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1000 4. (EC) 70.4g of hematite (iron ore) was placed in a flask whose maximum volume was 55.0 mL. The flask with hematite was then carefully filled with water too the maximum line and re-massed. The hematite and water had a mass of 109.3g. The density of the water was 0.996g/cm³. What was the density of the hematite?

1000 4. (EC) 70.4g of hematite (iron ore) was placed in a flask whose maximum volume was 55.0 mL. The flask with hematite was then carefully filled with water too the maximum line and re-massed. The hematite and water had a mass of 109.3g. The density of the water was 0.996g/cm³. What was the density of the hematite?

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Chem 105 Online POST LAB CONCLUSIONS (you may use a separate sheet of paper for calculations as well) 1. Why was something like a wooden block's density calculated by measuring the dimensions directly and calculating the volume instead of using the displacement in water method? The wood would have soaked the water up, therefore Creating inaccrrate totals. 2. Measure the mass of each coin separately and record the masses below. Why were several coins used for the density calculation instead of one. 5.00g 4.909 4.93g 4.979 5.05g 3. Ethyl acetate has a characteristic fruity odor and is used as a solvent in paint lacquers ethyl acetate. What volume is this (in and perfumes. An experiment requires 0.985 liters)? The density of ethyl acetate is 0.902g/mL. m=0.985 kg d=0.902g/mL m²0.985x10009=985g d=m/ dxv=m v=m 9859 V=0.902 ML v=1092.0ml V²1092-0 =1.092L 4. (EC) 70.4g of hematite (iron ore) was placed in a flask whose maximum volume was 55.0 mL. The flask with hematite was then carefully filled with water too the maximum line and re-massed. The hematite and water had a mass of 109.3g. The density of the water was 0.996g/cm³. What was the density of the hematite? 5. (EC) The land area of Greenland is 820,000 mi2, with only 134,000 mi² free of perpetual ice. The average thickness of this ice is 5,100 ft. Estimate the mass of the ice (assume two significant figures). The density of ice is 0.927g/cm³. greenland-820,000mi 2 Land of Thickness of ice=5100 ft volume of ice areax thickness 2686,000m: 2x5100 ft. 1² Area onder ice total area-area free of ice (820,000-134,000) mi ² = 686,000 mil Claaft mass mass of ice: density = volume mass-density x volum =0.9272x2.76 = 2.55852x10² 1 2 2 6 X10²!
Transcribed Image Text:Chem 105 Online POST LAB CONCLUSIONS (you may use a separate sheet of paper for calculations as well) 1. Why was something like a wooden block's density calculated by measuring the dimensions directly and calculating the volume instead of using the displacement in water method? The wood would have soaked the water up, therefore Creating inaccrrate totals. 2. Measure the mass of each coin separately and record the masses below. Why were several coins used for the density calculation instead of one. 5.00g 4.909 4.93g 4.979 5.05g 3. Ethyl acetate has a characteristic fruity odor and is used as a solvent in paint lacquers ethyl acetate. What volume is this (in and perfumes. An experiment requires 0.985 liters)? The density of ethyl acetate is 0.902g/mL. m=0.985 kg d=0.902g/mL m²0.985x10009=985g d=m/ dxv=m v=m 9859 V=0.902 ML v=1092.0ml V²1092-0 =1.092L 4. (EC) 70.4g of hematite (iron ore) was placed in a flask whose maximum volume was 55.0 mL. The flask with hematite was then carefully filled with water too the maximum line and re-massed. The hematite and water had a mass of 109.3g. The density of the water was 0.996g/cm³. What was the density of the hematite? 5. (EC) The land area of Greenland is 820,000 mi2, with only 134,000 mi² free of perpetual ice. The average thickness of this ice is 5,100 ft. Estimate the mass of the ice (assume two significant figures). The density of ice is 0.927g/cm³. greenland-820,000mi 2 Land of Thickness of ice=5100 ft volume of ice areax thickness 2686,000m: 2x5100 ft. 1² Area onder ice total area-area free of ice (820,000-134,000) mi ² = 686,000 mil Claaft mass mass of ice: density = volume mass-density x volum =0.9272x2.76 = 2.55852x10² 1 2 2 6 X10²!
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