Answer:
the initial concentration of SCN- in the mixture is 0.00588 M
Explanation:
The computation of the initial concentration of the SCN^- in the mixture is as follows:
As we know that

As it is mentioned in the question that KSCN is present 10 mL of 0.05 M
So, the total milimoles of SCN^- is
= 10 × 0.05
= 0.5 m moles
The total volume in mixture is
= 45 + 10 + 30
= 85 mL
Now the initial concentration of the SCN^- is
= 0.5 ÷ 85
= 0.00588 M
hence, the initial concentration of SCN- in the mixture is 0.00588 M
Answer:
pH = 2.66
Explanation:
- Acetic Acid + NaOH → Sodium Acetate + H₂O
First we <u>calculate the number of moles of each reactant</u>, using the <em>given volumes and concentrations</em>:
- 0.75 M Acetic acid * 50.0 mL = 37.5 mmol acetic acid
- 1.0 M NaOH * 10.0 mL = 10 mmol NaOH
We<u> calculate how many acetic acid moles remain after the reaction</u>:
- 37.5 mmol - 10 mmol = 27.5 mmol acetic acid
We now <u>calculate the molar concentration of acetic acid after the reaction</u>:
27.5 mmol / (50.0 mL + 10.0 mL) = 0.458 M
Then we <u>calculate [H⁺]</u>, using the<em> following formula for weak acid solutions</em>:
- [H⁺] =

Finally we <u>calculate the pH</u>:
Hybridisation influences the bond length and bond enthalpy strength in organic compounds. The sp hybrid orbital contains more s character and hence it is closer to its nucleus and forms shorter and stronger bonds than the sp3 hybrid orbital.
You can use the mass of neuron divided by the mass of conversion factor: 1.67*10^(-25)/(1.66054*10^(-24))≈1 amu. So the answer is 1 amu.
Carboxylic acid...........