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lapo4ka [179]
3 years ago
7

The atomic mass of chlorine is 35.45. Is it possible for any single atom of chlorine to have a mass number of exactly 35.45? Exp

lain
Chemistry
2 answers:
MatroZZZ [7]3 years ago
6 0
No because the atomic mass is the average of the different isotopes, so there is no isotope with a mass of 35.45. There are no isotopes that have decimal masses, they have to be a whole number.
ch4aika [34]3 years ago
4 0
<h3>Answer:</h3>

            No, Is it possible for any single atom of chlorine to have a mass number of exactly 35.45.

<h3>Explanation:</h3>

Average Atomic Mass of Chlorine:

                                                        Naturally Chlorine exists in many isotopes among which only two isotopes are stable. There atomic masses and natural abundance is given as,

            ³⁵Cl  Atomic Mass  =  35 amu

            ³⁵Cl natural Abundance  =  76 %

            ³⁷Cl  Atomic Mass  =  37 amu

            ³⁷Cl natural Abundance  =  24 %

                         Average Atomic Mass  =  (35 × 0.76) + (37 × 0.24)

                         Average Atomic Mass  =  26.6 + 8.88

                         Average Atomic Mass  =  ≈ 35.48

<h3>Conclusion:</h3>

                  Hence, it is cleared that a single Chlorine atom can have a whole number of atomic mass (sum of protons and neutrons). And therefore no single chlorine atom can have an atomic mass in decimals.


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<h3>Empirical formula </h3>

To calculate the empirical formula of a compound, the value of moles of each element is needed.

As we have the information of the mass value, we will use the molar mass expression, which corresponds to:

MM_O = 16g/mol\\MM_Fe = 55.8g/mol

                                              MM = \frac{m}{mol}

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                                                   16 = \frac{4.8}{x}

                                                   x = 0.3mol

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                                                    55.8=\frac{11.2}{x}\\x = 0.2

As the value of the empirical formula must be an integer, simply multiply the two values ​​by a common factor:

                                                O = 0.3 \times 10 = 3\\Fe = 0.2 \times 10 = 2

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So, the empirical formula of this compound is Fe_2O_3.

Learn more about empirical formula: brainly.com/question/1247523

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