If 42.1 mL of 1.02 M sodium hydroxide, measured using a graduated cylinder, is placed in a beaker filled with 300 mL of DI water
1 answer:
Answer:
Approximately 0.126 M
Explanation:
For the calculation of the dilution you take into account the moles of NaOH in the 42.1mL of the original solution and you use the new volume of 342.1 mL:
The standardization is necessary because a beaker is not not an instrument used to measure volumes and the marks on it only give an estimate of the volume of the solution, they are used to contain solutions and carry reactions among other things. If you would have measured the water with a graduated cylinder (an instrument designed to measure volumes) the standardization wouldnt be that necessary.
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Hello!
To solve this problem we are going to use the Henderson-Hasselbach equation and clear for the molar ratio. Keep in mind that we need the value for Acetic Acid's pKa, which can be found in tables and is 4,76:
![pH=pKa + log ( \frac{[CH_3COONa]}{[CH_3COOH]} )](https://tex.z-dn.net/?f=pH%3DpKa%20%2B%20log%20%28%20%5Cfrac%7B%5BCH_3COONa%5D%7D%7B%5BCH_3COOH%5D%7D%20%29%20)
![\frac{[CH_3COOH]}{[CH_3COONa}= 10^{(pH-pKa)^{-1}}=10^{(4-4,76)^{-1}}=5,75](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BCH_3COOH%5D%7D%7B%5BCH_3COONa%7D%3D%2010%5E%7B%28pH-pKa%29%5E%7B-1%7D%7D%3D10%5E%7B%284-4%2C76%29%5E%7B-1%7D%7D%3D5%2C75%20)
So, the mole ratio of CH₃COOH to CH₃COONa is 5,75
Have a nice day!
To solve this problem we are going to use the Henderson-Hasselbach equation and clear for the molar ratio. Keep in mind that we need the value for Acetic Acid's pKa, which can be found in tables and is 4,76:
So, the mole ratio of CH₃COOH to CH₃COONa is 5,75
Have a nice day!
Explanation:
For AX type ceramic material, the number of formula per unit cells is as follows.
or,
where, n' = no. of formula units per cell
= molecular weight of cation = 90.5 g/mol
= molecular weight of anion = 37.3 g/mol
= volume of cubic cell = 3.55
a = edge length of unit cell =
= Avogadro's number =
= density = 3.55
Now, putting the given values into the above formula as follows.
=
= 0.9
= 1 (approx)
Therefore, we can conclude that out of the given options crystal structure of cesium chloride is possible for this material.