Concentration of liquid foods is a fundamental operation in many food processes; it is completely different from dehydration. Usually‚ foods‚ which are concentrated‚ remain in the liquid state; while drying produces solid or semisolid foods with significantly lower water content. The concentration of liquid foods has three different methods; evaporation‚ membrane concentration‚ and freeze concentration. Evaporation usages gas liquid phase separation. It has the lowest capital cost and the maximum
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Objective: To find out the food substance in food sample X‚Y and Z. Biological Principle: Test for lipids using the grease spot test Test for glucose using Clinistix paper Test for reducing sugars using Benedict’s test Test for starch using the iodine test Test for proteins using Albustix paper Test for vitamin C using DCPIP solution A control using distilled water instead of the sample should be included. Independent: different types of food samples Dependent: different types of food
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Biology Controlled Assessment. Kenishia Pascal 10x3. Investigating How Different Concentration Effects The Rate Of Reaction. Strategy A Possible Factors * Source of catalase * Concentration * Surface Area of enzyme * Concentration of enzyme * pH * Temperature Chosen Factor We chose to investigate the concentration of enzyme as we had previously investigated the optimum temperature for catalase in the preliminary investigation. Concentration of enzyme
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A key component of the quantification of the protein concentration in the given samples was the generation of a standard curve. The data presented in Table 1 produced the graph in Figure 1 which was then used to calculate a line of best fit. The line of best fit yielded an equation y= 0.44976x which was in turn used to calculate unknown protein concentrations of given samples. The absorbances of the whole milk‚ cereal milk and muscle milk had been previously obtained and recorded via spectrophotometer
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11 ST Revision i) Multiply the following numbers and express in terms of a x 10k where 1 ≤a ≤ 10 and k ϵ Z: 3.2 x 104 5.6 x 10-2 ii) Add them and multiply by four. iii) Factorise x2 – 49 iv) Factorise 16x2 - 9 v) Factorise x2 – 7x + 12 vi) Solve x2 -7x + 12 = 0 vii) The first term of an arithmetic sequence is 8 and the common difference is 7. The nth term is 393. Find the value of n. viii) A geometric sequence has first term 2 and third term 32. Find the common Ratio. ix) Find
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I. Parts of a Lab Report 1. Introduction: a. Title b. Research Question c. Hypothesis d. Variables e. Control of Variables 2. Materials & Methods a. Materials b. Method 3. Data a. Data b. raw data c. uncertainty d. presentation e. processing data f. Graphs 4. Results/Conclusion a. Conclusion 5. Discussion a. Evaluation 6. References II. Other Help errors and uncertainty A. Design [pic] I. Research Question
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the Glucose into CO2 • Each step catalyzed by a specific enzyme. • It is a cycle because the product of step 8 is the reactant in step 1 (oxaloacetate). Substrate Level Phosphorylation 1 14-01-05 Electron Carriers (coenzymes) H Krebs Cycle H • The overall chemical equa%on is: 2 oxaloacetate + 2acetyl-‐coA + 2ADP + 2P + 6NAD+ + 2FAD à
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Biology EEI Effect of Inorganic and Organic Fertilisers on Yield and Growth of Tomatoes By Yash Teacher: Mrs. Elphick TABLE OF CONTENTS Abstract................................................................................3 Introduction..........................................................................4 Materials and Methods.........................................................6 Contents of fertilisers...........................................................6 Procedure....
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Online Biology Lab Report Evolution Name______ A. Natural Selection 1. Record the number of beetles present in each of the three generations from the lab website. Generation|Orange Beetles|Green Beetles| 1st Generation|3|13| Later Generation|6|10| Last Generation|16|0| 2. Nature selected for the _______orange__________ beetles. 3. Nature selected against the______green________ beetles. B. Speciation 1. Why were wasp species A and B no longer able to mate with each
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Purpose: To find the percent of sugar by mass in chewing gum Hypothesis: The percentage of chewing gum that is sugar for bubble gum is 30%‚ 20% for juicy fruit and 25% for stride gum. Mass Of Juicy Fruit Mass Of Bubble Gum Mass of Juicy Fruit unchewed | 7.12g- 1.72 = 5.4 g | 5.80g- 1.72g = 4.08 | 4.52g- 1.72g = 2.80g | chewed | 3.00g- 1.72g= 1.28g | 2.24g-1.72g =0.52g | 2.86g-1.72g= 1.14g | Percent of Sugar | 3.00g x 100/ 5.4 = 55.5%
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