mass of ascorbic acid = 176mg‚ mass of ascorbic acid used = 1.25mg Number of moles in 0.00125g of ascorbic acid = 1/176 x 0.00125 = 7.102 x 10-6 moles Mole ratio of DCPIP: Ascorbic acid = 1:1 Number of moles of DCPIP used = 7.102 x 10-6 moles If 20mL of DCPIP = 7.102 x 10-6 moles Then 1 mL of DCPIP = 7.102 x 10-6/20 = 3.55 x 10-7 moles Therefore 5 mL of DCPIP = 3.55 x 10-7 x 5 = 1.775 x 10-6 moles Since mole ratio of DCPIP: Ascorbic acid
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To determine the percentage by mass of Calcium carbonate in an egg shell Aim: is to find and work out the percentage of calcium carbonate by mass in eggshells by using acid base “back titration” Introduction: Calcium Carbonate‚ CaCO₃‚ is a compound which is polymorphic and therefore means that it can be found in a variety of different objects or organisms. For example: Rocks (limestone‚calcite)‚ sea snail shells‚ Eggshells‚ pearls and many more. This investigation will focus on finding out the
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large amount of dissolved solute. C¬1V1 = C2V2 Systems for Expressing Concentrations of Solutions Name of System Symbol Definition Molarity M moles of solute / liter of solution Molality M moles of solute / kg of solvent Formality F gram-formula weight of solute /liters of solution Mole Fraction N‚ X moles of solute / moles of solvent + moles of solute Normality N equivalents of solute / liters of solution Weight percent wt. % 100 x grams of solute / grams of solvent + grams of solute
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Lab: STOICHIOMETRY The reaction of Iron with Copper(II) Sulfate Purpose: In this experiment we will use stoichiometric principles to deduce the appropriate equation for the reaction between metallic iron and a solution of copper (II) sulfate. This reaction produces metallic copper‚ which is seen precipitating as a finely divided red power. Materials: Flask beaker Copper solution Balance Hot plate •150 ml beaker •1 gram of iron power
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control the CV | Concentration of acid | Rate of reaction | Vol of acid | 20 ml | Measure using a 50ml Measuring cylinder | 2 moles pr. L | Vol of gas | Temp of acid | Room temp. | Not controlled‚ only montored | 1 moles pr. L | | Amount of Mg | 5 cm or ribbon/ 0.07g | Ruler‚ and weight | 0.5 moles pr. L | | Precence of catalyst | No | Not have a catalyst | 0.25 moles pr. L | | Type of acid | Hydrocloric | Have same type of acid | Method: 1. Gather aparatus and materials‚ and set up as
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sides of the beaker. Na2CO3 + CaCl22H2O = CaCO3 + 2NaCl + 2H2O Convert 1g of CaCl22H2O to moles‚ 1g CaCl22H2O x 1mol = .00689Moles CaCl22H2O 144.9994 Consider the mole ratio of Na2CO3 to CaCl22H2O 1:1 Thus‚ 1 mole of CaCl22H2O will equal 1 mole of Na2CO3 Convert moles of Na2CO3 to grams: .00689moles Na2CO3 x 105.99 = .73027g Na2CO3 1 mole Na2CO3 How much CaCO3 can we expect? .00689 moles of CaCl22H2O x 1mole CaCO3 = .00689Moles CaCO3 1mole CaCl22H2O Convert .00689Moles CaCO3
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precipitation reaction. This is a reaction where two soluble salts Sodium Carbonate and Calcium Chloride are added together and the result is the precipitation of single Product while the other product remains in solution. This means by using the moles in this lab we could find the limiting reactant‚ percentage yield and the percent purity Materials: * Electronic Scale * Sodium Carbonate * Calcium Chloride * Wayne Paper
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the crucible with hydrate‚ and the crucible with an anhydrous and write down each weight‚ then repeat. Once we completed that task then we were to find the number of moles of water lost‚ the number of moles of anhydrous copper sulfate‚ percent comp of water in the hydrated copper sulfate‚ the mole ratio of moles of water and moles of anhydrous copper sulfate. Once this is all done‚ we found our percent error and compared it to the exact value. In the lab‚ we used the following tools: lab apron
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Objective 1. Use the analytical balance to weigh a given hydrated salt as accurate as possible in order to determine the weight of water in the salt Introduction Balances are important laboratory equipment as they are used to determine the mass of materials. Today‚ most balances used in the laboratory are analytical balances which can give readings up to four decimal places or higher. High accuracy is needed in certain experimental work such as material analysis or those involving small change
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math‚ however the calculations did. The harder math calculations included finding moles of the acid‚ moles of the base that was used to neutralize‚ and the molarity of the base. After all calculations‚ below is what we concluded. The molarity you had to be careful with because it was moles/liters and our readings were in milliliters so we had to convert first. In conclusion‚ even though the lab procedure is not about moles‚ most of the calculations are and this is one lab where the concept of
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