the atomic packing factor for HCP is 0.74. This problem calls for a demonstration that the APF for HCP is 0.74. Again‚ the APF is the ratio of the total sphere volume‚ VS‚ to the unit cell volume‚ VC. For HCP‚ there are the equivalent of six spheres per unit cell‚ and thus ⎛4πR3⎞ VS = 6⎜ 3 ⎟ = 8πR3 ⎝ ⎠ Now‚ the unit cell volume is the product of the base area times the cell height‚ c. The base area can be calculated as follows. The following figure shows an HCP unit cell and the basal plane
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has your data of KI masses‚ and the three volumes (from the three trials) needed to titrate the bleach Calculations (Data and Calculations Section in lab) 1. Use the equations given on the front page to determine the number of moles of sodium thiosulfate that are equivalent to one mole of sodium hypochlorite. 2. From your three trials‚ calculate the average volume of Na2S2O3 needed for the titration of 25.00mL of diluted bleach. 3. Use the average volume and the molarity of Na2S2O3 to determine
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Density Lab Report PURPOSE: a. To measure the masses and volumes of solids and liquids b. To calculate the densities of solids and liquids c. To calculate the specific gravities of solids and liquids d. To calculate the volume of a rectangular object and to express the volume in volume metric units e. To record data and calculate the values in the correct number of significant figures MATERIALS: 10 mL graduate balance test tube rack unknown liquid 50 mL graduate test
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types of volumetric glassware and the accuracy of measuring the volumes of liquids very precisely in quantitative laboratory work. The accuracy of the measurement the volumes is the degree of closeness of measurements of a quantity’s actual volumes while the precision of the volumes is the degree to which repeated measurements under unchanged conditions show the same results. Each of volumetric glassware is marked with its total volume‚ the notation of TD for ‘to deliver’ and TC for ‘to contain’ and
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described as the amount of matter (mass) squeezed into a given space (volume). The density of substance remains the same no matter the size of the sample at a given temperature. Quantitatively‚ density can be expressed as the mass of a substance per unit volume‚ and the volume of a cylinder can be expressed as π times the radius
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Pool Company Project * From Computer-Science: Software-Engineering Project Overview This document provide a sample of what is possible‚ and could lead to a project worth the maximum grade of 100% Something to think about!!! Do not think this is what you have to do to get the maximum number of points. It is just a sample! You are only bound by your programming ability. Please do not try and do more than you are capable of. Think‚ plan‚ design and code YOUR final project. This is your
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Trial 2 Trial 3 Mass of Empty 10 mL graduated cylinder (grams) 26.0 Volume of liquid (milliliters) 8.5 Mass od graduated cylinder and liquid (grams) 30.7 Part II: Density of Irregular-Shaped Solid Trial 1 Trial 2 Trial 3 Mass of solid (grams) 39.67 39.25 42.126 Volume of Water (milliliters) 50.1 50.0 50.09 Volume of water (milliliters) Volume of water and solid (milliliters) Part III: Density of Regular-Shaped
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strips‚ the area of the sphere and the area of the cylinder are approximately equal. This argument is in the spirit of how the Greeks compared slices to show that areas and volumes are the same. It is not only interesting reasoning in proportion‚ etc.‚ but it is a lesson in History. In the same spirit‚ you can compare the VOLUME of a hemi-sphere with that of the cylinder with an inverted cone removed. (I think of the hemisphere with the equator at the bottom and the
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Unknown Liquid Trial 1 Trial 2 Trial 3 Mass of Empty 10 mL graduated cylinder (grams) 25.55 25.56 25.55 Volume of liquid (milliliters) 8.8 8.65 8.5 Mass of graduated cylinder and liquid (grams) 30.65 30.62 30.565 Part II: Density of Irregular-Shaped Solid Mass of solid (grams) 39.537 38.515 40.975 Volume of water (milliliters) 50.01 49.9 52.4 Volume of water and solid (milliliters) 54.9 54 57 Part III: Density of Regular-Shaped Solid Mass of solid (grams)
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Formula Sheet Volume of prism area of cross section length – Area of trapezium = 1 (a + b)ha + b)h 2 Area of trapezium 1 ( 2 a h cross section Formula Sheet b length – Area of trapezium = 1 (a + b)h 2 a Formula Sheet h b – Area of trapezium = 1 (a + b)h 2 Volume of prism = area of cross section × length Volume of cone 1 πr 2 h 3 In anyCurved surface area of cone πrl triangle ABC – Area of triangle = 1 ab sin C 2 In any triangle ABC h b a Volume of prism = area
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