1. The partial molar volumes of acetone and chloroform are 74.16 cm3 mol-1 and 80.23 cm3 mol-1 respectively. Calculate the volume of a mixture containing 25.00 moles of acetone and 75.00 moles of chloroform. 2. Acetone has a molar mass of 58.1 g mol-1 and a density of 0.787 g cm-3. Chloroform has a molar mass of 119.4 g mol-1 and a density of 1.499 g cm-3. Estimate the volume of the above mixture using this information. 3. Calculate the molal freezing point depression constant Kf for
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Chemistry in the kitchen Teacher’s guidelines Lorena Payà Vayà Llicència C Curs 2007/2008 Index Unit 1. Can we do chemistry in the kitchen? 1.1. Strawberry smoothie. 1.2. Laboratory vs. Kitchen. 1.3. Safety in lab and in the kitchen. 1.4. Cutting onions or a scientific investigation. 3 Unit 2. How do we measure in the kitchen? 2.1. How do we measure the quantity of each ingredient in our cooking? 2.2. An Old Scottish Recipe or how to convert units. 2.3. Does one kilo of sugar
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will be verifying case 3 as explained by professor. We will measure the mass of the metal cylinder using the digital balance. After measuring the mass we also gather the diameter and height using the vernier caliper to calculate its volume. Using the mass and volume‚ we calculated its density. Now a beaker was filled with water and the mass was measured. The cylinder we originally used will now be suspended in the beaker full of water and the change of mass will be recorded. The difference between
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understand Boyle’s Law. In the experiment the pressure in the system under constant temperature and mass was used to confirm if the laws are true. Boyles law relates pressure and volume while all other factors are consistent and states: for a fixed amount of gas kept at constant temp‚ the product of the pressure of the gas and its volume will remain constant if either quantity is changed‚ or where k is constant. The experiment consisted of using a piston‚ or in this case a syringe. Weights were attached to
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Table 2.1: Data of equivalent radius‚ solid density‚ bulk density and porosity of granular materials. nj Red beans Green beans Ground nuts Weight of 400 beans‚ M (kg) 38 22 144 Volume of 400 beans + void spaces‚ V (ml) 48 26 232 Initial reading of cylinder V1 (ml) 300 300 300 Final reading of cylinder V2 (ml) 285 290 210 Volume of void spaces‚ v=v1-v2 (ml) 15 10 90 (V-v) x 10-6 m3 3.3 x 10-5 1.6 x 10-5 1.42 x 10-4 Equivalent radius‚ r (mm) 2.70 x 10-3 2.1216 x 10-3 4.3925 x 10-3 Solid density‚ ρs (kg/m3)
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Question 1 1 out of 1 points The volume of a regular cylinder is V = πr2h. Using the value 3.1416 for the constant π‚ the volume (cm3) of a cylinder of radius 2.34 cm and height 19.91 expressed to the correct number of significant figures is _________. Selected Answer: d. 342 Correct Answer: d. 342 Response Feedback: Correct Question 2 1 out of 1 points There are ____________ significant figures in the answer to the following computation: (29.2 - 20.0 ) (1.79 x 105)
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and creates a platform to discuss the implications of these changes. The four V model: This is a diagram representing the four V model for Beck’s Plc.’s current and previous operations. Low Volume High Low Volume High High Variety Low High Variety Low High Variation Low High
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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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take only mass value of the 10.0mL deionized water‚ the mass value of the graduated cylinder was subtracted from the mass value of the 10.0mL of deionized water + the cylinder. 6. Density of 10.0mL of deionized water was calculated by its’ mass and volume. (d=m/v) 7. The exact density of the 10.0mL of deionized water was recorded on the data sheet. 8. Deionized water was added up to the 30.0mL mark of the 50mL graduated cylinder and Steps 3 to 7 were repeated for 30.0mL. 9. Deionized water was
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mass of a volatile liquid. Data Table: Mass of Test Tube and Stopper (g) | 10.864 g | Barometric Pressure (mmHg) | 749.31 mmHg | Temperature of Boiling Water (C) | 97.1 C | Mass of Test Tube‚ Stopper‚ and Condensed Liquid (g) | 10.890 g | Volume of Flask (mL) | 9.90 mL | Calculations: 749.31 mmHG*1 atm760 mmHg= .98593 atm 9.90 mL*1 L1000 mL= .00990 L 97.1 C+273=370.1 K Substituting these values into the equation: .98593 atm* .00990 L=n* .0821 atm L mol-1 K-1*370.1 K n= .000321 moles
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