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The chapter sections to review are shown in parentheses at the end of each problem.
15.48 Which of the following are
a.
b.
c.
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Basic Chemistry
- 14.55 Heating solid sodium bicarbonate in a closed vessel establishes the following equilibrium: (15.5) 2NaHCO3(s) = N22CO3(s) + H20(g) + CO2(8) What would happen to the equilibrium position if (a) CO2 gas were added to the system, (b) Na2CO3 were removed from the system, (c) NaHCO3 were removed from the system. The temperature remains constant in each case.arrow_forward(10) * Which reaction is an example of both a precipitation and a neutralization? (A) H3PO4(aq) + 3 KOH(aq) –→ K3PO4(aq) + 3 H2O(1) (B) FeC13(aq) + 3 KOH(aq) → Fe(OH)3(s) + 3 KCI(aq) (C) (NH4)2CO3(s) → 2 NH3(g) + CO2(g) + H20(1) (D) H2SO4(aq) + Ba(OH)2(aq) → BaSO4(s) + 2 H2O(1) (E) 2 C(s) + 02(g) → 2 CO(g) (11) * All of the statements regarding redox reactions are true except (A) a reducing agent causes another substance to be reduced. (B) halogens usually behave as oxidizing agents because they readily gain electrons. (C) metal ions are produced when pure metals are oxidized. (D) when a substance is oxidized its charge (or oxidation number) decreases. (E) alkali metals often behave as reducing agents because they readily lose electrons. (12) In a typical oxidation-reduction reaction the electrons are transferred (A) from the oxidizing agent to the reducing agent. (B) from what is being oxidized to the substance being reduced. (C) from what is being reduced to the substance being…arrow_forwardCalculate the molar concentration HCl when 250 mL of 0.10 M HCl are mixed with 150 mL of 0.10 M NaOH in a beaker. (5) HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)arrow_forward
- 1.4 4 (a) Calculate the value of Kc for the reaction: PC15 (8) PC13 (g) + Cl2 (g) AH = Positive Given that when 8.4 mol of PC15 (g): is mixed with 1.8 mol of PC13 (g) and allowed to come to equilibrium in a 10 dm³ container the amount of PC15 (g) at equilibrium is 7.2 mol. Kc = (b) Explain the effect of the following changes below on the value of Kc: (i) Increasing temperature (ii) Lowering the concentration of chlorine (C1₂) (iii) Addition of a catalystarrow_forward(7.12) A technician plates a faucet with 0.86 g Cr metal by electrolysis of aqueous Cr2(SO4)3. If 12.5 min is allowed for the plating, what current is needed? Molar mass of Cr: 51.996 g/mol Analysis/Strategy: mass of Cr ® mol Cr mol e¯ (or Faraday) ® coulombs ® amperes Reduction half-reaction of chromium solution to chromium metal: Cr* ,+3, (aq) e¯ ® Cr(s) + chromium solution, Cr2(SO4)3, has the ions: [Cr*3 s04-²] Answer: The current needed is ampere. (2 sig fig)arrow_forward(7.2b) Complete and balance the following by ion-electron method. Reactions are taking place in alkaline solution. b) Cr(OH)a) + CIO (nq) - Cro, (aq) Analysis: Balancing under acidic condition first (adding H30 & H") Oxidation-Half reation: Balancing atoms & charges: Cr(OH)x) * H20 - Cro, ug * H' + Reduction-Half reation: Balancing atoms and charges: 10 taq H20 Balancing electrons lost & gained Multiply oxidation-half reaction by Multiply reduction-half reaction by e's Final balanced equation: Continue next page.arrow_forward
- (7.2b) Complete and balance the following by ion-electron method. Reactions are taking place in alkaline solution. b) Cr(OH)a) + CIO (aq) - Cro, (aq) + Clug Analysis: Balancing under acidic condition first (adding H30 & H*) Oxidation-Half reation: Balancing atoms & charges: Cr(OH)) * H20 - Cro, ug) + Reduction-Half reation: Balancing atoms and charges: e Clap H20 Balancing electrons lost & galned Multiply oxidation-half reaction by e's Multiply reduction-half reaction by e's Final balanced equation: Continue next page.arrow_forward4. 5. OH + HCI I strong acid catalystarrow_forward14.60 When iodine is oxidized in concentrated HNO3, it produces the white solid iodic acid, HIO3: L(s)+NO;(aq) HIO3(s)+NO2(g) Balance this equation, adding H*(aq) and H2O(1) as necessary.arrow_forward
- (7.2b) Complete and balance the following by ion-electron method. Reactions are taking place in alkaline solution. b) Cr(OH)a) + Clo (aq) - Cro, 2(aq) • Clzup Analysis: Balancing under acidic condition first (adding Hy0 & H") Oxidation-Half reation: Balancing atoms & charges: Cr(OH)y) * H20 - Cro, ng) + H' Reduction-Half reation: Balancing atoms and charges: CIo (aq) e Clp H20 Balancing electrons lost & gained Multiply oxidation-half reaction by e's Multiply reduction-half reaction by e's Final balanced equation: Continue next page.arrow_forwardmol/i. 22. Given the reaction: N2 (g) + O2 (g) 2NO (g) Ke = 10.1 at 2000°C If the reaction starts with 0.065 MN2 and 0.065 MO2, what is the equilibrium concentration (M) of NO? (5)arrow_forwardAt high temperatures nitrogen (N2) and oxygen (O2) will react to form NO.N2(g) + O2(g) ⇆ 2 NO(g) (4.1)The value for the equilibrium constant for reaction 4.1 is KC = 2.7 x 10^-17 at some temperature T.A system initially has [N2] = 0.0800 M and [O2] = 0.0500 M. There is no NO initially present inthe system. Find the value for [NO] that will be present at equilibrium.arrow_forward
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