Why is using Archimedes principle a more suitable way of obtaining the volume of a metal than say, by simple displacement of water in a graduate?
A metal with density 4.76g/ml and with a volume of 5 ml is immersed in an unknown liquid. it weighs 16.65 g in the unknown liquid. calculate the density of the liquid.
Suppose that the thread and knot billy used to attach the metal when weighing the metal submerged in the liquid were thicker and bulkier than a simple thin piece of thread. what would this do to the resulting metal density value?
If you could please help me it would greatly be appreciated :))
A metal with density 4.76g/ml and with a volume of 5 ml is immersed in an unknown liquid. it weighs 16.65 g in the unknown liquid. calculate the density of the liquid.
Suppose that the thread and knot billy used to attach the metal when weighing the metal submerged in the liquid were thicker and bulkier than a simple thin piece of thread. what would this do to the resulting metal density value?
If you could please help me it would greatly be appreciated :))
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1. The experiment is more accurate if you take the apparent weight loss due to displacement of water rather than reading two not so accurate volumes on the graduated cylinder and finding the volume of water displaced by subtraction.
2. The actual mass of the metal is 5 ml x 4.76g/ml = 23.80g
The apparent mass is 16.65g so the upthrust from the liquid (the mass of liquid displaced) = 7.15g
The volume of liquid displaced is equal to the volume of the metal = 5 ml.
The density of the unknown liquid = 7.15g/ 5ml = 1.43g/ml.
3. The knot and thread would displace more water producing a larger apparent loss in weight, and giving an apparent volume for the metal sample which would be greater than the actual volume thus giving a lower density. (The density of the thread and knot would be much less than an equivalent volume of metal and the density calculated would be lower as a result)
2. The actual mass of the metal is 5 ml x 4.76g/ml = 23.80g
The apparent mass is 16.65g so the upthrust from the liquid (the mass of liquid displaced) = 7.15g
The volume of liquid displaced is equal to the volume of the metal = 5 ml.
The density of the unknown liquid = 7.15g/ 5ml = 1.43g/ml.
3. The knot and thread would displace more water producing a larger apparent loss in weight, and giving an apparent volume for the metal sample which would be greater than the actual volume thus giving a lower density. (The density of the thread and knot would be much less than an equivalent volume of metal and the density calculated would be lower as a result)
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The weight of the 5 ml of liquid is equal to the difference between the weight of the metal and the apparent weight when submerged. Once you know the weight of the 5 ml of liquid you can calculate the density. Divide the weight of the 5 ml of liquid by the acceleration due to gravity and this will give you the liquids mass, then divide the mass by the volume of 5 ml and you get the density of the liquid.