Answer:
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Explanation:
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Have about 5 beakers all with different temperatures of water. Put in a teaspoon of salt at a time and when it stops dissolving stop adding and record how much salt it took. It should be more salt as the temperature rises. The independent variable is the waters temperature. The dependent variable is how much salt is used. Make sure that there is the same amount of water in each beaker. Or else it won’t work.
Answer:
Average atomic mass = 63.553 amu.
Explanation:
Given data:
Abundance of Y-63 = 69.17%
Abundance of Y-65 = 100 - 69.17 = 30.83%
Atomic mass of Y-63 = 62.940 amu
Atomic mass of Y-65 = 64.928 amu
Atomic mass of Y = ?
Solution:
Average atomic mass= (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass= (62.940×69.17)+(64.928×30.83) /100
Average atomic mass = 4353.560 + 2001.730 / 100
Average atomic mass = 6355.29 / 100
Average atomic mass = 63.553 amu.
