Answer: The mass percentage of 
is 5.86%
Explanation:
To calculate the mass percentage of in the sample it is necessary to know the mass of the solute ( in this case), and the mass of the solution (pesticide sample, whose mass is explicit in the letter of the problem).
To calculate the mass of the solute, we must take the mass of the 
precipitate. We can establish a relation between the mass of and using the stoichiometry of the compounds:
Since for every mole of Tl in there are two moles of Tl in , we have:
Using the molar mass of we have:
Finally, we can use the mass percentage formula:
Answer:
The atomic mass of element is 65.5 amu.
Explanation:
Given data:
Abundance of X-63 = 50.000%
Atomic mass of X-63 = 63.00 amu
Atomic mass of X-68 = 68.00 amu
Atomic mass of element = ?
Solution:
Abundance of X-68 = 100-50 = 50%
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (50×63)+(50×68) /100
Average atomic mass = 3150 + 3400 / 100
Average atomic mass = 6550 / 100
Average atomic mass = 65.5 amu.
The atomic mass of element is 65.5 amu.
Answer:
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Explanation:
As in relative abundance , one is take reference
So,
One is taken as 1:
Other is subtracted from it:
(1 - 0.6011)(atomic mass of Ga-71)
Equation can be written as:
<span>69.723 = (0.6011)(68.9256) + (1-0.6011)x </span>
<span>(1-0.6011) is the percentage abundance of Ga-71 expressed in percentage: </span>
<span>Solving for x </span>
<span>28.2918 = 0.3989 x </span>
<span>x= 70.9246.......</span>
Answer:
Answer:
Explanation:
"Magnesium is an alkaline earth metal, and forms a Mg2+ ion. And carbonate ion, CO2−3 , has an equal an opposite charge. And thus the ionic species is MgCO3"
