
Interpretation: The reason for placement of argon before potassium in spite of its greater atomic mass and the
Concept Introduction:
Periodic Table: The available chemical elements are arranged considering their
In periodic table the horizontal rows are called periods and the vertical column are called group.
In periodic table the horizontal rows are called periods and the vertical column are called group. There are seven periods and 18 groups present in the table and some of those groups are given particular name as follows,
Atomic Number: Atomic number of the element is equal to the number of protons present in the nucleus of the element which is denoted by symbol Z. The superscript presents on the left side of the
Atomic mass: It is generally expressed in terms of atomic mass unit
Mass number: It is given by the total number of protons and the neutrons present in the nucleus of the element and it is denoted by symbol A. The subscript that lies on the left side of the symbol of the element represents the mass number.

Explanation of Solution
The reason that supports the placement of
Analyze the position of argon and potassium in the periodic table.
Examining the periodic table shows that argon is placed before the potassium irrespective of high atomic mass of
The average atomic mass of
Determine the average atomic mass for the given elements.
The average atomic mass for the given elements is obtained by addition of individual atomic mass of the isotopes multiplied with their percent abundance of the respective isotope.
Therefore, the average atomic mass for argon is
The reason that support the position of argon before potassium and the average atomic mass of the element is determined by using the atomic number, atomic mass and the abundance of the isotopes for the element.
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Chapter 8 Solutions
Chemistry
- (2 pts) WSe2 is an ionic compound semiconductor that can be made to be p-type or n-type.What must happen to the chemical composition for it to be p-type? What must happen tothe chemical composition for it to be n-type?arrow_forward8. (2 pts) Silicon semiconductors have a bandgap of 1.11 eV. What is the longest photon wavelength that can promote an electron from the valence band to the conduction band in a silicon-based photovoltaic solar cell? Show all work. E = hv = hc/λ h = 6.626 x 10-34 Js c = 3.00 x 108 m/s 1 eV 1.602 x 10-19 Jarrow_forwardA solution containing 100.0 mL of 0.155 M EDTA buffered to pH 10.00 was titrated with 100.0 mL of 0.0152 M Hg(ClO4)2 in a cell: calomel electrode (saturated)//titration solution/Hg(l) Given the formation constant of Hg(EDTA)2-, logKf= 21.5, and alphaY4-=0.30, find out the cell voltage E. Hg2+(aq) + 2e- = Hg(l) E0= 0.852 V E' (calomel electrode, saturated KCl) = 0.241 Varrow_forward
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- Calculate the voltage of each of the following cells. a) Fe(s)/Fe2+ (1.55 x 10-2 M)//Cu2+ (6.55 x 10-3 M)/Cu(s) b) Pt, H2 (0.255 bar)/HCl (4.55 x 10-4 M), AgCl (sat'd)/Ag Fe2+ +2e- = Fe E0= -0.44 V Cu2+ + 2e- = Cu E0= 0.337 V Ag+ + e- = Ag E0= 0.799 V AgCl(s) + e- = Ag(s) + Cl- E0= 0.222 V 2H+ + 2e- = H2 E0= 0.000 Varrow_forwardA solution contains 0.097 M Ce3+, 1.55x10-3 M Ce4+, 1.55x10-3 M Mn2+, 0.097 M MnO4-, and 1.00 M HClO4 (F= 9.649 x 104 C/mol). a) Write a balanced net reaction that can occur between species in this solution. b) Calculate deltaG0 and K for the reaction. c) Calculate E and deltaG for the conditions given. Ce4+ + e- = Ce3+ E0= 1.70 V MnO4- + 8H+ + 5e- = Mn2+ + 4H2O E0= 1.507 Varrow_forward1. Provide a step-by-step mechanism for formation of ALL STEREOISOMERS in the following reaction. Na HCO3 (Sodium bicarbonate, baking soda) is not soluble in CH2Cl2. The powder is a weak base used to neutralize strong acid (pKa < 0) produced by the reaction. Redraw the product to show the configuration(s) that form at C-2 and C-4. Br2 OH CH2Cl2 Na* HCO3 Br HO OH + Na Br +arrow_forward
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- Chemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage Learning
