Please don't provide handwriting solution ElementB, AlCN, PO, SF, Cl, Br, IFC = -2XX(1 possibility)XXFC = -1++(4 possibilities)0:0(3 possibilities)(2 possibilities)(1 possibility)(1 possibility)(3 possibilities)FC = 0(4 possibilities)(3 possibilities)(2 possibilities)(1 possibility)H3CFC = +1H3CX(3 possibilities)(4 possibilities)(3 possibilities)Determine the formal charge of all atoms with non-zero formal charge in the following reaction.CH3HI. IH-N-CH3 +C.HHCH3(2 possibilities)34: N-CH3 +HH-C-HHNote: Formal charge gives us reactivity information. Atoms lacking electrons (positive) reactwith atoms with an excess of electrons (negative).

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Element B, Al N, P cx O, S FC = -2 F, Cl, Br, I x tar X (1 possibility) X FC = -1 X (4 possibilities) (3 possibilities) (2 possibilities) (1 possibility) FC = 0 (1 possibility) (3 possibilities) (4 possibilities) (3 possibilities) (2 possibilities) (1 possibility) FC = +1 X (3 possibilities) (4 possibilities) (3 possibilities) (2 possibilities)

Element B, Al N, P cx O, S FC = -2 F, Cl, Br, I x tar X (1 possibility) X FC = -1 X (4 possibilities) (3 possibilities) (2 possibilities) (1 possibility) FC = 0 (1 possibility) (3 possibilities) (4 possibilities) (3 possibilities) (2 possibilities) (1 possibility) FC = +1 X (3 possibilities) (4 possibilities) (3 possibilities) (2 possibilities)

General Chemistry - Standalone book (MindTap Course List)

11th Edition

ISBN:9781305580343

Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell

Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell

Chapter6: Thermochemisty

Section: Chapter Questions

Problem 6.93QP

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Given these reactions, where X represents a generic metal or metalloid 1) H, (g) + 0,(g) → H,0(g) AH1 = -241.8 kJ 2) X(s) + 2 Cl,(g) → XCl,(s) AH2 = +475.9 kJ 3) H,(g) + CL,(g) HCl(g) АНз 3 —92.3 kJ → 4) X(s) + O,(g) –→ XO,(s) AH4 = -748.7 kJ 5) Н, О(2) — н,00) AH5 = -44.0 kJ what is the enthalpy, AH, for this reaction? XCI, (8) + 2H,О() — ХО,(8) + 4 HCI(g)
Consider the following data. 2 H₂0 (1) 2 H₂(g) + O₂(g) 2 HNO3(1) = N₂05(9) + H₂O(1) 2 HNO30 (1) ⇒ N₂(g) + 3 O₂(g) + H₂(g) Use Hess's law to calculate AH for the reaction below. ΔΗ= |1083.8 AH = +571.7 kJ AH = +92.0 kJ AH = +348.2 kJ x kJ 2 N₂05(9) 2 N₂(g) + 5 O₂(g)
[References] Calculate AH for the reaction: 2NH3 (g) + O2(g) → N2H4 (1) + H2O(1) given the following data: 2NH3 (g) + 3N2O(g) → 4N2(g) + 3H2O(1) AH =–1010. kJ N2O(9) + ЗH2(9) — N;Ha (1) + Hа0(1) ДН %3D — 317 kJ N2H4 (1) + O2 (9) → N2(g)+2H20(1) ΔΗ--623 k.J На (9) + О2(9) — Н,0(1) AH = -286 kJ ΔΗ- kJ Submit Answer Try Another Version 3 item attempts remaining
(i) Identify the different enthalpy changes and construct the Born-Haber cycle for the formation of sodium choride (NaCl) using the information below. Calculate the missing enthalpy change value Na+(g) + Cl-(g) → NaCl (s) ∆Ho = ? Cl2 (g) → Cl (g) ∆Ho = +121 kJmol-1 Cl (g) + e- → Cl-(g) ∆Ho = -364 kJmol-1 Na (s) → Na (g) ∆Ho = +108 kJmol-1 Na (g) → Na+(g) + e- ∆Ho = +500 kJmol-1 Na (s) + Cl2 (g) → NaCl (s) ∆Ho = -411 kJmol-1
The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area (1 watt = 1 J/s). The plants in an agricultural field produce the equivalent of 21 kg of sucrose (C12 H22 O11) per hour per hectare (1 ha = 10,000 m Assuming that sucrose is produced by the reaction m²). 12CO2 (9) + 11H2O(1) → C12H22O11(s) + 120O2(g) AH=5640 kJ calculate the percentage of sunlight used to produce the sucrose – that is, determine the efficiency of photosynthesis in this field. Percent efficiency
Given the following data: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) ΔΗ -23 kJ 3FE2O3(s) + CO(g) → 2FE3O4(s) + CO2(g) AH=-39 kJ FezO4(s) + CO(g) → 3F€O(s) + CO2(g) AH = 18 kJ calculate AH value for the reaction FeO(s) + CO(g) → Fe(s) + CO2(g) ΔΗ kJ
e Schoology 8 Calendar | Schoology -> A bcps.schoology.com/common-assessment-delivery/start/3334253476?action=Donresume&submissionld-492882996 BCPS Links 9 Schoology 9 Schoology Which line on Graph 2 accurately shows the relationship between distance and the amount of potential energy stored in this system? Graph 2 A В DISTANCE BETWEEN MAGNETS WITH LIKE POLES FACING O Line A O Line B O Neither line A nor Line B is correct 2 4 7 8 10 11 Next » Sign out POTENTIAL ENERGY
4. (i) Identify the different enthalpy changes and construct the Born-Haber cycle for the formation of sodium choride (NaCl) using the information below. Calculate the missing enthalpy change value Na*(g) + Cl(g) → NaCI (s) AH° = ? Cl2 (g) → CI (g) AH° = +121 kJmol1 CI (g) + e¨ → Cl(g) AH° = -364 kJmol1 Na (s) → Na (g) AH° = +108 kJmol1 Na (g) → Na*(g) + e` AH° = +500 kJmol1 Na (s) + Cl2 (g) → NaCI (s) AH°= -411 kJmol1
Given the following data PA(s) + 6C12 (9) → 4PCI3 (g) АН — — 1225.6 kJ PA(s) + 502(9) → P,O10 (s) AH = -2967.3 kJ PCI3 (g) + Cl2 (g) → PCI5 (g) АН — — 84.2 kJ PCI, (9) + 02 (9) → CI3PO(9) AĦ = -285.7 kJ Calculate AH for the reaction P,O10 (s) + 6PCI5 (9) → 10C13PO(9) ΔΗ- kJ
2HBr(aq)+Ba(OH)2(aq)----> 2h20(l)+BaBr2(aq)C2H4(g)+3O2(g) -----> 2CO(g)+2H2O(l)Cu(s)+FeCl2(aq) ----> Fe(s)+CuCl2(aq)Na2S(aq)+2AgNO3(aq) ---> 2NaNO3(aq)+Ag2S(s) identify the type acid-base neutralizationprecipitationredox
The net change in the multistep biochemical process of photosynthesis is that CO₂ and H₂O form glucose (C₆H₁₂O₆)and O₂. Chlorophyll absorbs light in the 600 to 700 nm region.(a) Write a balanced thermochemical equation for formation of1.00 mol of glucose. (b) What is the minimum number of pho-tons with λ=680. nm needed to prepare 1.00 mol of glucose?
Consider the problem below: (Equation 1) CaCO3(s) → CaO(s) + CO2(g) AH = 178 kJ/mol (Equation 2) CO(g) + O3(g) → CO2(9) + 20(g) AH = -425.7 kJ/mol %3D (Equation 3) 2 O(a) → O2(a) AH = -498 kJ/mol (Equation 4) CaCO3(s) + O2(9) → CaO(s) + CO(g) + O3(9) AH = ??????? The AH for equation 4 is: (express in kJ) Screen reader version Consider the problem below: (Equation 1) C a CO3(s) arrow Ca Ols) + CO22) AH = 178 kJ/mol (Equation 2) CO(g) + O3(g) arrow CO2(g) + 20(2) AH = minus 425.7 kJ/mol (Equation 3) 2 O(2) arrow O2(e) AH = minus 498 kJ/mol (Equation 4) C a CO3(s) + O2(g) arrow C a O(s) + C Og) + O3(2) AH = ??????? %3D The AH for equation 4 is: _(express in kJ)

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General Chemistry - Standalone book (MindTap Cour…

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9781305580343

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Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell

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