of equation is equally balanced. The calculation for formula mass helps determine if you need to convert grams to a particular substance to moles‚ from a product. Moles are numbers that are in front of formulae. E.g.‚ 6NaCl‚ 6 is the equation for this formula. A mole would help you balance a skeleton equation‚ and also allows you to calculate how many moles are needed to take part in a chemical reaction. In the laboratory experiment‚ we are investigate the following equation and how they react:
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Table 1 to calculate the temperature at which a solution of 50 grams of sucrose (C12H22O11) in 400 grams of water will freeze. The molecular weight of sucrose is 12(12.01) + 22(1.01) + 11(16.00) = 342.34 g/mole so‚ the number of moles of sucrose is and the concentration of the solution in moles per kilogram of water is By taking the freezing point constant for water as 1.86 from Table and then substituting the values into the equation for freezing point depression‚ you obtain the change in freezing
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Final Exam Review Chapter 2 Study Questions 1. Define the following a) element b) compound c) pure substance 2. Classify each of the following as a pure substance or a mixture. For each pure substance‚ indicate whether it is an element or a compound. Which of the mixtures are solutions a) air b) titanium c) oak d) baking soda e) oxygen f) 7-Up g) wine h) carbon monoxide 3. Label each of the following drawings as element‚ compound‚ or mixture (Assume each type of circle
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shows the six points of attachment in red and green. Ca 2+ + EDTA 4- ® CaEDTA 2- This equation shows that the EDTA reacts with the metal ion‚ which in this case is calcium on a 1:1 ratio‚ therefore it can be worked out however many moles of EDTA is used the calcium will be the same. To work this
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=-mH2O×cH2O×△TH2O As the water has gained the heat produced by the reaction‚ the heat change of reaction is negative when the temperature of the water increases. As the heat change observed depends on the amount of reaction‚ for example the number of moles of fuel burned‚ enthalpy change reaction are usually expressed in kJmol-1. Chemicals Name of chemicals concentration Volume Water used for diluting Materials Name of materials Measuring cylinder Measuring cylinder Beaker Stirring rod Thermometer Balance
Free Thermodynamics Temperature Sodium hydroxide
The Ka and Molar Mass of a Monoprotic Weak Acid Chemistry Lab 152 Professor: James Giles November 7‚ 2012 Abstract: The purpose of this experiment was to determine the pKa‚ Ka‚ and molar mass of an unknown acid (#14). The pKa was found to be 3.88‚ the Ka was found to be 1.318 x 10 -4‚ and the molar mass was found to be 171.9 g/mol. Introduction Acids differ considerable as to their strength. The difference between weak and
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Vinegar Titration Lab Procedure: 1.) Obtain ~100 mL of NaOH. Record the molarity on the data table. 2.) Set up a ring stand with a buret. 3.) Place 5mL of vinegar in a 125 mL Erlenmeyer flask. Dilute the vinegar with 25 mL of water and add two drops of phenolphthalein. 4.) Fill the buret with NaOH. Record the initial volume of the buret in your data table. 5.) Titrate the vinegar sample until the first faint pink color does not disappear. 6.) Record the final volume of the buret in your data
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|28.00 |34.00 | Volume of acid = 50 cm3 ± 0.25 Acid concentration = 1.0 M Volume of base = 50cm3 ± 0.25 Base concentration = 1.0 M Number of mole of solution used = [pic] = [pic] = 0.05 mole ± 0.5 % Data Processing: Calculation: Temperature Change‚ [pic] = Final temperature – Average temperature of Acid and Base |Reaction |Average Temperature of Acid and Base‚ °C|Final
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N/A N/A Moles 6.75x10^-4 .032 Theoretical: 3.29x10^-4 N/A Final Mass N/A N/A Theoretical: 99mg (.099g) N/A Actual: Melting pt (or bp) 81-83 C N/A Literature: 312-315 C N/A Experimental: Percent Yield Calculation: Limiting Reagent: 1 mg= .001 g‚ 100 mg= .1 g vanillin Moles vanillin = .1g/152.15 g/mol= 6.57x10^-4 moles Density H2O2= 1.45 g/mL Mass H2O2= 1.45 g/mL x .75mL = 1.088g Moles H2O2= 1.088g/34.015 g/mol =.032 moles Limiting
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the molar mass of the metal that was present in the sample. The concentration of metal that was present in the TAP water was solved by the following calculations: 150 mL EDTA * (1 L/ 1000 mL) * (0.005 M) = 0.00075 moles of metal (0.00075 moles/ 0.050 L) = 0.015 moles/ L 0.00075 moles of metal * (32.192 g/ 1 mol) * (1000 mg/ 1 g) = 24.144 mg The concentration of metal that was present in the DI water was solved by the following calculations: 75 mL EDTA * (1L/1000 mL) * 0.005 M = 0.000375
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