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Gold is one of the few metals that can be obtained by panning, where a simple pan is used to separate gold from other deposits found in or near a stream bed. What two properties of gold make it possible to find gold, but not metals like copper, silver, lead, and aluminum, by panning?

Short Answer

Expert verified
The two properties of gold that make it possible to find gold by panning are its high density and its resistance to corrosion. Gold's high density of 19.3 grams per cubic centimeter allows it to settle at the bottom of the pan, making it easier to separate from lighter sediments. Its resistance to corrosion ensures that gold remains in its metallic form, while other metals like copper, silver, lead, and aluminum corrode over time and become more difficult to separate through panning.

Step by step solution

01

Understanding panning

Panning is a simple method of extracting gold from other deposits found in or near a stream bed. The process involves placing sediment and water from the stream bed into a pan and then gently swirling the water around. The denser gold particles will sink to the bottom and can be separated from the lighter material that is washed away. Step 2: Identifying the properties of gold
02

Identifying the properties of gold

Gold has several unique properties that make it suitable for panning. The two primary factors that help in finding gold while panning are its high density and its resistance to corrosion. Step 3: Explaining the importance of high-density
03

Explaining the importance of high-density

Gold is a very dense metal, with a density of 19.3 grams per cubic centimeter. Due to its high density, gold particles settle at the bottom of the pan, allowing them to be separated from the lighter sediments. Other metals like copper, silver, lead, and aluminum have lower densities, making them harder to separate using panning. Step 4: Explaining the importance of resistance to corrosion
04

Explaining the importance of resistance to corrosion

Gold is chemically inert and does not react with air, water, or most other chemicals. This makes it resistant to corrosion. In a stream bed, other metals like copper, silver, lead, and aluminum will corrode over time, forming oxides or other chemical forms that are more challenging to isolate through panning. Gold's corrosion resistance preserves its metallic form, making it easier to separate from other sediments in the panning process.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

High Density of Gold
Gold is renowned for its high density, which is significantly greater than many common materials and even most other metals. Density is defined as mass per unit volume, and for gold, this is approximately 19.3 grams per cubic centimeter. This remarkable density is key in methods like gold panning because gold particles will sink more rapidly than lighter materials when both are placed in a pan filled with water.

During the panning process, the heavy gold particles quickly move to the bottom of the pan, separating themselves from lighter sand, soil, and rocks, which can then be washed away. This natural physical separation is due to gravity acting more forcefully on the dense gold particles. To put it into perspective, metals like aluminum have a density of about 2.7 grams per cubic centimeter, which is why they would remain suspended in the mixture or be washed away more easily than gold during panning.
Resistance to Corrosion
Gold's resistance to corrosion is an intrinsic property that contributes to its enduring value and practical usage in panning. Unlike other metals that readily react with elements such as oxygen and sulfur to form oxides and sulfides, gold maintains its elemental state, free from corrosion, due to its extraordinary chemical inertness.

This non-reactivity means that even after extended periods submerged in water or buried in soil, gold will not corrode, tarnish, or lose its luster. Other metals, such as iron which rusts, or silver which can tarnish, are less resistant to environmental factors and thus are not typically found in metallic form through panning. Gold's long-lasting metallic state aids prospectors in identifying and collecting it amongst river sediments as the shiny, unaltered nuggets or flakes.
Properties of Metals
Metals are a class of elements characterized by a set of common properties that include high electrical and thermal conductivity, malleability, ductility, and usually a shiny appearance. They tend to lose electrons and form positive ions in chemical reactions. Metals often have high melting and boiling points due to strong metallic bonds, but these properties can vary widely among them.

Unique Metallic Characteristics of Gold

Gold, in particular, has some of these general metallic properties but is distinct for its softness, malleability, high density, and most notably, its exceptional resistance to corrosion. Understanding these properties helps elucidate why gold can be found naturally in nuggets or grains in stream beds, and why it has been so highly valued for millennia.
Separation Techniques in Chemistry
Chemistry employs various separation techniques to isolate components of a mixture or to remove impurities from a substance. These techniques are based on the unique physical and chemical properties of the substances involved.

Some common separation methods include filtration, distillation, crystallization, and chromatography. Panning, in the context of gold, is a form of gravity separation. The process capitalizes on the disparity in density between gold and other materials found in sediment. By utilizing the natural forces of gravity and the physical properties of gold, panners can effectively separate gold flakes or nuggets from surrounding debris. This technique is a prime example of applied separation science, demonstrating how understanding the fundamental properties of elements can yield practical methods for their extraction.

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Most popular questions from this chapter

Suppose you have a solution that might contain any or all of the following cations: \(\mathrm{Ni}^{2+}, \mathrm{Ag}^{+}, \mathrm{Sr}^{2+},\) and \(\mathrm{Mn}^{2+}\). Addition of HCl solution causes a precipitate to form. After filtering off the precipitate, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution is added to the resulting solution and another precipitate forms. This is filtered off, and a solution of \(\mathrm{NaOH}\) is added to the resulting solution. No precipitate is observed. Which ions are present in each of the precipitates? Which of the four ions listed above must be absent from the original solution?

(a) How many milliliters of a stock solution of \(6.0 \mathrm{M} \mathrm{HNO}_{3}\) would you have to use to prepare \(110 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{HNO}_{3} ?\) (b) If you dilute \(10.0 \mathrm{~mL}\) of the stock solution to a final volume of \(0.250 \mathrm{~L}\), what will be the concentration of the diluted solution?

Federal regulations set an upper limit of 50 parts per million (ppm) of \(\mathrm{NH}_{3}\) in the air in a work environment [that is, 50 molecules of \(\mathrm{NH}_{3}(g)\) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing \(1.00 \times 10^{2} \mathrm{~mL}\) of \(0.0105 \mathrm{M} \mathrm{HCl}\). The \(\mathrm{NH}_{3}\) reacts with \(\mathrm{HCl}\) as follows: $$\mathrm{NH}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q)$$ After drawing air through the acid solution for \(10.0 \mathrm{~min}\) at a rate of \(10.0 \mathrm{~L} / \mathrm{min},\) the acid was titrated. The remaining acid needed \(13.1 \mathrm{~mL}\) of \(0.0588 \mathrm{M} \mathrm{NaOH}\) to reach the equivalence point. (a) How many grams of \(\mathrm{NH}_{3}\) were drawn into the acid solution? (b) How many ppm of \(\mathrm{NH}_{3}\) were in the air? (Air has a density of \(1.20 \mathrm{~g} / \mathrm{L}\) and an average molar mass of \(29.0 \mathrm{~g} / \mathrm{mol}\) under the conditions of the experiment.) (c) Is this manufacturer in compliance with regulations?

Tartaric acid, \(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\), has two acidic hydrogens. The acid is often present in wines and precipitates from solution as the wine ages. A solution containing an unknown concentration of the acid is titrated with \(\mathrm{NaOH}\). It requires \(24.65 \mathrm{~mL}\) of \(0.2500 \mathrm{M} \mathrm{NaOH}\) solution to titrate both acidic protons in \(50.00 \mathrm{~mL}\) of the tartaric acid solution. Write a balanced net ionic equation for the neutralization reaction, and calculate the molarity of the tartaric acid solution.

A sample of \(5.53 \mathrm{~g}\) of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is added to \(25.0 \mathrm{~mL}\) of 0.200 \(\mathrm{M} \mathrm{HNO}_{3}\) (a) Write the chemical equation for the reaction that occurs. (b) Which is the limiting reactant in the reaction? (c) How many moles of \(\mathrm{Mg}(\mathrm{OH})_{2}, \mathrm{HNO}_{3},\) and \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) are present after the reaction is complete?

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