Answer:
The average atomic weight = 121.7598 amu
Explanation:
The average atomic weight of natural occurring antimony can be calculated as follows :
To calculate the average atomic mass the percentage abundance must be converted to decimal.
121 Sb has a percentage abundance of 57.21%, the decimal format will be
57.21/100 = 0.5721 . The value is the fractional abundance of 121 Sb .
123 Sb has a percentage abundance of 42.79%, the decimal format will be
42.79/100 = 0.4279. The value is the fractional abundance of 123 Sb .
Next step is multiplying the fractional abundance to it masses
121 Sb = 0.5721 × 120.904 = 69.169178400
123 Sb = 0.4279 × 122.904 = 52.590621600
The final step is adding the value to get the average atomic weight.
69.169178400 + 52.590621600 = 121.7598 amu
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Answer:
Explanation:
Hello,
In this case, it is widely known that when measurements with different significant figures are put under mathematical operations, the final result must be displayed with the same amount of significant figures of the shortest measurement, thus, due to the fact that 5.5 g has two significant figures only the result is consequently shown with two significant figures as well as shown down below:
By rounding the first six to seven due to the fact that the next six is greater than five, according to rounding rules, the result is:
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C, the amount of oxygen produced is the dependent variable since that is what is being measured and it is dependent on which substance is being tested
<u>Answer:</u> The molarity of Iron (III) chloride is 0.622 M.
<u>Explanation:</u>
Molarity is defined as the number of moles present in one liter of solution. The equation used to calculate molarity of the solution is:
Or,
We are given:
Mass of iron (III) chloride = 1.01 g
Molar mass of iron (III) chloride = 162.2 g/mol
Volume of the solution = 10 mL
Putting values in above equation, we get:
Hence, the molarity of Iron (III) chloride is 0.622 M.
