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Answer:
82.4 s
Explanation:
Find the NUMBEr of half lives...then multiply by 54.3
2.27 = 6.5 (1/2)^n
log (2.27/6.5) / log (1/2) = n = 1.52 half lives
1.52 * 54.3 = 82.4 s
Answer:
Molecular formula = C₁₂H₁₂O₄
Empirical formula is C₃H₃O.
Explanation:
Given data:
Mass of C = 91.63 g
Mass of H = 7.69 g
Mass pf O = 40.81 g
Molar mass of compound = 220 g/mol
Empirical formula = ?
Molecular formula = ?
Solution:
Number of gram atoms of H = 7.69 / 1.01 = 7.61
Number of gram atoms of O = 40.81 / 16 = 2.55
Number of gram atoms of C = 91.63 / 12 = 7.64
Atomic ratio:
C : H : O
7.64/2.55 : 7.61 /2.55 : 2.55/2.55
3 : 3 : 1
C : H : O = 3 : 3 : 1
Empirical formula is C₃H₃O.
Molecular formula:
Molecular formula = n (empirical formula)
n = molar mass of compound / empirical formula mass
Empirical formula mass = 3×12+ 3×1.01 +16 = 55.03
n = 220 / 55.03
n = 4
Molecular formula = 4 (empirical formula)
Molecular formula = 4 (C₃H₃O)
Molecular formula = C₁₂H₁₂O₄
There are two naturally occurring isotopes of gallium: mass of Ga-69 isotope is 68.9256 amu and its percentage abundance is 60.11%, let the mass of other isotope that is Ga-71 be X, the percentage abundance can be calculated as:
%Ga-71=100-60.11=39.89%
Atomic mass of an element is calculated by taking sum of atomic masses of its isotopes multiplied by their percentage abundance.
Thus, in this case:
Atomic mass= m(Ga-69)×%(Ga-69)+X×%(Ga-71)
From the periodic table, atomic mass of Ga is 69.723 amu.
Putting the values,
Thus,
Rearranging,
Therefore, mass of Ga-71 isotope is 70.9246 amu.
