(a)
Interpretation: The atomic symbols for the given set of values have to be stated.
Concept introduction: The atomic symbol for an isotope is written as
To determine: The atomic symbol
(a)
Answer to Problem 45E
Answer
The atomic symbol for the given isotope is
Explanation of Solution
Given
Atomic number,
Number of neutrons,
An atom is represented by the following symbol.
Where,
The atomic mass is equal to the sum of atomic number and number of neutrons.
Substitute the values of
Also, the atomic number 8 corresponds to oxygen
The given element is an isotope of oxygen which has the symbol
(b)
Interpretation: The atomic symbols for the given set of values have to be stated.
Concept introduction: The atomic symbol for an isotope is written as
To determine: The atomic symbol
(b)
Answer to Problem 45E
Answer
The atomic symbol for the given isotope of chlorine is
Explanation of Solution
Atomic number of chlorine is 17.
Therefore, the atom symbol is
The given isotope of chlorine has an atomic symbol as
(c)
Interpretation: The atomic symbols for the given set of values have to be stated.
Concept introduction: The atomic symbol for an isotope is written as
To determine: The atomic symbol
(c)
Answer to Problem 45E
Answer
The atomic symbol for the given isotope is
Explanation of Solution
Given
Mass number,
Atomic number,
The atomic number 27 corresponds to cobalt (Co) in the periodic table.
Thus, the atomic symbol is
The given element is an isotope of cobalt with the atomic symbol
(d)
Interpretation: The atomic symbols for the given set of values have to be stated.
Concept introduction: The atomic symbol for an isotope is written as
To determine: The atomic symbol
(d)
Answer to Problem 45E
Answer
The atomic symbol for the given isotope is
Explanation of Solution
Given
Number of protons,
Number of neutrons,
Number of protons in the atom is equal to the atomic number.
Atomic mass is equal to the sum of number of neutrons and number of protons.
The atomic number 26 corresponds to iron (Fe) in the periodic table.
The given element is an isotope of iron with the atomic symbol
(e)
Interpretation: The atomic symbols for the given set of values have to be stated.
Concept introduction: The atomic symbol for an isotope is written as
To determine: The atomic symbol
(e)
Answer to Problem 45E
Answer
The atomic symbol for the given isotope is
Explanation of Solution
Given
Isotope of iodine, with mass number,
Atomic number of an Iodine is 53.
Therefore, the symbol for isotope is
The given isotope of iodine has an atomic symbol as
(f)
Interpretation: The atomic symbols for the given set of values have to be stated.
Concept introduction: The atomic symbol for an isotope is written as
To determine: The atomic symbol
(f)
Answer to Problem 45E
Answer
The atomic symbol for the given isotope is
Explanation of Solution
Given
Atomic number,
Number of neutrons,
The atomic mass is equal to the sum of atomic number and number of neutrons.
Substitute the values of
The atomic number 3 corresponds to lithium
The given element is an isotope of lithium which has the symbol
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Chapter 1 Solutions
Chemistry: An Atoms First Approach
- Chlorine has two natural isotopes: 1737Cl and 1735Cl. Hydrogen reacts with chlorine to form the compound HCl. Would a given amount of hydrogen react with different masses of the two chlorine isotopes? Does this conflict with the law of definite proportion? Why or why not?arrow_forwardThe element europium exists in nature as two isotopes: 151Eu has a mass of 150.9196 amu, and 153Eu has a mass of 152.9209 amu. The average atomic mass of europium is 151.96 amu. a. Calculate the relative abundance of the two europium isotopes. b. Graph each fractional abundance value as a y-axis value in association with its corresponding mass value on the x-axis. Starting from each x-axis value, where y = 0, draw a vertical line up to the fractional abundance value. The result will approximate the type of visual graph a mass spectrometer would yield for europium in the 150155 amu range.arrow_forwardThe element europium exists in nature as two isotopes: 151Eu has a mass of 150.9196 u and 153Eu has a mass of 152.9209 u. The average atomic mass of europium is 151.96 u. Calculate the relative abundance of the two europium isotopes.arrow_forward
- A cube of sodium has length 1.25 in. How many atoms are in that cube? (Note: dNa=0.968g/cm3.)arrow_forwardChlorine has two isotopes, Cl-35 and Cl-37. Their abundances are 75.53% and 24.47%, respectively. Assume that the only hydrogen isotope present is H-1. (a) How many different HCI molecules are possible? (b) What is the sum of the mass numbers of the two atoms in each molecule? (c) Sketch the mass spectrum for HCI if all the positive ions are obtained by removing a single electron from an HCI molecule.arrow_forwardClick on the site (http://openstaxcollege.org/l/16PhetAtomMass) and select the Mix Isotopes tab, hide the Percent Composition and Average Atomic Mass boxes, and then select the element boron. Write the symbols of the isotopes of boron that are shown as naturally occurring in significant amounts. Predict the relative amounts (percentages) of these boron isotopes found in nature. Explain the reasoning behind your choice. Add isotopes to the black box to make a mixture that matches your prediction in (b). You may drag isotopes from their bins or click on More and then move the sliders to the appropriate amounts. Reveal the Percent Composition and Average Atomic Mass boxes. How well does your mixture match with your prediction? If necessary, adjust the isotope amounts to match your prediction. Select Nature’s mix of isotopes and compare it to your prediction. How well does your prediction compare with the naturally occurring mixture? Explain. If necessary, adjust your amounts to make them match Nature’s amounts as closely as possible.arrow_forward
- Two samples of different compounds of sulfur and oxygen have the following compositions. Show that the compounds follow the law of multiple proportions. What is the ratio of oxygen in the two compounds for a fixed amount of sulfur? Amount S Amount O Compound A l.210g 1.811 g Compound B 1.783 g 1.779 garrow_forwardThere are 1.699 1022 atoms in 1.000 g of chlorine. Assume that chlorine atoms are spheres of radius 0.99 and that they are lined up side by side in a 0.5-g sample. How many miles in length is the line of chlorine atoms in the sample?arrow_forwardThe following chart shows a general decline in abundance with increasing mass among the first 30 elements. The decline continues beyond zinc. Notice that the scale on the vertical axis is logarithmic, that is, it progresses in powers of 10. The abundance of nitrogen, for example, is 1/10,000 (1/104) of the abundance of hydrogen. All abundances are plotted as the number of atoms per 102 atoms of H. (The fact that the abundances of Li, Be, and B, as well as those of the elements near Fe, do not follow the general decline is a consequence of the way that elements are synthesized in stars.) (a) What is the most abundant main group metal? (b) What is the most abundant nonmetal? (c) What is the most abundant metalloid? (d) Which of the transition elements is most abundant? (e) Which halogens are included on this plot, and which is the most abundant?arrow_forward
- Write the chemical formula of each of the following: a The compound made up of a crystal with two particles coming from chromium atoms for every three particles coming from oxygen atoms. b The compound made up of a crystal with one particle coming from a barium atom for every two particles coming from chlorine atoms. c The compound made up of molecules with 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms. d The compound made up of molecules with three hydrogen atoms, one phosphorus atom, and four oxygen atoms.arrow_forwardA cube of sodium has length 1.25 in. How many atoms are in that cube? (Note: dNa=0.968 g/cm3.)arrow_forwardAverage Atomic Weight Part 1: Consider the four identical spheres below, each with a mass of 2.00 g. Calculate the average mass of a sphere in this sample. Part 2: Now consider a sample that consists of four spheres, each with a different mass: blue mass is 2.00 g, red mass is 1.75 g, green mass is 3.00 g, and yellow mass is 1.25 g. a Calculate the average mass of a sphere in this sample. b How does the average mass for a sphere in this sample compare with the average mass of the sample that consisted just of the blue spheres? How can such different samples have their averages turn out the way they did? Part 3: Consider two jars. One jar contains 100 blue spheres, and the other jar contains 25 each of red, blue, green, and yellow colors mixed together. a If you were to remove 50 blue spheres from the jar containing just the blue spheres, what would be the total mass of spheres left in the jar? (Note that the masses of the spheres are given in Part 2.) b If you were to remove 50 spheres from the jar containing the mixture (assume you get a representative distribution of colors), what would be the total mass of spheres left in the jar? c In the case of the mixture of spheres, does the average mass of the spheres necessarily represent the mass of an individual sphere in the sample? d If you had 80.0 grams of spheres from the blue sample, how many spheres would you have? e If you had 60.0 grams of spheres from the mixed-color sample, how many spheres would you have? What assumption did you make about your sample when performing this calculation? Part 4: Consider a sample that consists of three green spheres and one blue sphere. The green mass is 3.00 g, and the blue mass is 1.00 g. a Calculate the fractional abundance of each sphere in the sample. b Use the fractional abundance to calculate the average mass of the spheres in this sample. c How are the ideas developed in this Concept Exploration related to the atomic weights of the elements?arrow_forward
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