how many mL of a 3.5 M sodium hydroxide will be needed to neutralize 15mL of a 4.3 M hydrochloric acid solution?
1 answer:
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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)
= 4320
or, =
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
=
Hence, mass of displaced water is as follows.
mass = density × volume
=
=
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
= 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 =
=
=
Thus, we can conclude that the density of Styrofoam is .