A recent article in
Scientific American notes that life would be impossible on any planet close enough to the Sun to boil water, or far enough from the Sun that water froze. It also argues that the Earth is in one of the few parts of the galaxy where life could exist. Too close to the core, and either collisions with other objects would destroy the planet or cosmic radiation from neighboring stars would destroy life. Too far from the sun and there wouldnâ€t be enough of the elements needed to form a planet. Letâ€s assume that radiation becomes particularly dangerous to life when it carries enough energy to ionize a water molecule when it is absorbed.
2O(l) + h? ? H2O+ + eâ€“ ? = 1200 kJ/mol
Use Avogadroâ€s number and Planckâ€s constant to calculate the frequency (?) and wavelength (?) of a photon that has enough energy to ionize a single molecule of water. (Hint: Pay close attention to units!)
Some elements, such as lead and tin, have more than one common ion.
b. Predict the two most likely ions that can be formed by tin, we will call them ion #1 (the first ion that can be formed) and ion #2 (the second ion). Include the electron configurations of ions #1 and #2 in your answer and a brief explanation of how you chose these two ions.
c. Which of the following statements
best describes the resultant trend in atomic size when tin forms two ions. Explain your answer in terms of core charge, valence shell and/or number of electrons.
1. elemental tin is larger than ion #1 which is larger than ion #2
2. elemental tin is very much larger than ion #1 which is larger than ion #2
3. elemental tin is larger than ion #1 which is very much larger than ion #2
d. Draw a simple, representative photoelectron spectrum for tin. Assume that the subshells of n=3 and higher appear on the spectrum in the same orderly fashion as they did in n=1 and n=2, in other words, ignore any overlapping energies that might occur in shells n=3 and higher. Use this simple spectrum to illustrate and explain why tin does not go on to form a third cation. You do not need to include actual ionization energy values on your spectrum.
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