Chemistry, 02.10.2019 20:00 jaden2058

Apiece of high-density styrofoam measuring 24.0 cm by 36.0 cm by 5.0 cm floats when placed in a tub of water. when a 1.5 kg book is placed on top of the styrofoam, the styrofoam partially sinks at 3.0 cm. assuming that the density of water is 1.00 g/ml, what is the density of styrofoam?

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Answer from: Tonyang1742

Explanation:

The given data is as follows.

Width of Styrofoam = 24.0 cm

Length of Styrofoam = 36.0 cm

Height of Styrofoam = 5.0 cm

Therefore, volume of the Styrofoam will be calculated as follows.

Volume = length × width × height

= (36.0 × 24.0 × 5.0) $cm^{3}$

= 4320 $cm^{3}$

or, = $4.32 \times 10^{3} cm^{3}$

As Styrofoam partially sinks at 3.0 cm and total height of Styrofoam is 5.0 cm. Hence, height of Styrofoam above the water is (5.0 - 3 cm) = 2 cm.

So, volume of water displaced is as follows.

24.0 cm × 36.0 cm × 2.0 cm

= $1.73 \times 10^{3} cm^{3}$

Hence, mass of displaced water is as follows.

mass = density × volume

= $1.00 g/cm^{3} \times 1.73 \times 10^{3} cm^{3}$

= $1.73 \times 10^{3} g$

Since, book is placed on the Styrofoam. Therefore, mass of water displaced is also equal to the following.

Mass of water displaced = mass of book + mass of Styrofoam

$1.73 \times 10^{3} g$ = 1500 g + mass of Styrofoam

(1730 - 1500) g = mass of Styrofoam

mass of Styrofoam = 230 g

Therefore, calculate the density of Styrofoam as follows.

Density = $\frac{mass}{volume}$

= $\frac{230}{4.32 \times 10^{3} cm^{3}}$

= $53.24 \times 10^{-3} g cm^{-3}$

Thus, we can conclude that the density of Styrofoam is $53.24 \times 10^{-3} g cm^{-3}$ .

Answer from: Quest

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in $h_{2}s$ , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between $h_{2}s$ molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent $h_{2}s$ molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.

Answer from: Quest

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