Density Lab B

[YellowMamba]

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 Yeah on 2 I divided by 10000 instead of multiplying when converting to meters squared.
On #4 I probably forgot that a square meter is 10000 square cm rather than 1000. 

[YellowMamba]

1. A ball with diameter 8.7 x 10^5 micrometers has a mass of 1 kg. What is its density in mg/mL?
2. A weight made of lead is tied to a scale which reads 5 kg. When the weight is completely submerged in water, the scale reads 4.559 kg. What is the volume of the weight in cubic cm?
3. An iceberg (0.917 g/c^3) of volume 125,000,000 m^3 floats at an unknown height in the Atlantic Ocean (saltwater = 1.03 g/c^3). The iceberg then drifts into Hudson Bay (freshwater = 1 g/c^3) and floats at a different unknown height. Assuming the iceberg is a perfect cube, what is the difference between the height at which the iceberg floats in saltwater and the height at which the iceberg floats in freshwater?

[azboy1910]

1. A ball with diameter 8.7 x 10^5 micrometers has a mass of 1 kg. What is its density in mg/mL?
2. A weight made of lead is tied to a scale which reads 5 kg. When the weight is completely submerged in water, the scale reads 4.559 kg. What is the volume of the weight in cubic cm?
3. An iceberg (0.917 g/c^3) of volume 125,000,000 m^3 floats at an unknown height in the Atlantic Ocean (saltwater = 1.03 g/c^3). The iceberg then drifts into Hudson Bay (freshwater = 1 g/c^3) and floats at a different unknown height. Assuming the iceberg is a perfect cube, what is the difference between the height at which the iceberg floats in saltwater and the height at which the iceberg floats in freshwater?

Answers (Click to Show Hidden Text)

I've been pretty rusty, but I'll answer. 

1. Density in mg/mL is about 2.9, considering significant figures is 3. 
2. Volume in cubic cm is about 44.95, considering significant figures is 40. 
3. Difference in heights in m is about 13.35, considering significant figures is 10. 

[YellowMamba]

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 1. I got about 3.87:
8.7 x 10^5 micrometers = .87 meters
.87/2 = .435 (radius)
.435^3 = about .0823
.0823(pi) = about .259
The volume of the sphere is about .259 cubic meters
Therefore it is about 259,000 mL
The mass of the ball is 1 kg
Therefore it is 1,000,000 mg
1,000,000/259,000 = about 3.86, but taking out rounding on previous calculations gives us about 3.87
2. I got 441 cubic cm.
The weight weighs .441 kg less in the water
That is 441 grams
Water is about 1 gram/cubic cm
441 grams = 441 cubic Cm
3. Is correct 

[azboy1910]

Click to Show Answer

 1. I got about 3.87:
8.7 x 10^5 micrometers = .87 meters
.87/2 = .435 (radius)
.435^3 = about .0823
.0823(pi) = about .259
The volume of the sphere is about .259 cubic meters
Therefore it is about 259,000 mL
The mass of the ball is 1 kg
Therefore it is 1,000,000 mg
1,000,000/259,000 = about 3.86, but taking out rounding on previous calculations gives us about 3.87
2. I got 441 cubic cm.
The weight weighs .441 kg less in the water
That is 441 grams
Water is about 1 gram/cubic cm
441 grams = 441 cubic Cm
3. Is correct 

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I think you're right about #2, I made a mistake by dividing the buoyant force by the gravitational constant, which I don't know why I did lol. However, on #1, I noticed while you were trying to calculate the volume, you forgot to multiply your equation for the volume of the sphere by 4/3. 

[theprimegrinder]

Umm..I'm just gonna post another problem to keep things going.

1) The Van Der Waals Equation is shown in the link below. https://www.qsstudy.com/wp-content/uplo ... tion-1.jpg
a) In the equation, a & b are constants. What do they correct for in a gas?
b) What equation does it modify? What side of the equation/what variables are modified.
c) At what temperature (low/high) & pressure (low/high) are gases most ideal?

2) How large would a balloon filled with helium need to be in order to lift a 100kg man? The density of helium is ρHe=0.164kgm3. The density of air is ρair=1.29kgm3. The mass of the balloon is negligible.

[azboy1910]

Umm..I'm just gonna post another problem to keep things going.

1) The Van Der Waals Equation is shown in the link below. https://www.qsstudy.com/wp-content/uplo ... tion-1.jpg
a) In the equation, a & b are constants. What do they correct for in a gas?
b) What equation does it modify? What side of the equation/what variables are modified.
c) At what temperature (low/high) & pressure (low/high) are gases most ideal?

2) How large would a balloon filled with helium need to be in order to lift a 100kg man? The density of helium is ρHe=0.164kgm3. The density of air is ρair=1.29kgm3. The mass of the balloon is negligible.

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 1a) The constants a and b correct for the molecular forces between gas molecules and the volume of those molecules.  
1b) It modifies the ideal gas law. The left side of this gas law is modified, as pressure is corrected by the term an^2/V^2 and is added to the pressure (P) and volume is corrected by the term nb and is subtracted from the volume (V). (Hopefully that made sense)
1c) Gases are most ideal at high temperature and low pressure (relative).

2) about 88.8 m^3

[theprimegrinder]

All correct! Your turn!

[azboy1910]

1. Briefly explain what the compressibility factor of a gas is and how it affects its behavior.
2. What is the purpose of a ballast tank in a submarine in terms of buoyancy?
3. A man weighs 100. kg. What is the pressure he exerts on the floor, assuming both of his feet are flat on the ground and are represented by 0.3 m by 0.1 m rectangles?

[YellowMamba]

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1. Compressibility is the how much the volume of a gas can shrink under pressure. This affects the behavior of the gas by determining how tightly it can be packed.
2. A ballast tank in a submarine is used to make the submarine dive or rise. It changes the amount of water in it, which changes the density of the submarine, and can thus change the submarine’s density from positive to negative buoyancy or vice versa.
3. .161 atm