Answer:
4.26 %
Explanation:
There is some info missing. I think this is the original question.
<em>Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10
⁻⁴.</em>
<em />
Step 1: Given data
Initial concentration of the acid (Ca): 0.249 M
Acid dissociation constant (Ka): 4.50 × 10
⁻⁴
Step 2: Write the ionization reaction for nitrous acid
HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)
Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])
We will use the following expression.
Step 4: Calculate the percent ionization of nitrous acid
We will use the following expression.
Answer:
Solute = 5 mL; solution = 250 mL
Explanation:
The formula for percent by volume is
If you have 250 mL of a solution that is 2 % v/v,
If there is no change of volume on mixing,
Volume of solution = 250 mL
-Volume of solute = <u> </u><u>5</u><u> </u>
Volume of solvent = 245 mL
The average atomic mass of your mixture is 1.03 u
.
The average atomic mass of H is the weighted average of the atomic masses of its isotopes.
We multiply the atomic mass of each isotope by a number representing its relative importance (i.e., its % abundance).
Thus,
0.99 × 1.01 u = 0.998 u
0.002 × 2.01 u = 0.004 u
0.008 × 3.02 u = <u>0.024 u</u>
TOTAL = 1.03 u
The crust
1. divergent (moves away from each other)
2. convergent (moves towards each other)
3. transform (slides past each other)