93 g/mol? Not we get to use it! Yay! 93 g/mol / 31.06 g/mol = 3 (this is the multiplier) Multiply that whole number through the subscripts of the empirical formula. 3 x (C H5 N) = C3H15N3 Hydrated compounds Solving process: 1st- the difference between the initial mass and that of the dry sample is the mass of water that was driven off.
0.05mol/6M=8.3*10-3 L=8.3mL stock solution c. 100mL-8.3mL=91.7mLwater Add 91.7 water to 6M stock solution to prepare 0.5M acetic acid. Exercise 8: a. 42.35 - 0.55 = 41.8 mL b. The moles of EDTA4- : 0.0189M*(41.8*10-3)L=7.9*10-4mol c. Zn2+(aq)+EDTA4-(aq)—Zn(EDTA)2-(aq) The ratio Zn2+ and EDTA4- is 1:1 The moles of Zn2+= the moles of EDTA4-=7.9*10-4mol d. 7.9*10-4mol*65.39g/mol=0.0517g Zn e.
b) Calculate the enthalpy of combustion per gram of each sugar. C6H12O6 + 6O2 --> 6CO2 + 6H2O (-383.5 x 6 + -285.8 x 6) - (-1273) = -2742.8kJ C12H22O11 + 12O2 --> 12CO2 + 11H2O (-383.5x12 + -285.8x11) - (-2221) = -5524.8kJ -2742.8kJ / 180g/mole = -15.24kJ/g -5524.8kJ / 342g/mole = -16.14kJ/g 107. Three common hydrocarbons that contain four carbons are listed here, along wiht their standard enthalpies of formation: Formula Delta H in kJ per mol 1,3-Butadiene C4H6(g) 111.9 1-Butene C4H8(g) 1.2 n-Butane C4H10(g) -124.7 a. For each of these substances, calculate the molar enthalpy of combustion to CO2(g) and H2O(l). b.
25.0 cm3 of this solution was titrated against 0.1 moldm-3 HCl and 24.5 cm3 of the acid were required. Calculate the value of x given the equation: Na2CO3 + 2HCl → 2NaCl + CO2 + H2O 6. 25 cm3 of a sample of vinegar (CH3COOH) was pipetted into a volumetric flask and the volume was made up to 250 cm3. This solution was placed in a burette and 13.9 cm3 were required to neutralise 25 cm3 of 0.1 moldm-3 NaOH. Calculate the molarity of the original vinegar solution and its concentration in gdm-3, given that it reacts with NaOH in a 1:1 ratio.
Step 2. Heat test tube until all of the O2 has evolved Step 3. Record the amount of O2 produced Observations Mass of O2 produced: 3.915576 grams or .1224 moles Mass of KCl produced: 6.083363 grams or .0816 moles Calculations Chemical Reaction 2KClO3-> 2KCl + 3O2 Theoretical Yield (96/245.1)*10 grams= 3.916 grams of O2 Percent Yield 3.915576/3.916= .99989 *100= 99.998 percent Data Actual amount of Oxygen Produced |3.915576 grams | |Theoretical amount of oxygen Produced |3.916 grams of O2 | |The Percent Yield |99.998 percent | | Results The amount of oxygen produced was almost exactly what was expected with a percent yield of 99.998 percent. The Law of conservation of mass was also up held with there being a combined mass of 10 grams from the resulting O2 and KCl. Conclusion 10 Grams of Potassium chlorate when decomposed produces 3.915576 grams oxygen gas and 6.083363 grams potassium chloride Atomic Weight of Magnesium Introduction In this lab we will determine the atomic weight of magnesium by measuring the amount of hydrogen gas evolved when hydrochloric acid reacts with magnesium.
Chem 112 Assignment for Chapter 19 1. Calculate ΔS° values for the following reactions by using tabulated S° values from Appendix B (Petrucci text) a) N2H4 (g) + H2 (g) → 2NH3 (g) b) 2Al (s) + 3 Cl2 (g) → 2 AlCl3 (s) c) Mg(OH)2 (s) + 2 HCl (g) → MgCl2 (s) + 2 H2O (l) d) 2 CH4 (g) → C2H6 (g) + H2 (g) 2. Use data in your Appendix B to find ΔH° and ΔS° values at 25°C to calculate ΔG° where ΔG° = ΔH° - T ΔS° for each of the following reactions. a) Ni (s) + Cl2 (g) → NiCl2 (s) b) CaCO3 (s, calcite) → CaO (s) + CO2 (g) c) P4O10 (s) + 6 H2O (l) → 4 H3PO4 (aq) d) 2 CH3OH (l) + 3 O2 (g) → 2 CO2 (g) + 4 H2O (l) Note: the following values ΔHf° (kJ/mol) S° (J/K) Ni (s) 0 29.9 NiCl2
2 Subj. 3 Average 30 min (mL) 43.4 37.9 38.2 L/day 2.1 1.8 1.8 2.1 Dehydrated 30 min (mL) L/day 13 0.6 16.4 0.8 12.8 0.6 0.7 Water loaded 30 min (mL) 155.2 141.9 156.2 Graph 1. Average Daily Urine Production Under Different Hydration States 1. Does dehydration increase, decrease, or not change average urine production rate (L/day). Laboratory Report/ Marlinda Saintil/ Influence of Fluid Intake on Urine Formation/ Florence Vicil/ 05.04.2015/ Page  of  L/day 7.4 6.8 7.5 7.3 2.
(Make a rough titration first). 5. Calculate the moles and mass of acetylsalicylic acid in an aspirin tablet. Determine the mass percentage of acid in a tablet. Data collection VolumeV / cm3 | Titration trials | | | | 1 | 2 | 3 | 4 | 5 | Initial burette reading V / cm3±0.05 cm3 | 0.00 | 0.00 | 0.00 | 0.00 | 0.00 | Final burette reading V / cm3±0.05 cm3 | 13.6 | 13.9 | 14.0 | 13.9 | 14.1 | Total tire V / cm3 ±0.1 cm3 | 13.6 | 13.9 | 14.0 | 13.9 | 14.1 | Table 2 Volume of HNO3 required titrating NaOH in excess Mass of aspirin tablet ±0.001 [g] | 3.229 | 3.143 | 3.142
5 12WAB10835 Dioxane; 7 123-91-1 GC/FID; CSC 5 12WAB09523 Diphenylamine; 22; Fully Validated 122-39-4 HPLC/UV; Bubbler with Isopropyl Alcohol 5 12WAB10906 Diphenylamine; 78; Fully Validated 122-39-4 HPLC/UV; GFF coated with Sulfuric Acid 5 12WAB10073 Diphenyl; PV2022; Partially Validated 92-52-4 GC/FID; XAD-7 5 12WAB10850 Dipropyl Disulfide; PV2086; Partially Validated 629-19-6 GC/FID; Chromosorb 106 5 12WAB10450 Dipropylene Glycol Methyl Ether; 101; Fully Validated 34590-94-8 GC/FID; CSC 5 12WAB10905 Disulfoton; PV2105; Partially Validated 298-04-4 GC/FPD; OVS-2 5 12WAB11596 Diuron; PV2097; Partially Validated 330-54-1 HPLC/UV; OVS-2 5 12WAR11371 Divinylbenzene; 89; Fully Validated 108-57-6 GC/FID; CSC coated with TBC 5 12WAB10451 Dursban; 62; Fully Validated 2921-88-2 GC/FPD; OVS-2 5 12WAB10449 Endosulfan; PV2023; Partially Validated 115-29-7 GC/ECD; OVS-2 5 12WABI0801 Enflurane; 29; Fully Validated 13838-16-9 GC/FID; CSC 5 12WAB10448 Enflurane; 103; Fully Validated 13838-16-9 GC/FID; Anasorb CMS or Anasorb 747 5 12WAB09860 Epichlorohydrin; 7 106-89-8 GC/FID; CSC 5 12WAB09198 Estradiol; PV2001; Partially Validated 50-28-2 HPLC/UV; GFF 5 12WAB10447 Estriol; PV2001; Partially Validated 50-27-1 HPLC/UV;