What element in the second period has the largest atomic radius?
1 answer:
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Answer : The percent abundance of Li isotope-1 and Li isotope-2 is, 6.94 % and 93.1 % respectively.
Explanation :
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
.....(1)
Let the fractional abundance of Li isotope-1 be 'x' and the fractional abundance of Li isotope-2 will be '100-x'
For Li isotope-1 :
Mass of Li isotope-1 = 6.01512 amu
Fractional abundance of Li isotope-1 = x
For Li isotope-2 :
Mass of Li isotope-2 = 7.01600 amu
Fractional abundance of Li isotope-2 = 100-x
Average atomic mass of Li = 6.941 amu
Putting values in equation 1, we get:
By solving the term 'x', we get:
Percent abundance of Li isotope-1 =
Percent abundance of Li isotope-2 = 100 - x = 100-6.94 = 93.1 %
Answer:
Molarity of the solution? 0,262 M.
Concentration of the potassium cation? 0,786 M.
Concentration of the phosphate anion? 0,262 M.
Explanation:
Potassium phosphate (K₃PO₄; 212,27 g/mol) dissolves in water thus:
K₃PO₄ → 3 K⁺ + PO₄³⁻ <em>(1)</em>
Molarity is an unit of chemical concentration given in moles of solute (K₃PO₄) per liters of solution.
There are 250 mL of solution≡0,25 L
The moles of K₃PO₄ are:
13,9 g of K₃PO₄ × = 0,0655 moles of K₃PO₄
The molarity of the solution is:
= 0,262 M
In (1) you can see that 1 mole of K₃PO₄ produces 3 moles of potassium cation. The moles of potassium cation are:
0,0655 moles×3 = 0,1965 moles
The concentration is:
= 0,786 M
The moles of K₃PO₄ are the same than moles of PO₄³⁻, thus, concentration of phosphate anion is the same than concentration of K₃PO₄. 0,262 M
I hope it helps!