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Lemur [1.5K]
2 years ago
15

Density (D) is defined as the ratio of mass (m) to volume (V) and can be determined from the expression D = . Find the density o

f a sample of pure aluminum if a rectangular prism of volume 32.32 cm3 occupies a mass of 87.329 g. (Remember to put a space between your numerical value and the unit.)
Chemistry
1 answer:
Usimov [2.4K]2 years ago
7 0

That is a question about density.

To solve that question, it is important to know how we can obtain the Density of an object. Below, you can see the formula:

\boxed{D=\frac{m}{v} }

D = Density\\m = Mass\\V = Volume

Now, it is easy to answer, right? We just replace the values in the formula. The mass is 87.329g and the volume is 32.32cm³. So,

D=\frac{m}{v} \\\\D = \frac{87.329g}{32.32cm^3} \\\\D \approx 2.70g/cm^3

Therefore, the density of the the aluminium is approximately 2.7 g/cm³.

I hope I've helped. ^^

Enjoy your studies \o/

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Answer:

a)  Δπ = 1.264 atm

b) W = 128 joules

c)  ΔH >> W  ( a factor greater than 17,000 )

Explanation:

a) The osmotic pressure, π , is determined by :

π = nRT/V, where n= moles of solute

                          R= 0.0821 Latm/kmol

                          T = 300 K

calling π(sw) osmotic pressure for  for sea water and π (fw) for fresh water,

salinity of sea water = 3.5 g / 1L water   (assuming only NaCl for the salts)

salinity of fresh water = 0.5 parts per thousand (range: 0- 0.5 ppt)

πsw = (3.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 1.475 atm

πfw = (0.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 0.211 atm

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b) W = Δπ V = 1.426 atm x 1L = 1.43 L-atm

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W =  101.33 j/ Latm x  1.43 Latm = 128 joules

c) ΔH = Q₁ + nΔH vap, where

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            ΔH vap = heat of vaporization

Q = mCΔT = 1000 g x 4.186 j x 73 K = 305.6 j = 0.3056 kj

ΔH vap = (1000 g/ 18 g/mol ) 40.7 kj/mol = 2,261 kj

ΔH =  0.3056 kj + 2,261 kj = 2,261.3 kj

Note = Q << ΔH vap and we could have neglected it.

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