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- One of the few industrial-scale processes that produce organic compounds electrochemically is used by the Monsanto Company to produce1,4-dicyanobutane. The reduction reaction is 2CH2CHCH+2H++2eNC(CH2)4CN The NC(CH2)4CN is then chemically reduced using hydrogen gas to H2N(CH2)6NH2, which is used in the production of nylon. What current must be used to produce 150.kg NC(CH2)4CN per hour?Selenic acid, H2SeO4, is produced from elemental selenium in a two-stage reaction. First the selenium isoxidized to selenous acid by nitric acid. Then the selenous acid is oxidized by potassium permanganate toselenic acid. The equations are:3 Se (s) + 4 HNO3 (l) + H2O (l) → 3 H2SeO3 (aq) + 4 NO (g)8 H2SeO3 (aq) + 2 KMnO4 (aq) → 5 H2SeO4 (aq)+ K2SeO3 (aq)+ 2 MnSeO3 (aq) + 3 H2O (l)What mass of selenic acid (in kg) can be produced from 1.00 kg of Se?Molar masses (g/mol): Se 78.96 H2SeO4 144.97Identify the reactant that gets oxidized in the following reaction 5C2O4^2- (aq)+ 2MnO4^-(aq)+16H+(aq)---> 2Mn^2+(aq) +8H2O(l) +10CO2(g)
- Given the following half-reactions, 2 103 (aq) + 12H*(aq) + 10 e → 12(s) + 6H2O Fe2* (aq) → Fe3* (aq) + e" write the balanced overall reaction. 2103 - (aq) + 12 H* (aq) + Fe 2* (aq) → Fel2 (s) + 6 H2O o 2 103 (aq) + 10 Fe²* (aq) → l2 (s) + 10 Fe³+ (aq) 2 103 - (aq) + 12 H* (aq) + Fe 2* (aq) → 12 (s) + Fe3+ (aq) + 6 H20 2 103 - (aq) + 12 H* (aq) + 10 Fe 2* (aq) → 12 (s) + 10 Fe3+ (aq) + 6 H20Complete and balance the following redox reaction in basic solution 2- 3c₂²- 1 + 2 (s) H* CIO₂(g) →CIO₂ (aq) + CIO3(aq) 3 4 H₂O Reset ( (1) LO 5 OH- H 6 (g) H3O+ 7 8 CI • x H₂O 9 (aq) SComplete and balance the following redox reaction in acidic solution 10: (aq) + PO:³ (aq) → 10 (aq) + PO:³ (aq) 4- 2+ O3+ 4+ 1 3 4 7 8. |3 |4 Os (s) (1) (g) (aq) + H. OH- H3O+ e H+ H2O Reset • x H2O Delete 1L 3. 2.
- Q) Drinking water may contain several unwanted ions such as phosphate ions. In order to remove the phosphate ions from the drinking water, a solution of calcium hydroxide can be added. As a result, a solid Ca5OH(PO4)3 is formed and isolated by filtration and a hydroxide ion are also formed. The reaction for this process is below: 5 Ca(OH)2 (aq) + PO43- (aq) → Ca5OH(PO4)3 (s) + OH- (aq) a) If 3.00 mL of 0.100 M calcium hydroxide is mixed with 4.00 mL of an aqueous solution of 0.0800 M phosphate ions (PO43-). What is the mass of Ca5OH(PO4)3 that can be isolated from the reaction? b) How many moles of the excess reactant remains unreacted? c) What is the concentration of the hydroxide ions in this solution? d) What is the mass of calcium (in grams) that can be recovered?Enter the balanced complete ionic equation for K2SO4(aq)+CaI2(aq)→CaSO4(s)+KI(aq) Enter the balanced net ionic equation for K2SO4(aq)+CaI2(aq)→CaSO4(s)+KI(aq) Enter the balanced complete ionic equation for NH4Cl(aq)+NaOH(aq)→H2O(l)+NH3(g)+NaCl(aq) Enter the balanced net ionic equation for NH4Cl(aq)+NaOH(aq)→H2O(l)+NH3(g)+NaCl(aq) Enter the balanced complete ionic equation for AgNO3(aq)+NaCl(aq)→AgCl(s)+NaNO3(aq) Enter the balanced net ionic equation for AgNO3(aq)+NaCl(aq)→AgCl(s)+NaNO3(aq) Enter the balanced complete ionic equation for HCl(aq)+K2CO3(aq)→H2O(l)+CO2(g)+KCl(aq) Enter the balanced net ionic equation for HCl(aq)+K2CO3(aq)→H2O(l)+CO2(g)+KCl(aq)For the reaction below, Kc = 1.10 x 104. Note Kc is sometimes called K. What is the equilibrium concentration of C if the reaction begins with 0.200 M A and 0.400 M B? 2 A (aq) + B (aq) = C (aq)
- Write the balanced molecular chemical equation for the reaction in aqueous solution for cesium carbonate and magnesium nitrate. If no reaction occurs, write NR. CS2CO3 + Mg(NO3)2 →2 CSNO3 + M9CO3 04 02+03+ 4+ 3 6. 7 8. 6. 0. 口3||□4 Os 19 10 (s) (1) (g) (aq) Mg Mn N NR Cs 69°F O pe here to search 4- 2.Balance the equation and give the coefficient of the question mark (?). кон + НзРОд K3PO4 + H20 (? )KOH + (x)H3PO, → (x)K3PO4 + (x)H20The balanced molecular equation for complete neutralization of H 2SO 4 by KOH in aqueous solution is H2SO4 (aq) + 2КОН (аq) — 2H20 (I) + К2S04 (s) B H2SO4 (aq) + 20H- (aq) → 2H2O (I) + SO42- (aq) (C) 2H+ (aq) + 20H- (aq) → 2H2O (I) D H2SO4 (aq) + 2KOH (aq) – 2H2O (I) + K2SO4 (aq) 2H+ (aq) + 2KOH (aq) → 2H2O (I) + 2K+ (aq) E.