Oumou Fofana Period: 1 Mrs. Furlong December 19‚ 2012 Gummy Bear Lab Question- If the percent of the concentration of a solution varies‚ will the amount of water gained or lost vary in a linear fashion? Hypothesis- If the concentration of a solution varies‚ then the amount of water lost or gain will not vary in a linear fashion Prediction- I think that the amount of water lost will not vary in a linear fashion. I think the water will vary like one gummy bear would lose 4 grams while
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chemical equations contain important information about the amount of reactants required to produce given products. These amounts are represented by ____________________. Coefficients Solutions (10 marks) 5. A ____________________ is defined as a mixture of two or more substances that are evenly distributed. Solution 6. The shape of the water molecule‚ combined
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cautious when handing the precipitate as the solid separated into smaller sup unit which increased the surface area enabling the photodecomposition of the precipitate to occur more dramatically. We also added our precipitating agent to the supernatant solution too quickly‚ making all the chlorine unable to bond with the Ag+ as other‚ smaller ions co-precipitate instead. The precipitate may not have undergone a complete reaction as our test for more precipitate was not as thorough as it could have
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Colorado Northwestern Community College Science of Biology Mrs. Farrow Lab 3 – Slime Time Submitted by Chase Kenemer 22 February 2015 Abstract Polar solvents dissolve‚ or pick-up‚ polar substances and non-polar solvents dissolve‚ or pick-up‚ non-polar substances. In the conducted experiment‚ the polarity of molecules and their properties are explored. The results of using two solvents on both polar and non-polar inks‚ further verify this to be true. The student conducted the experiment
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water and add 20 drops of blue dye #1 and stir. Add more drops of blue dye to solution if it is not darker than your commercial dye. (I added an extra 20 to make mine darker) This gave me a concentration of 5.2x10-4. Record concentration Step #2:Place a 12 well strip on a white sheet of paper and number 1-10 starting from the left. Step #3: Using the 1mL fine tip pipet add the appropriate number of drops of blue dye solution to each well. (refer to data table 1 in lab assistant section) Rinse pipet
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big is osmosis‚ and how it had the overall impact in this excitement a little on osmosis. Osmosis takes place when two solutions of different concentrations are separated by a semi-permeable membrane in which the solvent can pass through but not the solute. In our experiment‚ we used a sucrose solution that will be a hypotonic concentration of solute. This tells us that the solution has a lower concentration of water than does the cells. Therefore‚ due to osmosis‚ the cells will gain water weight also
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which consisted of ten drops of fish blood in a test tube containing 10mL of 0.7% NaCl. Eleven other solutions‚ (erythritol‚ xylose‚ monacetin‚ diacetin‚ triacetin‚ urea‚ thiourea‚ glycerol‚ ethylene glycol‚ glucose and fructose) all isosmotic but not necessarily isotonic with the cytoplasm of the erythrocyte‚ were combined with a 0.2mL of well-mixed stock suspension were added to 0.27M of each solution‚ one at a time‚ and timed how long it took for hemolysis to occur with a stop watch. As soon as the
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exactly 3 minutes as instructed in the lab manual (Scott et al‚ 2016) and we tried avoid denaturing the enzyme. The benzoquinone had a greater absorption at acidic pH levels because the acidic solution denatures catechol oxidase‚ causing it to lose its secondary and
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The first solution was tube 1 which was made with 1 mL of EDTA‚ 1 mL of CO‚ and 5 drops of CAT. Tube 2 was made with 1 mL of PTU‚ 1 mL of CO‚ and 5 drops of CAT. Tube 3 was made with 1mL of distilled water‚ 1 mL of CO‚ and 5 drops of CAT. After mixing each solution and putting a piece of Parafilm on Tube 1‚ we set tube 3 as our blank and then measured the change in absorption at 400nm of
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engine. We have to use a liquid that can absorb all that heat. A homogeneous mixture consisting of a solute and a solvent based on colligative properties‚ which means solution’s properties will differ depend on the proportion of solute present. Solutions have both a lower freezing point and a higher boiling point than pure solvent. The more solute is present the bigger the difference between the freezing point and the boiling point. To explain furthermore‚ we need to understand that temperature is
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