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Chapter 20: Problem 75

Although He is the second most abundant element in the universe, it is very rare on the earth. Why?

Short Answer

Expert verified
Helium is rare on Earth due to its lightweight nature, its inability to be retained in minerals and rocks, its tendency to escape Earth's gravitational pull, and its limited production as a byproduct of radioactive decay within the Earth's crust. These factors, combined with its inert properties, result in helium's scarcity on our planet, despite its abundance in the universe.

Step by step solution

01

Understand the properties of helium

Helium is an inert gas with an atomic number of 2. It is a light element, and because it does not readily bond with other elements, it remains mostly in its atomic form. Helium is primarily formed through nuclear fusion processes in stars, like our Sun. This is a key property of helium to remember as we explore its rarity on Earth.
02

Examine helium's origin on Earth

Helium can be found on Earth in trace amounts, mainly formed as a byproduct of nuclear decay from heavier radioactive elements, like uranium and thorium, found inside the Earth's crust. As these elements decay, they release helium atoms, which eventually make their way to the surface and become part of the Earth's atmosphere. This process is slow, and the amount of helium produced is very small compared to its abundance in the universe.
03

Discuss helium's escape from Earth's atmosphere

Due to its lightweight nature and relatively high energy, helium atoms can easily escape Earth's gravitational pull when they reach the upper layers of the atmosphere. Once they escape, helium atoms drift off into space, further decreasing the amount of helium available on Earth.
04

The role of helium in Earth's geological processes

Helium is poorly retained in minerals and rocks; therefore, the geological processes, like the formation of mountains or the movement of tectonic plates, do not incorporate helium into Earth's crust. This property of helium also contributes to its low abundance on Earth.
05

Conclude the explanation of helium's rarity on Earth

Factors that cause helium to be rare on Earth include its lightweight nature, its poorly-retained property in minerals and rocks, its escape from Earth's gravitational pull, and its limited production through the decay of radioactive elements within the Earth's crust. Despite being the second most abundant element in the universe, these unique properties and processes result in helium's scarcity on our planet.

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Most popular questions from this chapter

Captain Kirk has set a trap for the Klingons who are threatening an innocent planet. He has sent small groups of fighter rockets to sites that are invisible to Klingon radar and put a decoy in the open. He calls this the 鈥渇ishhook鈥� strategy. Mr. Spock has sent a coded message to the chemists on the fighters to tell the ships what to do next. The outline of the message is Fill in the blanks of the message using the following clues. (1) Symbol of the halogen whose hydride has the second highest boiling point in the series of HX compounds that are hydrogen halides. (2) Symbol of the halogen that is the only hydrogen halide, HX, that is a weak acid in aqueous solution. (3) Symbol of the element whose existence on the sun was known before its existence on earth was discovered. (4) The Group 5A element in Table 20.13 that should have the most metallic character. (5) Symbol of the Group 6A element that, like selenium, is a semiconductor. (6) Symbol for the element known in rhombic and monoclinic forms. (7) Symbol for the element that exists as diatomic molecules in a yellow-green gas when not combined with another element. (8) Symbol for the most abundant element in and near the earth鈥檚 crust. (9) Symbol for the element that seems to give some protection against cancer when a diet rich in this element is consumed. (10) Symbol for the smallest noble gas that forms compounds with fluorine having the general formula \(\mathrm{AF}_{2}\) and \(\mathrm{AF}_{4}\) (reverse the symbol and split the letters as shown). (11) Symbol for the toxic element that, like phosphorus and antimony, forms tetrameric molecules when uncombined with other elements (split the letters of the symbol as shown). (12) Symbol for the element that occurs as an inert component of air but is a very prominent part of fertilizers and explosives.

The heaviest member of the alkaline earth metals is radium (Ra), a naturally radioactive element discovered by Pierre and Marie Curie in \(1898 .\) Radium was initially isolated from the uranium ore pitchblende, in which it is present as approximately 1.0 g per 7.0 metric tons of pitchblende. How many atoms of radium can be isolated from \(1.75 \times 10^{8} \mathrm{g}\) pitch- blende \((1 \text { metric ton }=1000 \mathrm{kg})\) ? One of the early uses of radium was as an additive to paint so that watch dials coated with this paint would glow in the dark. The longest-lived isotope of radium has a half-life of \(1.60 \times 10^{3}\) years. If an antique watch, manufactured in \(1925,\) contains 15.0 \(\mathrm{mg}\) radium, how many atoms of radium will remain in 2025\(?\)

What is the hybridization of the central atom in each of the following molecules? a. \(\mathrm{SF}_{6}\) b. \(\mathrm{ClF}_{3}\) c. \(\mathrm{GeCl}_{4}\) d. \(\mathrm{XeF}_{4}\)

The resistivity (a measure of electrical resistance) of graphite is \((0.4 \text { to } 5.0) \times 10^{-4}\) ohm \(\cdot \mathrm{cm}\) in the basal plane. (The basal plane is the plane of the six-membered rings of carbon atoms.) The resistivity is 0.2 to 1.0 ohm \(\cdot \mathrm{cm}\) along the axis perpendicular to the plane. The resistivity of diamond is \(10^{14}\) to \(10^{16} \mathrm{ohm} \cdot \mathrm{cm}\) and is independent of direction. How can you account for this behavior in terms of the structures of graphite and diamond?

Write the Lewis structure for \(\mathrm{O}_{2} \mathrm{F}_{2}\) . Predict the bond angles and hybridization of the two central oxygen atoms. Assign oxidation states and formal charges to the atoms in $\mathrm{O}_{2} \mathrm{F}_{2} .\( The compound \)\mathrm{O}_{2} \mathrm{F}_{2}$ is a vigorous and potent oxidizing and fluorinating agent. Are oxidation states or formal charges more useful in accounting for these properties of $\mathrm{O}_{2} \mathrm{F}_{2} ?$
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