(d)
Answer to Problem 8H.1E
The chemical equation for oxidation of water at the anode of an electrolytic cell is,
Explanation of Solution
The oxidation of water in anode of electrolytic cell results in the formation of gaseous oxygen and hydrogen ions. The chemical equation for oxidation of water at the anode of an electrolytic cell is,
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Chapter 8 Solutions
Chemical Principles: The Quest for Insight
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- The amount of sodium hypochlorite in a bleach solution can be determined by using a given volume of bleach to oxidize excess iodide ion to iodine; ClO- is reduced to Cl-. The amount of iodine produced by the redox reaction is determined by titration with sodium thiosulfate, Na2S2O3; I2 is reduced to I-. The sodium thiosulfate is oxidized to sodium tetrathionate, Na2S4O6. In this analysis, potassium iodide was added in excess to 5.00 mL of bleach (d=1.00g/cm3) . If 25.00 mL of 0.0700 M Na2S2O3 was required to reduce all the iodine produced by the bleach back to iodide, what is the mass percent of NaClO in the bleach?arrow_forwardUse data from Appendix J to calculate the enthalpy change and the Gibbs free energy change for the reduction of chromium(III) oxide by aluminum.arrow_forward
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