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91Ó°ÊÓ

To four significant figures, all of the following masses are possible for an individual titanium atom except one. The exception is (a) 45.95 u; (b) 46.95 u; (c) 47.87 u; (d) 47.95 u; (e) 48.95 u; (f) 49.94 u.

Short Answer

Expert verified
The impossible atomic mass for an individual titanium atom is (e) 48.95 u.

Step by step solution

01

Understanding taken values

Each number given is atomic mass to four significant figures, Therefore we can easily compare them. We examine \((a) 45.95 \,u;\) \((b) 46.95\, u;\) \((c) 47.87 \,u;\) \((d) 47.95 \,u;\) \((e) 48.95 \,u;\) and \((f) 49.94 \,u.\) atomic weights to current approx weight of titanium which is 47.87 u.
02

Identifying impossible mass

Compare every given atomic weight with the atomic weight of titanium, which is 47.87 u, obviously the value that exceeds the most from this weight will be the impossible mass for an individual titanium atom.
03

Select the right option

An analysis shows that options (a), (b), (c), (d), (e) and (f) lies close to the atomic weight of Titanium except (e) 48.95 u, which seems to be exceeding the atomic weight of Titanium most when compared with other options.

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Key Concepts

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

Atomic Mass
Atomic mass is a measure of the mass of an atom, typically expressed in atomic mass units (u). This unit is essential for understanding the properties of elements and how they behave in chemical reactions. It closely relates to the concept of significant figures, which help in representing the precision of the atomic mass values.
The atomic mass of an element considers both protons and neutrons found in the nucleus of an atom, as electrons have negligible mass. This value is crucial for determining the stoichiometry of chemical equations and allows scientists to make predictions about how elements will interact.
  • Protons and Neutrons: The sum of protons and neutrons gives the mass number, critical for calculating atomic mass.
  • Significant Figures: Use these to express atomic mass accurately, determining the level of precision in measurements.
Recognizing the atomic mass of titanium is especially useful in exercises where differing atomic masses are analyzed to identify exceptions, just as in the discussed problem.
Isotopes
Atoms of the same element can have varying numbers of neutrons, leading to different isotopes. Isotopes are fundamental to understanding atomic mass, as they affect the average atomic mass of an element.
An isotope of an element has the same number of protons but a different number of neutrons. For example, the element titanium has several isotopes, each contributing to its average atomic mass.
  • Isotopic Composition: Determines the different masses due to varying neutron numbers.
  • Averaging Isotope Masses: The atomic mass listed on the periodic table reflects weighted averages of all isotopes present.
In practical terms, understanding isotopes is necessary when deciphering exercises involving atomic masses. They help clarify which mass values are plausible for a given element.
Comparison of Atomic Weights
When comparing atomic weights of different atoms or isotopes, precision is key. Atomic weights differ due to isotopes and must be compared carefully to understand the variations completely. This process involves comparing the atomic masses carefully and observing how they relate to known values on the periodic table such as titanium's atomic mass in this exercise.
Representing atomic weights with four significant figures provides clarity in discrepancies.
  • Comparing Precisions: Ensures accurate identification of atomic weight discrepancies.
  • Expected Variances: Values can vary due to isotopic abundances, affecting comparisons like those seen with titanium's atomic weight.
Understanding atomic weights through comparison helps in identifying anomalies, precisely like picking the impossible mass of a titanium atom in puzzles such as the one in the original problem.

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Most popular questions from this chapter

Within the limits of experimental error, show that the law of conservation of mass was obeyed in the following experiment: \(10.00 \mathrm{g}\) calcium carbonate (found in limestone) was dissolved in 100.0 mL hydrochloric acid \((d=1.148 \mathrm{g} / \mathrm{mL}) .\) The products were \(120.40 \mathrm{g}\) solution (a mixture of hydrochloric acid and calcium chloride) and 2.22 L carbon dioxide gas \((d=1.9769 \mathrm{g} / \mathrm{L})\)

For the atom \(^{108}\) Pd with mass 107.90389 u, determine (a) the numbers of protons, neutrons, and electrons in the atom; (b) the ratio of the mass of this atom to that of an atom of \(^{12}_{6} \mathrm{H}\)

In one experiment, 2.18 g sodium was allowed to react with \(16.12 \mathrm{g}\) chlorine. All the sodium was used up, and 5.54 g sodium chloride (salt) was produced. In a second experiment, 2.10 g chlorine was allowed to react with \(10.00 \mathrm{g}\) sodium. All the chlorine was used up, and 3.46 g sodium chloride was produced.Show that these results are consistent with the law of constant composition.

What is the total number of atoms in (a) 15.8 mol \(\mathrm{Fe}\); (b) \(0.000467 \mathrm{mol} \mathrm{Ag} ;\) (c) \(8.5 \times 10^{-11} \mathrm{mol} \mathrm{Na} ?\)

From the densities of the lines in the mass spectrum of krypton gas, the following observations were made: \bullet Somewhat more than \(50 \%\) of the atoms were krypton-84. \(\bullet\) The numbers of krypton- 82 and krypton- 83 atoms were essentially equal. \(\bullet\) The number of krypton-86 atoms was 1.50 times as great as the number of krypton- 82 atoms. \(\bullet\) The number of krypton-80 atoms was \(19.6 \%\) of the number of krypton- 82 atoms. \(\bullet\) The number of krypton- 78 atoms was \(3.0 \%\) of the number of krypton- 82 atoms. The masses of the isotopes are \(^{78} \mathrm{Kr}, 77.9204 \mathrm{u} \quad^{80} \mathrm{Kr}, 79.9164 \mathrm{u} \quad^{82} \mathrm{Kr}, 81.9135 \mathrm{u}\) \(^{83} \mathrm{Kr}, 82.9141 \mathrm{u} \quad^{84} \mathrm{Kr}, 83.9115 \mathrm{u} \quad^{86} \mathrm{Kr}, 85.9106 \mathrm{u}\) The weighted-average atomic mass of \(\mathrm{Kr}\) is \(83.80 .\) Use these data to calculate the percent natural abundances of the krypton isotopes.

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