c) 0.08 g of oxygen reacted with the magnesium (0.22 g – 0.14 g = 0.08 g). d) Refer to “Determining the Empirical Formula of Magnesium Oxide Lab Solutions” sheet e) Element | % | m (g) | M (g/mol) | n (m÷M) | ÷ by | Ratio | Magnesium | 63.63 | 63.63 | 24.31 | 2.617441382 | 2.273125 | 1 | Oxygen | 36.37 | 36.37 | 16.00 | 2.273125 | 2.273125 | 1 | Since the ratio is 1:1, the empirical formula is MgO. 2. Refer to “Determining the Empirical Formula of Magnesium Oxide Lab Solutions” sheet 3. You need to round the empirical formula to a whole number ratio because you cannot have decimals in the subscripts, which means that you cannot have a fractional amount of molecules in a substance.
Water (H2O) is colorless liquid that’s the basis of life on Earth. It contains two hydrogen bonds and is a polar molecule; it has a dipole moment of 1.85D[2]. Volume percent is a method used when preparing solution of liquids. It is defined as Concentration solute (v/v%) = [(volume of solute)/(volume of solution)] x 100% [5] This experiment is to use volume percent to create different concentration of 3M nitric acid solutions and investigate the effect of different concentration, which creates different strength of intermolecular forces, on the volume of a drop of nitric acid. Research question: What is the effect of different concentration of nitric acid on its volume of its drop?
KSCN ionizes into K+ and SCN-, and in the presence of the H+ ion supplied by nitric acid, the H+ and SCN- will combine to form the weak acid HSCN. Since there is a large excess of nitric acid compared to KSCN, we can assume that all of the SCN- will be in the form of HSCN. Procedure part 2: Six test tubes were filled with various amounts of 0.003M sodium thiocynate. A constant amount of 10 ml of 0.003 M Fe(NO3)3. The first tube had no sodium thiocynate.
a) mass c. color b) density d. boiling point 4) An example of an element is . a) sugar c. water b) soil d. oxygen 5) An example of a heterogeneous mixture is . a) salt c. mud b) nitrogen d. air 6) Which process is a chemical change? a) heating to boiling b) dissolving in alcohol c) burning in air d) slicing into two pieces 7) Water boiling at 100°C is an example of a(n) . a) chemical property c. physical property b) extensive property d. chemical change 8) Every pure chemical compound consists of two or more elements .
|is less than the mass of the reactants | |b. |is greater than the mass of the reactants | |c. |is equal to the mass of the reactants | |d. |has no relationship to the mass of the reactants | ____ 23. Calculate the number of moles of Al[pic]O[pic] that are produced when 0.60 mol of Fe is produced in the following reaction.
How is the formula of a conjugate acid similar to its conjugate base, and how is it different? (2) They all transform to same molecules, vut Conjugate acid donates an H+ ion and conjugate base gains an H+ ion 10. Show how ammonia, NH3, is a base according to the Bronsted-Lowry definition, but not according to the Arrhenius definition. (Hint: See #8a.) (2) Although NH3 forms OH- with water, it did not donate any proton to form OH- but only gain.
Title : Reaction of Carboxylic Acids Objective : To Determine The Reaction of Carboxylic Acids Procedure : As referred to Lab manual. Results: A. Salt Formation |Compound |Solvent |Solubility | |Benzoic Acid |Cold Water |Partially soluble | |Benzoic Acid |10 % NaOH |Soluble | B. Salt Hydrolysis Sodium acetate solution changed the color of litmus from red to blue. C. Reaction With Sodium Carbonate Observation : A lot of gas bubbles was released.
The apparent reaction rate was calculated using the equation, kapp= (1/∆t) x ([S2O32-]/[S2O82-]) which resulted in apparent rate constants of 5.66 x 10-5 s-1, 6.1958 x 10-5 s-1, 6.0356 x 10-5 s-1. The actual concentration was calculated using the basic chemical equation, C1V1 = C2V2. In order to find the order of reaction a a graph of log rate vs. log [S2O82-] was drawn, and was found that the results gave a zero order reaction But in reality the reaction order in [I-] and [S2O82-] is in first order each, although [I-] is kept at a constant volume throughout the reaction therefore the overall reaction is pseudo- first order. -d [S2O82-] = kapp [S2O82-] Dt In the second part the rate constant was found using the equation k = (1/∆t) x ([S2O32- ]/[Iodine][S2O82-]). Where it resulted to values of 3.990602 x 10-3 s-1, 4.653278 x 10-3 s-1, 5.944044 x 10-3 s-1, 7.499958 x 10-3 s-1, 7.499958 x 10-3 s-1, 9.84554 x 10-3 s-1, for flasks 4, 5, 6, 7, 8.
Main – group metals usually for one cation (positive ion). In a binary ionic compound the metal (cation) is named first. Then the nonmetal (anion) is named, and the suffix -ide is added. To create the formula, you switch the charges, and that tells you how many of each element you will need. For example: Cation Anion Formula Name of Compound Ba2+ I- Ba2+ I- Barium Ion Iodide Ion BaI2 Barium Iodide Type 2: Binary Ionic Type 2 Binary Ionic compounds consist of a metal and a nonmetal.
Si, P, S and Cl are increasingly electronegative. They cannot form oxide ions from oxygen at all, so they behave as acids. The trend is from strongly basic oxides on the left-hand side to strongly acidic ones on the right, via an amphoteric oxide (aluminium oxide) in the middle. In addition, when going across the period 3 the acid-base nature of the oxides change from base to amphoteric and finally to become acid. This is because the less electronegative sodium has a weak Na-O bond and the oxygen is more easily given up to reacts with H+.