Problem 106
Question
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.
Step-by-Step Solution
Verified Answer
The impossible atomic mass for an individual titanium atom is (e) 48.95 u.
1Step 1: 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.
2Step 2: 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.
3Step 3: 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.
Key Concepts
Atomic MassIsotopesComparison of Atomic Weights
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.
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.
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.
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.
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.
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.
Other exercises in this chapter
Problem 103
The properties of magnesium will most resemble those of which of the following? (a) cesium; (b) sodium; (c) aluminum; (d) calcium; (e) manganese.
View solution Problem 105
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View solution Problem 107
The mass of the isotope \(\frac{84}{36} \mathrm{Xe}\) is 83.9115 u. If the atomic mass scale were redefined so that \(\frac{84}{36} \mathrm{Xe}=84 \mathrm{u},\)
View solution Problem 108
A 5.585-kg sample of iron (Fe) contains (a) \(10.0 \mathrm{mol} \mathrm{Fe}\) (b) twice as many atoms as does \(600.6 \mathrm{g} \mathrm{C}\) (c) 10 times as ma
View solution