.1-amEr
(c) Gram molecular mass of HNO3 = Mass of 1-molecule of HNO3x NA
= 63 amu x NA = 63 gm X NA = 63 gram
NA
Solved Example-4:Find out the mass of carbon -12 that would contain 1.0 x1019carbon-12 atoms. Solution : Mass of 6.022 x1023 carbon-12 atoms = 12 g
Mass of 1.0 x1019 carbon-12 atoms = 12x1x10 19 g
6.022 x1023 = 1.99 x10-4g
Solved Example-5: How many molecules are present in 100 g sample of NH3?
Solution : Molar mass of NH3 = (14 + 3) g mol-1 = 17 g mol-1
17 g sample of NH3 contains 6.022 x1023 molecules
Therefore, 100 g sample of NH3 would contain = 6.022x10 23 x 100g molecule
I 7g '
= 35.42x1023 g molecules.
= 3.542x1024 g molecules.
Solved Example -6 : Molar mass of 0 is 16 g mot-1. What is the mass of one atom and one molecule of oxygen? Solution : Mass of 1 mol or 6.022 x1023 atoms of 0 = 16 g
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(b) Calculate the volume occupied by 8 g of sulphur dioxide at STP[S = 32; 0 =
Next, we determined the mass of the penny by placing it on a balance. The mass of the penny was 2.47 grams. Afterwards, we placed the penny in a beaker filled with 20 mL of 6 M HCl. In the end we put the beaker in the fume hood and allowed it to sit overnight. During day two of the penny lab, we removed the penny skin from the beaker using tweezers.
Question3: Experiment 3 The unknown acid sample was 1 • Monoprotic Acid Trails Initial NaOH solution (mL) final NaOH solution (mL) The volume of NaOH to titrate the acid (mL) Amount of Unknown Acid sample 1 (g) The moles of the Unknown Acid (mol) Molar mass of the Unknown Acid (g/mol) A 3.38 28.31 24.93 0.150 0.0026 57.69 B 0.18 29.32 29.14 0.175 0.0029
Atomic weight = mass of protons + mass of neutrons. Question # 1: Part C Calculate the density of vanadium. The structure is BCC.
In order to begin this experiment, first one must find the balanced chemical equation for the reaction which occurs between the aluminum and copper (II) chloride. This balanced equation being 2Al(s)+3CuCl2 (aq)3Cu(s)+2AlCl3 (aq). After finding this equation, one must use the process of stoichiometry in order to find how many grams of aluminum are needed in order to produce 0.15 grams of copper. In this experiment, the purpose was to produce between 0.1 and 0.2 grams of copper, so one should attempt to produce 0.15 grams of copper seeing as it is the average of those two numbers. The first step in the stoichiometric process which one has to complete is finding how many grams of copper are in one mole of copper.
In addition, for calculating the mass of the pennies and the unknown substance, zero the balance and place it in the weigh boat to receive the data. Lastly, to calculate the density of the substances, use the formula D=MV, in which dividing the mass by the volume allows to do so. In order to be certain of the data that is collected, running multiple trials could help be accurate. A method to get an average value of the density can be expressed by the formula D1+D22. This formula basically allows one to add the data of all the trials and divide it by the number of trials that was performed.
Which of the following is the appropriate prefix to measure the amount of space taken up? A. Mili B.Deca C.Kilo D.Micro 38. In addition to oxygen and carbon dioxide, the circulatory system is the primary delivery system for? A.cerebrospinal fluid B.exocrine secretions C.biliary fluids D.endocrine hormones 39.
Bradley Trotter & John Fussell Chem 1121-42 March 29th, 2018 Identity of an Unknown Weak Acid Lab Report Introduction: Being able to determing the molar mass of an unknown solution is one of the many ways to identify what the unknown solution is. The use of titrations and LoggerPro, make extrapolating data more precise and will produce more accurate results than if done manually. By comparing the caculated molar mass, and pKa values, with accepted values a hypothesis can be made to the identity of that unknown.
Post Lab Questions: 2Mg + O2 →2MgO 0.29gMg 1molMg24.312molMgO2molMg100= 4.40molMgO ←theoretical yield 2Mg + O2 → 2MgO →2/24 =
Molar mass is the quantity in grams of one mole of a substance. The molar mass is found by adding the number of grams of each element to find the total amount of grams per mole. For example CaH2 has the elements of Calcium which has the molar mass of (40.078g/mole) and two Hydrogen which is (2 x 1.0079 g/mol).When the mass of both elements are added the molar mass of CaH2 is 42.0938g/mol. Moles are really important because it allows chemist to be able to have a standardize meausment which gives them the chance to compare compounds.
Weight of KCl heated = (Weight of test tube + Mn+ KCl) – (Weight of test tube + Mn) = 46.428g – 45.490g = 0.938g 2. Determine the weight of oxygen gas lost using the constant weight obtained from repeated heatings? Weight of oxygen gas lost =
Next, there is mass this contains 5 sections in it these are: • challenge
Stoichiometry is a method used in chemistry that involves using relationships between reactants and products in a chemical reaction, to determine a desired quantitative data. The purpose of the lab was to devise a method to determine the percent composition of NaHCO3 in an unknown mixture of compounds NaHCO3 and Na2CO. Heating the mixture of these two compounds will cause a decomposition reaction. Solid NaHCO3 chemically decomposes into gaseous carbon dioxide and water, via the following reaction: 2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g). The decomposition reaction was performed in a crucible and heated with a Bunsen burner.
There are many gravimetric analysis methods, but the one used in this lab, measured mass via precipitation. The ion, analyte, Strontium Carbonate was precipitated in a Büchner funnel apparatus with 150 mL of deionized water and a calculated amount of Strontium Chloride
Procedure A. Preparation of NaOH solution The molarity of a solution is the ratio of the number of solutes dissolved in a liter of solution. To figure out the needed mass (in grams) of NaOH pellets to be dissolved in a 0.25 L of water, remember that a mole is equivalent to the quotient of mass over the molar mass of the substance. This was used to rearrange the base formula and to derive the mathematical equation of mass in terms of molarity. mass (g) =
℃^(-1)×6.40℃±3.1 %=1337.6 J±4.06 % ∆H=(-1337.6 J±4.06 %) /(0.025 mol ±0.16 %)= -53504 J m〖ol〗^(-1)±4.22 % ∆H=-53504 J m〖ol〗^(-1)±4.22 %÷1000=-54 kJ m〖ol〗^(-1)±4.22 % Conclusion and