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- One day in lab, while adding a gnarled root to a dark liquid bubbling in an iron cauldron, your friend Carmen (an expert chemist) says this: "Metal sulfides roasted with oxygen produce the corresponding oxide and sulfur dioxide gas." Using Carmen's statement, and what you already know about chemistry, predict the products of the following reaction. Be sure your chemical equation is balanced! olo Ag,s (s) → [ + 0,(8) 2 Ar NO O+0 REACTIONR = 8.314 mol·K F = 96,485 mol AG° + RT · In(Q) AE° () · In(Q) AG ΔΕ %3D Half Reaction (Note: All given as reduction) E° (V) 02 (g) + 4 H*(aq) + 4 e → 2 H20 (I0) 1.229 Z2 (s) + 2 e - → 2Z (aq) 0.426 (aq) + 3 e A (s) 0.292 2 H20 (1) + 2 e G2+ (aq) + 2 e M2+ (aq) + 2 e Н2 (в) + 2 ОН" (aq) - 0.828 G (s) - 1.245 M (s) - 1.893 Using a U-tube, a student sets up a non-spontaneous electrochemical cell with a battery connected to two carbon electrodes that are submerged in 1 M MZ2 (aq) solution (M is a metal and Z is an anion composed of the newly discovered element Z). Use the reference information given in the table above to answer the following three questions. Question 10 Oxidation will occur at the while reduction will occur at the Possible answers are "anode" and "cathode".Iron can be extracted from the iron(III) oxide found in iron ores (such as haematite) via an oxidation-reduction reaction with carbon. The thermochemical equation for this process is: 2 Fe2O3(s) + 3 C(s) → 4 Fe(l) + 3 CO2(9) ΔΗ = +467.9 kJ
- Calculate AG for the reaction 2 H2S(g) + SO2(g) 53 Srhombie(s) + 2 H2O(g) For the following conditions at 25°C: PH,8=0.0002 atm Pso, =0.0129 atm PH,0=0.0503 atm AG°F H2S(g) = -34 kJ/mol SO2(g) = -300. kJ/mol H20(g) = -229 kJ/mol AG kJ %3DUsing the equation 2Mn(OH)2(s)+O2(aq)->2MnO(OH)2(s), Determine the number of moles of MnO(OH)2 Formed for every one mole of O2 consumed. Show work.1
- help please answer in text form with proper workings and explanation for each and every part and steps with concept and introduction no AI no copy paste remember answer must be in proper format with all workinghelp please answer in text form with proper workings and explanation for each and every part and steps with concept and introduction no AI no copy paste remember answer must be in proper format with all workingCalculate AG for the reaction 2 H2S(g) + SO2(g) 53 Sthombic(s) + 2 H2O(g) For the following conditions at 25°C: PH,s=0.0001 atm Pso; =0.0290 atm PH,0=0.0219 atm AG°? H2S(g) = -34 kJ/mol SO2(g) = -300. kJ/mol H20(g) = -229 kJ/mol AG = kJ
- [References] Cryolite (Na3AIF6) is used in the commercial production of aluminum from ore. Cryolite itself is produced by the following reaction: 6 N2OH + Al203 + 12 HF → 2 Na3AlF6 + 9 H2O A mixture containing 440.0 kg of NaOH, 236.9 kg of Al203, and 600.0 kg of HF is heated to 950 °C until it reacts to completion. What is the maximum mass of Na3AIF6 formed? kg Na3AIF6Calculate AG for the formation of one mole of N₂O4 from its elements using AH and ASfº for N₂O4. Use data in the table below: Substance Ag(s) AgCl(s) Al(s) Al₂O3(s) C(s) (graphite) (8) CO₂(g) CH₂(g) CH₂Cl(g) CH₂OH(1) C₂H₂(g) C₂H4(g) C₂H6(g) C8H18 (7) C₂H₂OH(1) Ca(s) CaCO3(s) CaCl₂(s) 197.9 213.6 186.2 234.2 126.8 CO(NH,),(s) 104.6 CO(NH,),(aq) 173.8 200.8 219.8 229.5 466.9 161 CO(g) CO₂(g) CH4(g) Substance Ag(s) AgBr(s) AgCl(s) Al(s) Al₂O3(s) C(s) (graphite) Standard Entropies of Some Typical Substances at 298.15 K Sº (J mol-¹ K-¹) 40 76.1 107 C₂H₂(g) C₂H4(g) C₂H6(g) C₂H,OH(1) Ca(s) CaBr₂(s) CaCO3(s) CaCl₂(s) CaO(s) Cl₂(g) Fe(s) Sᵒ (J mol-¹ K-¹) Fe₂O3(s) H₂(g) H₂O(g) H₂O(1) AGO = CH₂Cl(g) CH₂I(g) CH₂OH(1) CO(NH₂)2(s) (urea) CO(NH,),(aq) kJ mol-1 42.55 96.2 28.3 51.0 Ca(OH)₂(s) CaSO4(s) CaSO4 H₂O(s) CaSO4-2H₂O(s) 5.69 41.4 92.9 114 Substance CaO(s) Ca(OH)₂(s) CaSO4(s) CaSO4+H₂O(s) 131 CaSO4-2H₂O(s) 194.0 223.0 27 Cl₂(g) Fe(s) Fe₂O3(s) H₂(g) H₂O(g) H₂O(1) HCI(g) HNO3(1) H₂SO4(1)…20. Use the data given below to find the equilibrium constant for the following reaction at 46 °C: Br2(1) + 21(aq) = 2 Br (aq) + 2(s) E° Br2 /Br= 1.07 V E° I2 /1 = 0.53 V (R = 8.314 J/mol.K = 0.0821 L.atm/mol.K, Faraday constant = 96500 C/mol.e) %3D 6.9 x 1017 O 9.1 x 1017 O 3.1 x 10*17 O 5.3 x 10*17 O 1.1 x 10*17
