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    in the rope (tension) is 94.0N. How much work is done on the sled? Solution: W= Fd W= 94.0N x 35.0m W= 3290 Nm or J 2. The cable of a large crane applies a force of 2.2x10^4N to a demolition ball as it lifts it vertically a distance of 7.6m. a) How much work is done on the ball? b) Is the work positive or negative? Why? Solution: A.) W=( 2.2x10^4N) ( 7.6 m) W= 1.6x10^5 Nm or J B.) Positive‚ because the force on the system is positive

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    pBlu lab

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    February 20th‚ 2014 Lab report 4 Abstract This pBlu lab had for purpose to present the changes of the strain of E. coli bacteria due to new genetic information being introduced into the cell. In this experiment we are freezing and heat shocking the E. Coli bacteria that is then forced to take the plasmid DNA. The E. coli then transforms the pBLu plasmid‚ which carries the genes coding for two identifiable phenotypes. After following the Carolina Biological steps our lab worked well and we able

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    Osmosis Lab

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    Osmosis of Sucrose Solutions of Different Molarities Through Dialysis Tubing (a Semi-Permeable Membrane) I. DESIGN A. PROBLEM/RESEARCH QUESTION 1. How does increasing molarity of sucrose affect osmosis through dialysis tubing? B. VARIABLES 1. The independent variable in this lab is the molarity of sucrose each dialysis bag is filled with. The time (30 minutes)‚ the temperature (23C) and the type of dialysis tubing used are all constants. 2. The dependent

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    Crystallization Lab

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    Partner: Camille Dupiton Lab #1 Purification of a Solid: Crystallization and Melting Point Section: A61 Laboratory Exercise #1 Purification of a Solid: Crystallization and Melting Point Introduction In this lab exercise‚ we will be learning experimental techniques using glassware and other apparatuses. In order to successfully complete this lab‚ we will use techniques 1.0‚ 1.1‚1.2‚2.0‚2.1‚2.2‚ and 2.3 that are described in the Lab Manuel. In addition‚ we will

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    density lab

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    Lab # 4 Determination of Density of Liquids Name: Lab Partner: Period: 3 Date Completed: 9/23/2014 Date Submitted: 9/29/2014 Data TABLE 4 DENSITY OF SALT SOLUTIONS-INDIVIDUAL GROUP’S RESULTS Concentration (%) Mass (g) Volume (mL) Density (g/mL) 0 9.9522 10.00 0.9952 4 10.1291 10.00 1.013 8 10.5233 10.00 1.052 12 10.7487 10.00 1.075 16 11.0297 10.00 1.103 Unknown # 10.6234 10.00 1.062 Calculations 1. Show all density

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    Saponification Lab

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    Abstract: For the first part of this lab we refluxed different Carboxylic acids and alcohols in the presence of a acid catalyst in order to form Esters by Fischer Esterification. These Esters had different pleasant smells that we then evaluated. In the second part of the experiment‚ we broke the ester bonds of a triglyceride in order to form glycerol and carboxylate salts. This process is known as Saponification because it produces amphiphilic molecules that allow soap to remove dirt from the surface

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    5 9

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    T&P 5-9 1. The patient has a pH of 6.96‚ pO2 of 12‚ and pCO2 of 54. 2. There is a 1.0 x 0.5-cm area of avulsed tissue and a 3-cm gaping deep laceration of the chin. 3. Lungs: Clear to P&A. Heart: Not enlarged; A2 is greater than P2. There was a grade 1/6 decrescendo early diastolic high-frequency murmur. 4. Cycloplegic refraction: OD = +3.25 + 0.75 x 125 =20/30-1. 5. The patient received a 6000-gamma roentgen dose. 6. Iodipamide sodium I-131 was used. 7. We used a concentration of 5 x 105/mL.

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    Antifreeze Lab

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    antifreeze solutions Data | Freezing Point (°C) | Distilled Water | 0.0°C | 10% Antifreeze | -3.5°C | 20% Antifreeze | -8.0°C | Calculations Questions: 1. Using the equation ΔT = Kf m‚ calculate the molality of the 10% antifreeze solution. 2. Use the formula molality = moles solute/kg solvent to find the number of moles in 10% antifreeze solution. 3. Using the formula molar mass = grams/moles calculate the molecular mass of antifreeze in the 10% solution. 4.

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    Measurement Lab

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    Measurements Lab Name Length Measurements – Follow the Instructions in the Lab Manual and fill in your data in the tables provided. Data Table 1 – Length measurements |Object |Length (cm) |Length (mm) |Length (m) | |CD or DVD |12.00 |120.0 |.1200 | |Key

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    potato lab

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    HYPOTHESIS: Increase in salt concentration results in a hypotonic solution where the mass of the potato increases due to the movement of water into the cell. On the other hand‚ decrease in salt concentration results in a hypertonic solution where the mass of the potato decreases due to the movement of water outside the cell. MATERIALS: Potato Forceps Stopwatch Scalpel Test tubes Knife Mass balance Graduated cylinder ruler salt solution test tube rack PROCEDURE: 1. Cut 18 potato cores from

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