Answer : The volume of 6M NaOH stock solution is, 16.7 mL
Explanation :
To calculate the volume of NaOH stock solution, we use the equation given by neutralization reaction:
where,
are the molarity and volume of NaOH stock solution.
are the molarity and volume of NaOH.
We are given:
Putting values in above equation, we get:
Thus, the volume of 6M NaOH stock solution is, 16.7 mL
E represents units of energy, m represents units of mass, and c2 is the speed of light squared, or multiplied by itself
So I THINK that A is the right answer but i may be wrong.
I hope this helped :)
Phosphorus!!!! Hope this helps
The pH of a solution is 9.02.
c(HCN) = 1.25 M; concentration of the cyanide acid
n(NaCN) = 1.37 mol; amount of the salt
V = 1.699 l; volume of the solution
c(NaCN) = 1.37 mol ÷ 1.699 l
c(NaCN) = 0.806 M; concentration of the salt
Ka = 6.2 × 10⁻¹⁰; acid constant
pKa = -logKa
pKa = - log (6.2 × 10⁻¹⁰)
pKa = 9.21
Henderson–Hasselbalch equation for the buffer solution:
pH = pKa + log(cs/ck)
pH = pKa + log(cs/ck)
pH = 9.21 + log (0.806M/1.25M)
pH = 9.21 - 0.19
pH = 9.02; potential of hydrogen
More about buffer: brainly.com/question/4177791
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