*Note: there seems to be a "typo" in the given problem. The "1000" must probably be "100" only since summing up the total number of atoms given for the different isotopes would not equal 1000 but 100 instead. Using 1000 as the basis for the mass fractions would give an answer which is 10 times less than should be.
To determine the average atomic mass of halfnium, the mass fractions of the isotopes multiplied by their respected atomic masses must all be added.
1. Determining the mass fractions (m1, m2...m5) with 100 atoms total as the basis:
m1 = 5/100 = 0.05
m2 = 19/100 = 0.19
m3 = 27/100 = 0.27
m4 = 14/100 = 0.14
m5 = 35/100 = 0.35
2. Multiplying the mass fractions with the atomic masses of the respective isotopes.
Average atomic mass of Halfnium
= (m1*176) + (m2*177) + (m3*178) + (m4*179) + (m5*180)
= (0.05*176) + (0.19*177) + (0.27*178) + (0.14*179) + (0.35*180)
= 178.55 amu
Thus, the average atomic mass of Halfnium based on the given data for its isotopes is 178.55 amu.
The correct answer is A. PH value
If a substance is very acidity it has a high PH, if it is less acidic (alkalinity) is has a lower PH. For instance, water has a PH of 7 because it is in between of being acidic and alkaline.
<span>To draw an acceptable lewis structure we must first determine the number of valance elections of the molecule in the ground state, and then adjust the molecule to refect the molecular radial cation after one or more of the electrons have been removed from its outter most shell. The lewis structure should refect the best configuation with good resonance and stability.</span>
Answer:
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