Homework Solution: PrintCalculatorPeriodic Table 24 of 30 apling Learning On another planet, the isotopes of…

    PrintCalculatorPeriodic Table 24 of 30 apling Learning On another planet, the isotopes of titanium have the following natural abundances. Isotope Abundance Mass (amu) 46Ti | 78.700% | 45.95263 ATi | 17.000% | 4794795 Ti | 4.30096 | 49.94479 What is the average atomic mass of titanium on that planet? Number amu O Previous ⓧ Give Up & View Solution nt
    PrintCalculatorPeriodic Table 24 of 30 apling Learning On another planet, the isotopes of titanium have the following natural abundances. Isotope Abundance Mass (amu) 46Ti | 78.700% | 45.95263 ATi | 17.000% | 4794795 Ti | 4.30096 | 49.94479 What is the average atomic mass of titanium on that planet? Number amu O Previous ⓧ Give Up & View Solution nt

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    on another planet the isotopes of titanium enjoy the subjoined consistent absences
    PrintCalculatorPeriodic Table 24 of 30 apling Learning On another planet, the isotopes of titanium enjoy the subjoined consistent wealths. Isotope Wealth Body (amu) 46Ti | 78.700% | 45.95263 ATi | 17.000% | 4794795 Ti | 4.30096 | 49.94479 What is the mediocre ultimate body of titanium on that planet? Number amu O Previous ⓧ Give Up & View Solution nt

    PrintCalculatorPeriodic Table 24 of 30 apling Learning On another planet, the isotopes of titanium enjoy the subjoined consistent wealths. Isotope Wealth Body (amu) 46Ti | 78.700% | 45.95263 ATi | 17.000% | 4794795 Ti | 4.30096 | 49.94479 What is the mediocre ultimate body of titanium on that planet? Number amu O Previous ⓧ Give Up & View Solution nt

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    Defination of mediocre ultimate body: “The mediocre ultimate body of an atom is the complete of the bodyes of its isotopes, each divers by its consistent wealth developed in percentage”.

    Thus, Avg. ultimate body = (a1times m1) + (a2times m2) + (a3times m3) +….+ (antimes mn)

    where, a = wealth each isotop, and m = ultimate body of relative isotope

    In consecrated Q. : a1= 0.787 and m1= 45.95263, a2 = 0.17 and m2= 47.94795, a3 = 0.043 and m3 = 49.94479

    Thus, Avg. ultimate body of Ti on that planet = (0.787 times 45.95263)+(0.17 times 47.94795)+(0.043 times 49.94479)

    Avg. ultimate body of Ti on that planet = 46.46350 amu.