amounts of a substance in terms of moles? The compound I was using in my experiment was NaOH. The first thing I needed to work out was the number of moles I needed. To do this I used the formula: Number of moles = Mass ÷ RFM. The mass is the amount that I weighed on the balance which was 1g. I then worked out the RFM of NaOH which is 40. I next put my mass and the RFM into the formula to find out the number of moles. Number of moles = 1 ÷ 40 Number of moles = 0.025
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number of moles of water molecules per mole of salt. Sample Calculations: Mass of hydrate= 35.232g - 32.005g =3.227g Mass of anhydrous salt= 33.583g - 32.005g =1.581g Mass of water liberated= 35.232g – 33.586g =1.646g Mass of h2O in hydrate= 1.646g/3.227g X 100 = 51% % anhydrous salt in hydrate= 1.581g/3.227g X 100 = 49% Moles of anhydrous salt in 100g hydrate= 49% X 1mol/120.38= 0.407moles Moles of water in 100g of hydrate= 51% X 1mol/18.02= 2.83moles Moles of water per mole of anhydrous
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After running multiple tests in the three different procedures‚ the Crystallization method proved to be the best method for determining the charge of the metal ion by using mole to mole ratio. Introduction The objective of the Get Charged Up lab was to determine the charge on a metal ion reacting with HCl by determine the mole ratio in a reaction by determining the amount of excess reactant‚ amount of product‚ and amount of hydrogen gas produced as well as finding the best research method to determine
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concentration that are commonly used. Some of these are listed below. Molarity‚ M = moles solute/liter of solution Normality‚ N = equivalents of solute/liter of solution Weight %‚ Wt % = (mass of solute/mass of solution) x 100% Parts per million‚ ppm = (mass of solute/mass of solution) x 106 Mass per volume‚ mg/L = mass of solute/liter of solution molality‚ m = moles of solute/mass of solvent mole fraction‚ χ = moles of solute/total moles Concentrations expressed as ppm and N are less familiar to most students
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81 10.01x+11.01-11.01x=10.81 -1.00x=0.2 X=0.2 The Mole and Avogadro’s constant Mole: the SI unit that is used to measure how much of a substance. (balanced eqn coefficient are moles) Converting Moles -> number of particles Number of Moles = number of particles / Avogadro’s constant Avogadro’s constant: 6.02214179 x 10^23 Converting Moles to Mass/molar mass Molar Mass: how much grams per mol of an element = atomic mass Number of Moles = mass / Molar Mass Percentage Composition Laws of
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Synthesis Procedure 1 -1.0 Introduction A Cobalt-Amine-Halide compound is synthesized from cobalt (II) chloride hexahydrate. An orange-tinted solid is produced and is considered to be unknown since the specific ligand amounts are unknown. By determining the percent composition of various elements and compounds in the unknown‚ its true identity can be predicted. Chloride‚ ammonia‚ and cobalt are three examples of percent compositions determined to help narrow the selection of possible unknowns
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60108 x 68.92558amu .39892 x 70.9247005 average atomic mass = 69.723amu The Mole (Section 3.2) In chemistry‚ we use the counting unit of a mole (mol). This is defined as the amount of particles that are in exactly 12.00 g of the isotope 12C. This number is 6.022 x 1023. Using this same relationship‚ it can be reasoned that 6.022 x 10 23 amu = 1.000 g. Therefore‚ the average mass of one mole of any element in grams is equal to the average atomic weight‚ also called the average
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Eggshell Lab Lab Set-Up: Materials: * pipette with pipette bulb * conical flask * 1 beaker * 1 molar sodium hydroxide solution * 2 molar hydrochloric acid solution * 1 funnel * 1 piece circular filter paper * crushed poultry eggshell * crushed farm eggshell * phenolphthalein * Distilled water * White tile * Paper tray * burette in burette stand * electronic scale Procedure: Step 1: Standardization of the NaOH solution using
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(a) simplete ratio of atoms of each element in a compound (1) 1 (b) (1) × 12 (1) (or definition in terms of moles) 2 (c) CHO = 29 = 6 (1) C6 H6 O6 (1) 2 (d) CH4O+ O2 CO2+2H2O (1) 1 [6] 3. (a) (i) (1) = 1.64(1) allow 1.63 to 1.64 PV = nRT (1) V = (1) if no × 3 CE allow use of p = 100 if answer in dm3 (1) allow 0.162 to 0.166 allow conseq on moles CH2NO2 5 (ii) V = V1 × (1) allow conseq on vol of gas products in (i) = 0.410 (m3) (1) allow 0.4 to
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followed: Mg + 2HCL H2 + Mg2+(aq) + 2Cl-(aq) In this experiment there is a one to one relationship between the number of moles of hydrogen gas evolved and the moles of magnesium metal consumed in the reaction. Therefore in the finding of the experiment moles of H2 evolved is equal to the moles of Mg consumed‚ and atomic weight of Mg is equal to the weight of Mg consumed per moles of H2 evolved. Procedure 1st. Obtain a 600ml beaker‚ add 300ml of water 2nd. add 30ml of HCl (2M) to the beaker
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