Introduction: Lactose intolerance is defined as the incapacity of the human body to digest the disaccharide sugar molecule‚ lactose. Lactose is a common carbohydrate found primarily in milk and associated dairy products. This widely known gastrointestinal disorder is typified by symptoms like nausea & vomiting‚ bloating‚ flatulence‚ abdominal pain and even diarrhoea. Lactose intolerance is a non-fatal disorder of the gastrointestinal tract. It is simply a nuisance and relatively harmless in and
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wont dissolve. Solubility of a solute = mass of a solid required to a saturate 100g of water at a particular temperature. Calculating Solubility 2g potassium chlorate dissolves in 20g water at 28oC what is its solubility? 2 x 100/20 = 10.0g potassium chlorate/100g water 4g potassium sulphate dissloves in 30g water at 50oC what is its solubilty? 4 x 100/30 = 13.33g potassium sulphate/100g water 30g sodium chloride dissolves in 75g water at 10oC what is its solubility? 30 x 100/75
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n this lab experiment factors that affected solubility are temperature‚ liquids‚ concentration‚ pressure‚ polarity‚ and molecule. In this experiment we use temperature‚ centrifuge‚ and concentration solutions to be able to achieve the results that we wanted. Step one of the experiment was to mix HCl‚ to cause a precipitation form‚ after that we had to decant the solution‚ then we used a centrifuged to separate the the precipitate of the supernatant liquid. After the decant was done‚ then the process
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Marcela Rodriguez Ricardo Sarabia January 16‚ 2013 Mariana Mendoza Alvaro Puccini Maria Quin Solubility Lab Purpose: The main purpose of doing this lab is to learn how to interpret solubility graphs and how the temperature does affects the solubility of different substances. There are other objectives of the lab which are learning what are concentrated‚ diluted‚ supersaturated‚ and saturated substances and how can they be identified. The hypothesis
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Name SOLUBILITY CURVES Answer the following questions based on the solubility curve below. Which salt is least soluble in water .. at 2O° C? 2. How many grams of potassium chloride can be dissolved in 200 g of water at 80° C? IO 3. At 40° C‚ how much potassium _ __nitrate coin be dissoiu$tl ^n 30D.g of water? ------W- ’1 80 70 ...- O --60 0 5© 40 4. Which salt shows the least change 30 In solubility from 0° - 100° C? 20 10 At 30° C‚ 90 g of sodium
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References: Buchanan‚ G.W. "Solubility Tests and Recrystallization." CHEM 2203/2204 Organic Chemistry Laboratory Manual. By C.A. White. N.p.: n.p.‚ n.d. 21-31. Print. "IR Spectrum." Research Gate. N.p.‚ n.d. Web. 08 Oct. 2014. Wagner‚ Kathryn. "Like Dissolves Like." LIKE DISSOLVES LIKE (n
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Nutrition: Barriers to Digestion Lactose Intolerance: - Lactase is one of the many enzymes required for complete digestion of lactose (a disaccharide). - Lactose intolerance is not an allergy‚ and is not to be confused with a milk allergy‚ which initiates an immune reaction when milk is ingested. Lactose intolerance instead is an enzyme deficiency (lactase). - S/s including gas‚ bloating‚ cramping‚ nausea‚ and diarrhea. Some people can ingest small amounts‚ and others none. Some can take
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Title: Study of Solubility Equilibrium Abstract The effect of temperature on the solubility product constant‚ Ksp‚ of potassium hydrogen tartrate in water was investigated in the temperature range of 285K to 318K at normal atmospheric pressure. It was found that the solubility of potassium hydrogen tartrate decreases with a decrease in temperature and consequently a smaller volume of sodium hydroxide is needed to neutralize it. The molar solubility of potassium hydrogen tartrate was calculated
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Mateo Castro April 3‚ 2013 Lab Partner: Unur Abdul Kader T.A: Katie Experiment 22: Molar Solubility‚ Common-Ion Effect Abstract The purpose of this experiment was to determine the molar solubility‚ the solubility constant‚ and the effect of a common ion on the molar solubility of calcium hydroxide. To accomplish this the experiment was split into two parts; part A and Part B. in Part A of the experiment a standardized 0.05 M solution of HCl was titrated into a 25 mL solution of saturated
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laboratory is to calculate the solubility of a substance under a variety of temperatures and construct a solubility curve based on experimental data to see the effect of temperature on solubility. Data collection and quantitative observation VOLUME H2O/ml/±0.05 TEMPERATURE/°C/±0.1 5.00 71.3 6.00 59.3 7.00 52.4 8.00 47.2 9.00 41.8 Mass of NaCl: 3.0±0.05 g Data Processing To calculate the solubility‚ we need to use the formula: Solving for x‚ the solubility of salt‚ we get the formula:
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