The molarity of the resulting solution obtained by diluting the stock solution is 3 M
<h3>Data obtained from the question </h3>
- Molarity of stock solution (M₁) = 15 M
- Volume of stock solution (V₁) = 500 mL
- Volume of diluted solution (V₂) = 2.5 L = 2.5 × 1000 = 2500 mL
- Molarity of diluted solution (M₂) =?
<h3>How to determine the molarity of diluted solution </h3>
M₁V₁ = M₂V₂
15 × 500 = M₂ × 2500
7500 = M₂ × 2500
Divide both side by 2500
M₂ = 7500 / 2500
M₂ = 3 M
Thus, the volume of the resulting solution is 3 M
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We are given
0.2 M HCHO2 which is formic acid, a weak acid
and
0.15 M NaCHO2 which is a salt which can be formed by reacting HCHO2 and NaOH
The mixture of the two results to a basic buffer solution
To get the pH of a base buffer, we use the formula
pH = 14 - pOH = 14 - (pKa - log [salt]/[base])
We need the pKa of HCO2
From, literature, pKa = 1.77 x 10^-4
Substituting into the equation
pH = 14 - (1.77 x 10^-4 - log 0.15/0.2)
pH = 13.87
So, the pH of the buffer solution is 13.87
A pH of greater than 7 indicates that the solution is basic and a pH close to 14 indicates high alkalinity. This is due to the buffering effect of the salt on the base.
Answer:
A working model used to test a design is called a
stop
Explanation:
1mm = 1/10 cm
5.43 x 10^-6 mm = 5.43 × 10^-7 cm
Energy is made of everything and it can’t be created for destroyed. force is not made out of everything
