<span>Dr. Garcia should compare fly cells to human cells.
</span>
Answer:
The average atomic mass is closer to Si- 28 because this isotope is present in more percentage in the sample.
Explanation:
Given data:
Atomic mass of silicon= ?
Percent abundance of Si-28 = 92.21%
Atomic mass of Si-28 = 27.98 amu
Percent abundance of Si-29 = 4.70%
Atomic mass of Si-29 = 28.98 amu
Percent abundance of Si-30 = 3.09%
Atomic mass of Si-30 = 29.97 amu
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)+(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (92.21×27.98)+(4.70×28.98)+(3.09×29.97) /100
Average atomic mass = 2580.04 +136.21+92.61 / 100
Average atomic mass = 2808.86 / 100
Average atomic mass = 28.08amu.
The average atomic mass is closer to Si- 28 because this isotope is present in more percentage in the sample.
As mentioned above, phosphoric acid has 3 pKa values, and after 3 ionization it gives 3 types of ions at different pKa values:
H₃PO₄(aq)
+ H₂O(l) ⇌ H₃O⁺(aq) + H₂PO₄⁻ (aq) pKₐ₁
<span>
</span>H₂PO₄⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + HPO₄²⁻ (aq) pKₐ₂
HPO₄²⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + PO₄³⁻ (aq) pKₐ₃
At the highest pKa value (12.4) of phosphoric acid, the last OH group will lose its hydrogen. On the picture I attached, it is shown required protonated form of phosphoric acid before reaction whose pKa value is 12.4.
Answer:
0.21 M. (2 sig. fig.)
Explanation:
The molarity of a solution is the number of moles of the solute in each liter of the solution. The unit for molarity is M. One M equals to one mole per liter.
How many moles of NaOH in the original solution?
,
where
is the number of moles of the solute in the solution.
is the concentration of the solution.
for the initial solution.
is the volume of the solution. For the initial solution,
for the initial solution.
.
What's the concentration of the diluted solution?
.
is the number of solute in the solution. Diluting the solution does not influence the value of
.
for the diluted solution.- Volume of the diluted solution:
.
Concentration of the diluted solution:
.
The least significant number in the question comes with 2 sig. fig. Keep more sig. fig. than that in calculations but round the final result to 2 sig. fig. Hence the result: 0.021 M.
Answer:
The isotopic mass of 41K is 40.9574 amu
Explanation:
Step 1: Data given
The isotopes are:
39K with an isotopic mass of 38.963707u and natural abundance of 93.2581%
40K with an isotopic mass of 39.963999u
41K wit natural abundance of 6.7302 %
Average atomic mass =39.098 amu
Step 2: Calculate natural abundance of 40 K
100 % - 93.2581 % - 6.7302 %
100 % = 0.0117 %
Step 3: Calculate isotopic mass of 41K
39.098 = 38.963707 * 0.932581 + 39.963999 * 0.000117 + X * 0.067302
39.098 = 36.33681 + 0.0046758 + X * 2.067302
X = 40.9574 amu
The isotopic mass of 41K is 40.9574 amu