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On a mission to a newly discovered planet, an astronaut finds chlorine abundances of 13.85 % for 35cl and 86.15 % for 37cl. What is the atomic mass of chlorine for this location? The mass of 35cl is 34.9700 amu . The mass of 37cl is 36.9700 amu . Express your answer to two decimal places, and include the appropriate u

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transitionstate

Answer:

              36.70 amu

Solution:

Data Given:

                       Atomic Mass of ³⁵Cl  =  34.9700 amu

                       Natural Abundance of ³⁵Cl  =  13.85 %

                       Atomic Mass of ³⁷Cl  =  36.9700 amu

                       Natural Abundance of ³⁷Cl  =  86.15 %

Formula Used:

Average Atomic Mass  =  [(Atomic Mass of ³⁵Cl × Natural Abundance of ³⁵Cl) + (Atomic Mass of ³⁷Cl × Natural Abundance of ³⁷Cl)] ÷ 100

Putting values,

Average Atomic Mass  =  [(34.9700 × 13.85) + (36.9700 × 86.15)] ÷ 100

Average Atomic Mass  =  [(484.3345) + (3184.9655)] ÷ 100

Average Atomic Mass  =  3669.3 ÷ 100

Average Atomic Mass  =  36.70 amu

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