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Chapter 1: Problem 81

Water has a density of 0.997 \(\mathrm{g} / \mathrm{cm}^{3}\) at \(25^{\circ} \mathrm{C} ;\) ice has a density of 0.917 \(\mathrm{g} / \mathrm{cm}^{3}\) at \(-10^{\circ} \mathrm{C}\) (a) If a soft-drink bottle whose volume is 1.50 \(\mathrm{L}\) is completely filled with water and then frozen to \(-10^{\circ} \mathrm{C},\) what volume does the ice occupy? (b) Can the ice be contained within the bottle?

Short Answer

Expert verified
The mass of the water in the bottle is approximately 1495.5 g. When frozen, the ice occupies a volume of approximately 1630.4 cm³. Since the volume of the ice is greater than the volume of the bottle (1500 cm³), the ice cannot be contained within the bottle.

Step by step solution

01

Calculate the mass of the water

First, we need to find the mass of the water in the bottle before freezing. We can do this by using the formula for finding mass, which is density times volume. We are given the volume of the bottle, 1.50 L, and the density of the water, 0.997 g/cm³. First, we need to convert the volume from liters to cubic centimeters, as follows: \[1.50\, \text{L} \times \frac{1000\, \text{cm}^3}{1\, \text{L}} = 1500\, \text{cm}^3\] Now, we can find the mass of the water: \[m = \rho \times V = 0.997\, \frac{\text{g}}{\text{cm}^3} \times 1500 \, \text{cm}^3 \approx 1495.5\, \text{g}\] The mass of the water is approximately 1495.5 g.
02

Calculate the volume of the ice

Now, we need to find the volume occupied by the ice by using the mass and the density of the ice at -10°C. We are given the density of the ice as 0.917 g/cm³. We can find the volume by rearranging the formula for finding mass: \[V_\text{ice} = \frac{m}{\rho_\text{ice}} = \frac{1495.5\, \text{g}}{0.917\, \frac{\text{g}}{\text{cm}^3}} \approx 1630.4\, \text{cm}^3\] The volume of the ice is approximately 1630.4 cm³.
03

Can the ice be contained within the bottle?

Finally, we need to determine if the bottle can contain the ice. We do this by comparing the volume of the ice with the volume of the bottle. \[V_\text{ice} = 1630.4\, \text{cm}^3 > 1500\, \text{cm}^3 = V_\text{bottle}\] Since the volume of the ice is greater than the volume of the bottle, the ice cannot be contained within the bottle.

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Most popular questions from this chapter

Judge the following statements as true or false. If you believe a statement to be false, provide a corrected version.(a) Air and water are both elements. (b) All mixtures contain at least one element and one compound.(c) Compounds can be decomposed into two or more other substances; elements cannot. (d) Elements can exist in any of the three states of matter. (e) When yellow stains in a kitchen sink are treated with bleach water, the disappearance of the stains is due to a physical change. (f) A hypothesis is more weakly supported by experimental evidence than a theory. (g) The number 0.0033 has more significant figures than 0.033 . (h) Conversion factors used in converting units always have a numerical value of one. (i) Compounds always contain at least two different elements.

Using your knowledge of metric units, English units, and the information on the back inside cover, write down the conversion factors needed to convert (a) \mum to mm, (b) ms to ns, (c) mi to km, (d) \(\mathrm{ft}^{3}\) to \(\mathrm{L}\) .

What type of quantity (for example, length, volume, density) do the following units indicate? (a) \(\mathrm{mL},(\mathbf{b}) \mathrm{cm}^{2}\) ,\((\mathbf{c}) \mathrm{mm}^{3},(\mathbf{d}) \mathrm{mg} / \mathrm{L},(\mathbf{e}) \mathrm{ps},(\mathbf{f}) \mathrm{nm},(\mathrm{g}) \mathrm{K}\)

Perform the following conversions: (a) 5.00 days to s, (b) 0.0550 \(\mathrm{mi}\) to \(\mathrm{m},(\mathbf{c}) \$ 1.89 / \mathrm{gal}\) to dollars per liter,(d) 0.510 in. \(/ \mathrm{ms}\) to \(\mathrm{km} / \mathrm{hr},\) (e) 22.50 \(\mathrm{gal} / \mathrm{min}\) to \(\mathrm{L} / \mathrm{s}\) (f) 0.02500 \(\mathrm{ft}^{3}\) to \(\mathrm{cm}^{3}\) .

Gold can be hammered into extremely thin sheets called gold leaf. An architect wants to cover a 100 \(\mathrm{ft} \times 82\) ft ceiling with gold leaf that is five-millionths of an inch thick. The density of gold is \(19.32 \mathrm{g} / \mathrm{cm}^{3},\) and gold costs \(\$ 1654\) per troy ounce \((1\) troy ounce \(=31.1034768 \mathrm{g}) .\) How much will it cost the architect to buy the necessary gold?
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