Acetic Acid‚ when dissolved in H2O CH3COOH (aq) + H2O (l) (ACID) (BASE) CH3COO (aq) + H3O(l) (CONJ BASE) (CONJ ACID) CH3COOH (aq) + H2O(l) ↔ CH3COO (aq) + H3O(l) *ONLY weak acid and back can be reversible Equation for Ka : Keq= [CH3COO(aq)] [H3O(l)][CH3COOH(aq)] [H2O(l)] Keq x [H2O (l)] = Ka Ka= [CH3COO(aq)] [H3O(l)] [CH3COOH(aq)] Example Question: Calculate the pH at equilibrium for a 1.0 mol/L Ka= 1.8x10-5 CH3COOH (aq) + H2O(l) ↔ CH3COO (aq) + H3O(l)
Premium Acid dissociation constant PH Acetic acid
The Diaspora has been an integral part of the African culture for many >years. Since slaver‚ people of African decent have dealt with the imposed >"veil"; it represents society’s view of African Americans as a problem and >it’s refusal to accept the African Americans as contributive members to >their society. Moreover‚ the "Diaspora" deals with the "double >consciousness‚" the looking of self through the eyes of others while being >graded by the society around the African American. The
Premium African American Black people
The Effect of Temperature on Solubility By Aviraj Singh Rogers 2 Background: The solubility of most solid substances is generally said to increase as the temperature of the solvent increases. However‚ some substances‚ such as ytterbium sulfate‚ do the opposite. This can be explained through the Second Law of Thermodynamics which states that “in all energy exchanges‚ if no energy enters or leaves the system‚ the potential energy of the state will always be less than that of the initial state”
Premium Energy Kinetic energy Temperature
The percent yield is a ratio of the ____________ yield to the _____________ yield‚ multiplied by 100%. 20. 0.1156 g of a compound reacted with oxygen gives 0.1638 g CO2 and 0.1676 g of H2O‚ rest is nitrogen. What is the formula of the compound? CO2 = 44.01 g/mole H2O=18.02 g/mole Fraction of C in CO2: mass of C/total mass of CO2= 12.01g of C/44.01g of CO2 0.1638 g of CO2 x12.01g of C/44.01g of CO2 = 0.0447 g of C 0.0447g of C/0.1156 g compound x 100 = 38.67%
Premium Oxygen Chlorine Carbon dioxide
Molar Mass of H2O = 2 x (1.01) + 16.00 = 18.02gmol1- How many moles of H2O evaporated Trial 1: 0.72g±0.04g/18.02gmol1- = 0.03995… ≈ 0.040mol±6.25% = 0.040mol±0.003mol Molar Mass of CuSO4 Cu: 63.55 gmol1- S=32.07 gmol1- O x 4= (16.00) x 4=64.00 gmol1- CuSO4= 159.62 gmol1- Moles of anhydrous part (CuSO4) 1.35g±0.02g/159.62 gmol1-=0.0084575… ≈ 0.00846mol±1.48%=0.00846mol±0.00013mol Moles of H2O in 1 mole of CuSO4 0.00846mol of CuSO4-----------> 0.040mol of H2O 1mol of CuSO4 ------------------>
Premium Gender Mole
there would likely be no water for consumption. Thus‚ the study and importance of evaporation in medications would be of little concern if humans were void water or it became scarce. That would mean humans couldn’t evolve and survive on Earth. In conclusion‚ the researcher intends to proceed with experimentation as it relates to various common liquids‚ including H2O‚ and its evaporation effects over time. The
Premium Water Evaporation Humidity
Year 11 Module 1 – The Chemical Earth The Earth is made up of a large number of different substances: elements‚ compounds & mixtures We can classify elements & compounds as pure substances. These are always HOMOGENEOUS. Mixtures are not pure substances and can be either homogeneous or HETEROGENEOUS. Pure substances have a fixed composition eg copper metal‚ sulfur‚ carbon dioxide (CO2)‚ methane (CH4). Mixtures have variable composition and can also be separated into its components relatively
Free Atom Chemical bond Chemistry
limiting reactant is high. In order to increase the percent yield of product‚ increasing the limiting reactant‚ possibly‚ is the most effective. In this experiment we were able to calculate limiting reactants from the reaction of CaCl2. 2H2O + K2C2O4.H2O(aq). As a group‚ we obtained our salt mixture of calcium chloride and potassium oxalate‚ and weighed the mixture. We were able to make an aqueous solution from the mixture and distilled water. We boiled and filtered off the solution‚ leaving the precipitate
Premium Stoichiometry Chemical reaction Chemistry
University of Petroleum and Minerals Mechanical Engineering Department ME203 Thermodynamics Quiz # 2 (Term 072) Saturday 22-3-2008 35 Minutes Student Name ID # A Fill the following table Substance H2O H2O H2O R134a R134a T‚ oC 250 105 340 Substance H2O H2O H2O R134a R134a P‚ kPa 1800 v‚ m3/kg u kJ/kg x Phase description 1 0.025 50 8200 360 2000 T‚ oC 250 105 340 5.82 50 P‚ kPa 1800 120.8 8200 360 2000 v‚ m3/kg 0.125 1
Premium Pressure Thermodynamics
AP Chemistry Analysis of Alum * Purpose: * In this lab we performed several tests to determine if our crystals were actually aluminum potassium sulfate. * Procedures: * Materials: * Chemicals: * Aluminum potassium sulfate‚ 2.5 g * Equipment – Part 1: * 150 mL beaker * Bunsen burner * 2 capillary tubes * Mortar and pestle * Notched stopper to hold thermometer *
Premium Chemistry Water Alum