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Chapter 19, Problem 80E
Interpretation Introduction

Interpretation: The nuclear equation for the most likely mode of decay for each of the given unstable nuclide is to be written.

Concept Introduction: Type of radioactivity can be explained by nuclear stability after the decay, which is determined by the ratio of neutrons to protons. The NZ ratio for a stable nucleus with an atomic number around 40 is 1.25 and for a stable nucleus with an atomic number above 83 is 1.5, where N is the number of neutrons and Z is the atomic number.

There are two possibilities:

When NZ is too high, the nucleus tends to have beta decay as there are more neutrons than protons and they tend to convert to protons.

However, when NZ is too low, the nucleus tends to undergo positron emission or electron capture as there are more number of protons and they tend to convert to neutrons.

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The emission data in cps displayed in Table 1 is reported to two decimal places by the chemist. However, the instrument output is shown in Table 2. Table 2. Iron emission from ICP-AES Sample Blank Standard Emission, cps 579.503252562 9308340.13122 Unknown Sample 343.232365741 Did the chemist make the correct choice in how they choose to display the data up in Table 1? Choose the best explanation from the choices below. No. Since the instrument calculates 12 digits for all values, they should all be kept and not truncated. Doing so would eliminate significant information. No. Since the instrument calculates 5 decimal places for the standard, all of the values should be limited to the same number. The other decimal places are not significant for the blank and unknown sample. Yes. The way Saman made the standards was limited by the 250-mL volumetric flask. This glassware can report values to 2 decimal places, and this establishes our number of significant figures. Yes. Instrumental data…
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Chapter 19 Solutions

Chemistry: A Molecular Approach & Student Solutions Manual for Chemistry: A Molecular Approach, Books a la Carte Edition Package

Ch. 19 - Prob. 1ECh. 19 - Prob. 2ECh. 19 - Prob. 3ECh. 19 - Prob. 4ECh. 19 - Prob. 5ECh. 19 - Prob. 6ECh. 19 - Prob. 7ECh. 19 - Prob. 8ECh. 19 - Prob. 9ECh. 19 - Prob. 10ECh. 19 - Prob. 11ECh. 19 - Prob. 12ECh. 19 - Prob. 13ECh. 19 - Prob. 14ECh. 19 - Prob. 15ECh. 19 - Prob. 16ECh. 19 - Prob. 17ECh. 19 - Prob. 18ECh. 19 - Prob. 19ECh. 19 - Prob. 20ECh. 19 - Prob. 21ECh. 19 - Prob. 22ECh. 19 - Prob. 23ECh. 19 - Prob. 24ECh. 19 - Prob. 25ECh. 19 - Prob. 26ECh. 19 - Prob. 27ECh. 19 - Prob. 28ECh. 19 - Prob. 29ECh. 19 - Prob. 30ECh. 19 - Prob. 31ECh. 19 - Prob. 32ECh. 19 - Prob. 33ECh. 19 - Prob. 34ECh. 19 - Prob. 35ECh. 19 - Prob. 36ECh. 19 - Prob. 37ECh. 19 - Prob. 38ECh. 19 - Prob. 39ECh. 19 - Prob. 40ECh. 19 - Prob. 41ECh. 19 - Prob. 42ECh. 19 - Prob. 43ECh. 19 - Prob. 44ECh. 19 - Prob. 45ECh. 19 - Prob. 46ECh. 19 - Prob. 47ECh. 19 - Prob. 48ECh. 19 - Prob. 49ECh. 19 - Prob. 50ECh. 19 - Prob. 51ECh. 19 - Prob. 52ECh. 19 - Prob. 53ECh. 19 - Prob. 54ECh. 19 - Prob. 55ECh. 19 - Prob. 56ECh. 19 - Prob. 57ECh. 19 - Prob. 58ECh. 19 - Prob. 59ECh. 19 - Prob. 60ECh. 19 - Prob. 61ECh. 19 - Prob. 62ECh. 19 - Prob. 63ECh. 19 - Prob. 64ECh. 19 - Prob. 65ECh. 19 - Prob. 66ECh. 19 - Prob. 67ECh. 19 - Prob. 68ECh. 19 - Prob. 69ECh. 19 - Prob. 70ECh. 19 - Prob. 71ECh. 19 - Prob. 72ECh. 19 - Prob. 73ECh. 19 - Prob. 74ECh. 19 - Prob. 75ECh. 19 - Prob. 76ECh. 19 - Prob. 77ECh. 19 - Prob. 78ECh. 19 - Prob. 79ECh. 19 - Prob. 80ECh. 19 - Prob. 81ECh. 19 - Prob. 82ECh. 19 - Prob. 83ECh. 19 - Prob. 84ECh. 19 - Prob. 85ECh. 19 - Prob. 86ECh. 19 - Prob. 87ECh. 19 - Prob. 88ECh. 19 - Prob. 89ECh. 19 - Prob. 90ECh. 19 - Prob. 91ECh. 19 - Prob. 92ECh. 19 - Prob. 93ECh. 19 - Prob. 94ECh. 19 - Prob. 95ECh. 19 - Prob. 96ECh. 19 - Prob. 97ECh. 19 - Prob. 98ECh. 19 - Prob. 99ECh. 19 - Prob. 100ECh. 19 - Prob. 101ECh. 19 - Prob. 102ECh. 19 - Prob. 103ECh. 19 - Prob. 104ECh. 19 - Prob. 105ECh. 19 - Prob. 106ECh. 19 - Prob. 107ECh. 19 - Prob. 108ECh. 19 - Prob. 109E
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