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. 7. 2.5 g of a sample of ethanedioic acid, H2C2O4.nH2O, was dissolved in water and the solution made up to 250 cm3. This solution was placed in a burette and 15.8 cm3 were required to neutralise 25 cm3 of 0.1 moldm-3 NaOH. Given that ethanedioic acid reacts with NaOH
Computer Additivity of Heats of Reaction: Hess’s Law 18 (1) Solid sodium hydroxide dissolves in water to form an aqueous solution of ions. (2) Solid sodium hydroxide reacts with aqueous hydrochloric acid to form water and an aqueous solution of sodium chloride. NaOH(s) + H+(aq) ) + Cl–(aq) → H2O(l) + Na+(aq) + Cl–(aq) ∆H2 = ? OBJECTIVES • • • • In this experiment, you will Combine equations for two reactions to obtain the equation for a third reaction. Use a calorimeter to measure the temperature change in each of three reactions.
The Ksp of Magnesium Oxalate Abstract The Ksp for the acid catalyzed titration of the saturated oxalate is 1.8 x 10-3. Introduction In this experiment, the solubility equilibrium for the salt magnesium oxalate must be found in order to determine a solubility product constant. Solubility equilibrium is a type of dynamic equilibrium which exists when a chemical compound in the solid state is in chemical equilibrium with a solution of that compound. At the point of equilibrium the solution becomes saturated. The chemical reaction used to find this constant is as follows: MgC2O4 (s) ↔Mg(aq)2++ C2O4 (aq)2- Kc= Mg2+[C2O42-][MgC2O4] Ksp=Mg2+[C2O42-] The solid salt magnesium oxalate is prepared through the following precipitation reaction: Mg(SO4)(aq)+NaC2O4 (aq) → MgC2O4 (s)+NaSO4 (aq) Next, the concentration of the Mg2+ and C2O42- ions is found through a redox titration.
The following data were obtained when a sample of barium chloride hydrate was analyzed as described in the Procedure section. Calculate (a) the mass of the hydrate, (b) the mass of water lost during heating, and (c) the percent water in the hydrate. Mass of empty test tube 18.42 g Mass of test tube and hydrate (before heating) 20.75 g Mass of test tube and anhydrous salt (after heating) 20.41 g. Mass of the Hydrate is 2.33g. Loss (H2O) is 0.34g. Percent H2O in Hydrate is equal 0.34/2.33=14.6% 3.
How do you go from calculated/measured values to accomplishing stated purpose? (4 pts) The purpose of this lab is to determine the exact concentration of an unknown acetic acid solution. Titrations for two different reactions will be performed. Controlled volumes of one reagent are added to a flask containing the other reagent until the equivalence point of the reaction is reached, one between sodium hydroxide (NaOH) and potassium hydrogen phthalate (KHP) and the other between NaOH and acetic acid (CH3COOH). NaOH (aq) + KHP (aq) —› Na+ (aq) + K + (aq) + P2- (aq) + H2O (l) NaOH (aq) + CH3COOH (aq) —› Na+ (aq) + CH3COO- (aq) + H2O (l) The titration of NaOH with KHP will identify the concentration of the NaOH provided.
The specific heat constant for water, 4.184 J/g C, is used for this equation. The specific heat can be found by using The Law of Dulong and Petit: Eq. 3 Cs(aluminum) = slope x 1/atomic weight This equation is used to find specific heat from the graph that will be drawn based on the results of the metal specific heats. II. Materials and Procedure See General Chemistry 101/102 Laboratory Manual (pg.
The purpose of this lab was to find the molecular weight of two unknown substances by analyzing the freezing points in cyclohexane and to provide a visual representation of the freezing point depression effect. The theory of this lab is; by using measurements of mass of the unknown substances (solute) in correlation with the mass of the cyclohexane (solvent) and the freezing point constant of the solvent, you could determine the molecular weight of the solute by using the same math involved in deciding the freezing point of the solution/ The theory behind the visual of the FPD effect is that if one was to record the freezing point of a solution and two solutions of the same substances with more solute, one would see a visible drop in freezing point. The equations you needed for this lab were the freezing point formula for organic substances (ΔFp=(m)Kf) and the
1) Jeffrey Cox CHE111-DL01 Lab number 10 Stoichiometry of a Precipitation Reaction 2) Purpose/ Intro. In this lab we will be able to calculate the actual, theoretical, and percent yield of the product from a precipitation reaction. We will thusly learn the concepts of solubility and the formation of a precipitate. A precipitate reaction is a reaction in which soluble ions in separate solutions are mixed together to form an insoluble compound that settles out of the combined solution as a solid. The solid then is the insoluble compound, called a precipitate.
In this experiment, the amount of energy (heat) involved in a chemical change will be determined. When alcohol burns it produces carbon dioxide and water as products. Energy is also released in the reaction. The alcohols that will be used are, methanol CH3OH, ethanol CH3CH2OH, propanol CH3(CH)2OH, butanol C4H9OH, octanol C8H18O, and paraffin wax C25H52. The heat obtained when a known mass of alcohol or paraffin wax burns will be used to warm a measured volume of water.