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- Would you help me whit this example (3) Thank you.ΔH° f (kJ/mol) SO 2Cl 2( g) + 2 H 2O( l) → 2 HCl( g) + H 2SO4( l) ΔH° rxn = ? SO 2Cl 2( g) -364H 2O( l) -286HCl( g) -192H 2SO 4( l) +314Combustion of a compound containing just C, H, N, and O showed that it contains 46.21 wt% C, 9.02 wt% H, 13.74 wt% N, and, by difference, 100 - 46.21 - 9.02 - 13.74 = 31.03 wt% O. That is, 100 g of unknown contain 46.21 g of C, 9.02 g of H, and so on. Find the atomic ratio C:H:N:O. Divide each stoichiometry coefficient by the smallest one and express the composition in the lowest reasonable integer ratio (CxHyNzOw, where x, y, z, and w are integers and one of them is 1).
- The following results are obtained from combustion studies of an unknown hydrocarbon: C S2. 14 percent, H 13.13 percent, 034.737. For mula, actermine is the empiracle.General formula of a polymer is-(CH2CHCHCN)n-, where n is typically greater than 10,000. Then take a sample of this polymer weighs 755.9 g and contains 3.112 x 1020 molecules of -(CH2CHCHCN)n-, so calculate n?Mass of sample (g ) 1.94g Mass of empty crucible (g) 84.248g Crucible and sample after heating (g) 85.376g Mass of residue ( KCl ) (g) Mass of O2 Percent O2
- A 0.235 mL sample of carbon tetrachloride (density = 1.59 g/mL, MM = 153.81 g/mol) contains how many AMU of Cl? Note: Use 6.022 x 1023 and/or 1.661 x 10-24 as appropriate. 5.18 x 1022 2.42 x 10-3 2.07 x 1020 1.30 x 1022 2.07 x 1023 0.333 g of a hydrocarbon (CxHy) was analyzed by combustion analysis and 1.009 g of CO2 (MM = 44.01 g/mol) and 0.516 g of H2O (MM = 18.02 g/mol). Which of the following is the correct empirical formula of the hydrocarbon? C2H4 C2H5 C3H7 C2H3 CH3Covert 0.618g H2O to molesHow many ,mew mg are there in 1 mg?
- Calcium ΔHof (kJ/mol) ΔGof (kJ/mol) So (J/mol K) Ca (s) 0 0 41.4 Ca (g) 178.2 144.3 158.9 Ca2+ (g) 1925.9 CaC2 (s) -59.8 -64.9 70.0 CaCO3 (s, calcite) -1206.9 -1128.8 92.9 CaCl2 (s) -795.8 -748.1 104.6 CaF2 (s) -1219.6 -1167.3 68.9 CaH2 (s) -186.2 -147.2 42.0 CaO (s) -635.1 -604.0 39.8 CaS (s) -482.4 -477.4 56.5 Ca(OH)2 (s) -986.1 -898.5 83.4 Ca(OH)2 (aq) -1002.8 -868.1 -74.5 Ca3(PO4)2 (s) -4126.0 -3890.0 241.0 CaSO4 (s) -1434.1 -1321.8 106.7 CaSiO3 (s) -1630.0 -1550.0 84.0 Carbon ΔHof (kJ/mol) ΔGof (kJ/mol) So (J/mol K) C (s, graphite) 0 0 5.7 C (s, diamond) 1.9 2.9 2.4 C (g) 716.7 671.3 158.1 CCl4 (l) -135.4 -65.2 216.4 CCl4 (g) -102.9 -60.6 309.9 CHCl3 (l) -134.5 -73.7 201.7 CHCl3 (g) -103.1 -70.3 295.7 CH4 (g) -74.8 -50.7 186.3 CH3OH (g)…Mass of empty beaker=16.26Volume of CuSO4 5H2O=30mLMolarity of CuSO4 5H2O used=0.542Moles of CuSO4 5H2O=0.016Mass of aluminum foil used=0.26gMass of beaker and copper product=52.12gMass of copper metal product=35.86gMoles of Al used = 0.009636 molMoles of CuSO4 5H2O used=0.016mol Find the moles of copper product based on moles of Al.Find the moles of copper product based on moles of CuSO4 5H2OFind the limiting reactant in gramsWrite the empirical formula for the hydrated KAl(SO4)2, based on moles of anhydrous KSI(SO4)2= 0.046 mol molar mass of H2O= 18g/ mol moles of H2O= 0.0444mol ratio pf moles H20 to moles of anhydrous KAI(SO4)2= 9.65/1 . Show all work including units. Hint: if the ratio of moles of H2O to moles of anhydrous KAl(SO4)2 was 4, then the empirical formula would be: KAl(SO4)2•4H2O.