1) The density of lead metal is 11.35 g/cm^3. If 12.49g of lead metal is added to a 10-mL graduated cylinder containing 5.72 mL of water, what will be the final volume reading of the water in the cylinder?
2) The mass of a beaker is 5.944g. After 5.00 mL of alcohol is pipetted into the beaker, the combined mass in 9.891g. From the data determine the density of the alcohol.
Volume Question? Chemistry?
2015-08-30 6:07 pm
回答 (2)
2015-08-30 6:27 pm
1)
Volume of lead metal
= (Mass of lead metal) / (Density of lead metal)
= (12.49 g) / (11.35 g/cm³)
= 1.10 cm³
Final volume reading
= (5.72 mL) + (1.10 cm³)
= (5.72 mL) + (1.10 mL)
= 6.82 mL
2)
Mass of alcohol
= (9.891 - 5.944) g
= 3.947 g
Density of alcohol
= (Mass of alcohol) / (Volume of alcohol)
= (3.947 g) / (5.00 mL)
= 0.789 g/mL
Volume of lead metal
= (Mass of lead metal) / (Density of lead metal)
= (12.49 g) / (11.35 g/cm³)
= 1.10 cm³
Final volume reading
= (5.72 mL) + (1.10 cm³)
= (5.72 mL) + (1.10 mL)
= 6.82 mL
2)
Mass of alcohol
= (9.891 - 5.944) g
= 3.947 g
Density of alcohol
= (Mass of alcohol) / (Volume of alcohol)
= (3.947 g) / (5.00 mL)
= 0.789 g/mL
2015-08-30 6:23 pm
1) Volume of 12.49g of lead metal = 12.49 g/11.35 g/cm^3 = 1.10 cm^3 = 1.10 mL
Thus the final volume reading of the water in the cylinder = 5.72 + 1.10 = 6.82 mL
2) Mass of alcohol = 9.891 - 5.944 = 3.947 g
Volume of alcohol = 5.00 mL
Hence density = 3.947 g/5.00 mL = 0.789 g/mL
Thus the final volume reading of the water in the cylinder = 5.72 + 1.10 = 6.82 mL
2) Mass of alcohol = 9.891 - 5.944 = 3.947 g
Volume of alcohol = 5.00 mL
Hence density = 3.947 g/5.00 mL = 0.789 g/mL
收錄日期: 2021-04-18 00:07:44
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