Lab 2 Measurements: Accuracy and Precision A. Data Tables (36 points) Place your completed data tables into your report here: Data Table 1 Measurement | Data | Length of aluminum plastic packet | 4.50cm | Height of aluminum plastic packet | 7.50 cm | Temperature of faucet water | 26.0 degrees Celsius | Temperature of ice water | 10.0 degrees Celsius | Volume of water in 10-mL graduated cylinder | 10.0ml | Volume of water in 50-mL graduated cylinder | 9.0ml | Data Table 2 Measurement | Data | Inside diameter of 50-mL graduated cylinder | 2.50 cm | Height of 50-mL graduated cylinder | 10.0 cm | Water temperature | 25.0 degrees Celsius | Initial volume of water in 50-mL graduated cylinder | 10.0 mL | Mass of water in the 50-mL graduated cylinder (remember, 1 g of water weights 1 mL since its density is 1 g/mL) | 10.0gm | Volume of water and aluminum shot in 50-mL graduated cylinder | 18.0ml | Mass of aluminum shot (given on outside of packet) | 20.0gm | B. Follow-Up Questions (Show all calculations) Part I (Each question is worth 10 points.) 1. Convert the length and height measurements for the packet that contains the aluminum shot from units of cm to units of mm using the unit-factor method.
Background: For the entire explanation of this demo see Shakhashiri's Chemical Demonstrations, Vol. 4, pp 42-43. The following is a excerpt from this volume. " The sudden change from colorless to deep blue solutions in this demonstration can be explained with the following sequence of equations[2]: 3 I-(aq) + H202 + 2 H+ (aq) = I3- (aq) + 2 H20 (l) (1) I3- (aq) + 2 S2O3 2- (aq) = 3 I- (aq) + S4O6 2- (aq) (2) 2 I3- (aq) + starch = starch-I5- complex + I- (aq) (3) The first equation indicates that, in an
The fork was held horizontally over the pipe and the pipe was moved up and down in the water. At the highest pitch of sound the pipe was held in place and the distance between the surface of the water and the top of the pipe was recorded. Data: Data Table 1 | Tuning fork frequency(ƒ),Hz | Length, L(water level to top of pipe) | Diameter of pipe, d | λ=4(L+0.3d) | ExperimentalV= ƒ λ | Room Temperature, Celsius | 384 | | 0.025cm | | | 23 degrees Celsius | Sample Calculation: Theoretical Speed of Sound: v = 331.4 + 0.6TC m/s v= 331.4+0.6(23) v=345.2 Percent Error: % error = experimental value – theoretical value × 100/theoretical value % error = 149.76 – 345.2 × 100/345.2 % error =56.6 Results: Data Table 1 | Tuning fork frequency(ƒ),Hz | Length, L(water level to top of pipe) | Diameter of pipe, d | λ=4(L+0.3d) | ExperimentalV= ƒ λ | Room Temperature, Celsius | 384 | 0.09 m | 0.025m | λ=4(L+0.3d)=4(0.09+0.3x0.025)=4(0.0975)=0.39 | V= ƒ λ=384(0.39)=149.76 | 23 degrees Celsius | Conclusion: There could be many errors that could lead to the percent error calculated. One of which could be the position of either the tuning fork or the pipe. Another could have been the movement of the pipe up and down in the
Friction Objectives: To provide an understanding of the concept of friction. To calculate the coefficient of friction of an object by two methods. Materials: Ramp board: 3 - 4 feet long, 10 cm wide Can of soft drink or item of similar weight Friction block set-PK Protractor Scale-Spring-500-g Tape measure, 3-m Lab notes: Using the wooden block provided in LabPaq, a long board, a can of beans and the 500-g spring scale I will try and determine the force of kinetic friction, N, and the force of static friction, N while pulling the block at a constant speed. I will convert kg-mass to Newtons by multiplying the kg-weight by 9.8 m/s2, i.e., 100 g = 0.1 kg = 0.1 x 9.8 = .98 N. Observations: Mass of block (with can): 3995 kg Weight: 3.91 N Data Table 1: Flat board Flat board Force of Kinetic Friction, N Force of Static Friction, N Trial 1 1.1 0.6 Trial 2 1 0.7 Trial 3 1 0.9 Average 1.03 0.73 Data table 2: Flat board - Block Sideways Mass of block (with can) 3995 kg Weight: 3.91 N Flat Board - Block sideways Force of Kinetic Friction, N Force of Static Friction, N Trial 1 1.3 1.4 Trial 2 1.1 1.5 Trial 3 1.1 1.1 Average 1.2 1.5 Data Table 3: Different surfaces Surfaces tried: Glass surface Force of Kinetic Friction, N Force of Static Friction, N Trial 1 0.4 0.1 Trial 2 0.4 0.1 Trial 3 0.4 0.2 Average 0.4 0.13 Data Table 4: Different Surfaces Surfaces tried: Sandpaper Force of Kinetic Friction, N Force of Static Friction, N Trial 1 2.2 1.5 Trial 2 2.1 1.7 Trial 3 2 1.1 Average 2.1 1.43 Data Table 5: Different Surfaces Surfaces tried: Wood on Carpet Force of Kinetic Friction, N Force of Static Friction, N Trial 1 1.4 1.9 Trial 2 1.5 1.6 Trial 3 1.5 1.7 Average 1.47 1.73 Data Table 6: Raised Board Height Base Length θ max μs Trial 1 .44196 m .71120 m 60 deg 0.62143 Trial 2
What does this result mean? The t ratio of -0.65 represents the smallest relative difference between the pretest and 3 months outcomes. This ratio does not have an asterisk next to it in the table which according to the footnotes the asterisk is said to represent p < 0.05 the least stringent acceptable value for statistical significance. 4. What are the assumptions for conducting a t-test for dependent groups in a study?
Laboratory Techniques and Measurements Peter Jeschofnig, Ph.D. Version 42-0165-00-01 Lab RepoRt assistant Data Table 1: Length measurements. | Object | Length (cm) | Length (mm) | Length (m) | CD or DVD | 12 | 120 | O.12 | Key | 5.3 | 53 | 0.053 | Spoon | 17.3 | 53 | 0.173 | Fork | 18.7 | 187 | 0.187 | Data Table 2: Temperature measurements. | Water | Temperature (°C) | Temperature (°F) | Temperature (K) | Hot from tap | 43 | 109.4 | 316 | Boiling | 100 | 212 | 373 | Boiling for 5 minutes | 103 | 217.4 | 376 | Cold from tap | 19 | 66.2 | 292 | Ice water – 1 minute | 12 | 53.6 | 285 | Ice water – 5 minutes | 6 | 42.8 | 279 | Data Table 3: Mass measurements | Object | Estimated Mass (g) | Actual Mass (g) | Actual mass (kg) | Pen or pencil | 10 | 11.1 | 0.0111 | 3 Pennies | 5 | 7.6 | 0.0076 | 1 Quarter | 4 | 5.6 | 0.0056 | 2 Quarters, 3 Dimes | 16 | 18.6 | 0.0186 | 4 Dimes, 5 Pennies | 18 | 19.7 | 0.0197 | 3 Quarters, 1 Dime, 5 Pennies | 30 | 31.7 | 0.0317 | Key | 5 | 8.1 | 0.0081 | Key, 1 Quarter, 4 Pennies | 18 | 23.9 | 0.0239 | Data Table 4: Liquid measurements. | | | Mass A | Mass B | Mass B - A | | | Liquid | Volume(mL) | GraduatedCylinder (g) | GraduatedCylinderwith liquid (g) | Liquid (g) | Densityg/mL | %Error | Water | 5 | 15.8 | 20.8 | 5 | 1 | 0 | Isopropyl alcohol | 5 | 15.8 | 19.5 | 3.9 | 0.78 | 0.76 | .76 Data Table 5: Magnet – Measurement Method.
3. Place the “R” tube in a beaker and get a large test tube, another boiling tube and a pipet. Label the large test tube “C” and the boiling tube “P”. Place one boiling chip into the “P” tube. 4.
10. Repeat the procedure for a second metal. Analysis: Our data | Trial #1 | Trial | Mass of zinc | 1.99g | 4.01g | Mass of water in Calorimeter | 45g | 45g | Temp. of water in Calorimeter | 20°C | 21°C | Temp. of boiling water | 100°C | 100°C | Peak temp.
The following formula was used: cm=mwcwT3-T1+mcccT3-T1 mmT2-T3 Where cw is the specific heat capacity of water and cc is the specific heat capacity of calorimeter. Substituting the values, we get an answer of 998.81 J/kg·C0 with a percent error of 10.98% determined by the formula: % error=theoretical-experimantalexperimental ×100 Table 1. Specific Heat of Metal Mass of Sample (mm) | 0.01655 kg | Mass of Inner Vessel of Calorimeter (mc) | 0.04723 kg | Mass of Inner Vessel of Calorimeter with Water | 0.24242 kg | Mass of Water inside the Inner Vessel of Calorimeter (mw) | 0.19519 kg | Initial Temperature of Water and Inner Vessel of Calorimeter (T1) | 330C | Initial Temperature of Sample (T2) | 860C | Equilibrium Temperature of Sample, Water, and Inner Vessel of Calorimeter (T3) | 290C | Calculated Specific Heat of Sample (cm) | 998.81 J/kg·C0 | Accepted Value of Specific Heat | 900 J/kg·C0 | % Error | 10.98% | Possible sources of error for this experiment may be due to changes in the temperature due to heat transferring to the environment and human error (wrong readings for temperature and mass). In the second activity, the latent heat of fusion of ice was computed (Table 2). Latent heat is associated with a phase change.
c) 0.08 g of oxygen reacted with the magnesium (0.22 g – 0.14 g = 0.08 g). d) Refer to “Determining the Empirical Formula of Magnesium Oxide Lab Solutions” sheet e) Element | % | m (g) | M (g/mol) | n (m÷M) | ÷ by | Ratio | Magnesium | 63.63 | 63.63 | 24.31 | 2.617441382 | 2.273125 | 1 | Oxygen | 36.37 | 36.37 | 16.00 | 2.273125 | 2.273125 | 1 | Since the ratio is 1:1, the empirical formula is MgO. 2. Refer to “Determining the Empirical Formula of Magnesium Oxide Lab Solutions” sheet 3. You need to round the empirical formula to a whole number ratio because you cannot have decimals in the subscripts, which means that you cannot have a fractional amount of molecules in a substance.