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kupik [55]
3 years ago
12

Calculate the molarity of each aqueous solution: (a) 25.5 ml of 6.25 m hcl diluted to 0.500 l with water (b) 8.25 ml of 2.00×10–

2 m ki diluted to 12.0 ml with water
Chemistry
1 answer:
Gnoma [55]3 years ago
6 0
1. Using the dilution equation;
M1V1=M2V2
M1 = 6.25 m, v1= 25.5 ml or 0.0255 liters
Therefore; (6.25 ×0.0255)= M (0.5)
                 M = 0.159375/0.5
 Therefore Molarity is 0.31875 M Hcl

2. From the same equation 
 M1v1=m2v2
m1 = 0.02 m, v1=8.25 ml or 0.00825 l, while v2= 0.012 l
Thus, 0.02 ×0.00825 = M (0.012)
          M = 0.000165/0.012
              = 0.01375 m
Thus molarity is 0.01375 M KI
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Answer:

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Explanation:

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There are many formulas but the most used are:

LOD = X + 3σ

LOQ = X + 10σ

Where X is average and σ is standard desvation

For the blanks readings the average X is 0,0105 and σ is 0,0024

Thus:

<em>LOD = 0,0177</em>

<em>LOQ = 0,0345</em>

I hope it helps!

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3 years ago
For the vaporization reaction Br2(l) → Br2(
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    The  temperature  at   which  the  process  be   spontaneous  is  calculated  as  follows

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During translation, the new polypeptides are often directed to specific parts of the cell by the presence or absence of short se
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The strong base (BOH) is completely dissociated in water:

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The resulting conjugate acid (OH⁻) is a weak acid, so it remains in solution as OH⁻ ions.

By other hand, the weak acid (HA) is only slightly dissociated in water:

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The resulting conjugate base (A⁻) is a weak base. Thus, it reacts with H⁺ ions from water to form HA, increasing the concentration of OH⁻ ions in the solution.

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8 0
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