Abstract: All the materials were measured and weighed. It was found in the experiment that the yield of copper hydroxide in 40%. Introduction: The copper (II) sulphate is then placed in 100 mL of distilled water. Then 20 mL of CuSO4 is measured and placed 100 mL of distilled water. This can later be weighed to determine the mol of CuSO4 and the mol/L concentration. Then this was used to find out how many mL of 0.5 NaOH solution is needed to react completely with all the copper (II)
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Write the complete balanced equation for the reaction that occurred in this lab. Hint: H2CO3 is not a final product of the double-replacement reaction; it breaks down (decomposes) immediately into two products. (3 points) NaHCO3 + HCl → CO2 + H2O + NaCl 3. The NaHCO3 is the limiting reactant and the HCl is the excess reactant in this experiment. Determine the theoretical yield of the NaCl product‚ showing all of your work in the space below. (5 points)
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Ideally‚ the percent yield should be 100%‚ as this means that you have recovered 100% of that material. A yield over 100% would mean that the substance still has some traces of another material that is adding additional mass. Ex. The iron filings having some sand particles leftover. A yield under 100% would mean that some of the substance was not recovered‚ it could have been lost (spilled) or found in another substance (not separated completely). The percent yields may give some insight into what
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Percent Error occurs due to many factors. In this lab by the graph I can state that there was a 100% error in all three unknown substances‚ Pure Leaf Lemon Tea‚ Sprite‚ and Tree Ripe Lemonade. This could be a result of not having properly zeroed the scale or the scale glitching when it came to zeroing it‚ which would have added extra mass and skewed the results. Another factor that could have resulted in a high percent error would have been the scale itself glitching which would have caused the wrong
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Percent Copper in Brass Background The scientific concepts involved in this lab include Spectroscopy‚ Beer’s law‚ Calibration curve‚ concentration‚ and electronic transitions. The main objective of this experiment is to see how the percent composition of brass can be determined to verify the properties influenced by copper and zinc. Brass is a generic term for alloys of copper and zinc. The main technique used in this experiment is Spectroscopy. The three equations used in this lab are: Y = mx
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Determining the Limiting Reactant and Percent Yield in a Precipitate Reaction (SMG 6D) AP Chemistry One example of a double replacement (metathesis) reaction is the mixing of two solutions resulting in the formation of a precipitate. In solution chemistry‚ the term precipitate is used to describe a solid that forms when a positive ion (cation) and a negative ion (anion) are strongly attracted to one another. In this experiment‚ a precipitation reaction will be studied. Stoichiometry
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Determining the Limiting Reactant and Percent Yield in a Precipitation Reaction Objectives: Observe the reaction between solutions of sodium carbonate and calcium chloride. Determine which of the reactants is the limiting reactant and which is the excess reactant. Determine the theoretical mass of precipitate that should form. Compare the actual mass with the theoretical mass of precipitate and calculate the percent yield. Materials: Balance 0.70 M sodium carbonate solution‚ Na2CO3(aq)
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we can then calculate the water evolved. The first method we use to determine the percent composition is Gravimetric. With this method we use the mass of the reactant and the mass of the product. Another way to acquire the percent composition is by the Volumetric Method. This method requires measuring the water displaced by the O2 gas. If the experiment is done correctly‚ we should be able to calculate the percent composition of KClO3 by using both methods. Theory: This experiment requires us
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Yield of CuCl2.2DMSO Formula weight (Mr) of CuCl2 = 63.55 + (35.45 x 2) =134.45g/mol Formula weight of product CuCl2.2DMSO = 134.45 + 2[16 + 32.06 + (12.01 x 2) + (1.0079 x 6)] = 290.704g/mol Mass of CuCl2= 0.850g Equation for reaction CuCl2 + 2DMSO -> CuCl22DMSO Mole ratio between CuCl2 and CuCl22DMSO = 1:1 Mole of CuCl2 = Mass/ Mr = 0.850/134.45 = 0.00632 moles Since the ratio between CuCl2 and CuCl22DMSO = 1:1‚ mole of CuCl2DMSO is also 0.0063 moles. To find theoretical yield of CuCl2
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with a 12 percent coupon. Bond D is a 6 percent coupon bond currently selling at a discount. Both bonds make annual payments‚ have a YTM of 9 percent‚ and have five years to maturity. The current yield for Bonds P and D is percent and percent‚ respectively. (Do not include the percent signs (%). Round your answers to 2 decimal places. (e.g.‚ 32.16)) | If interest rates remain unchanged‚ the expected capital gains yield over the next year for Bonds P and D is percent and percent‚ respectively
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