Clay and organic matter generally have higher capacity for retaining nutrients than silt and sand.
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<u>Answer:</u> The amount of water required to prepare given amount of salt is 398.4 mL
<u>Explanation:</u>
To calculate the volume of solution, we use the equation used to calculate the molarity of solution:
We are given:
Molarity of solution = 0.16 M
Given mass of manganese (II) nitrate tetrahydrate = 16 g
Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol
Putting values in above equation, we get:
Volume of water = Volume of solution = 398.4 mL
Hence, the amount of water required to prepare given amount of salt is 398.4 mL
Hello!
First, we need to determine the pKa of the base. It can be found applying the following equation:
Now, we can apply the Henderson-Hasselbach's equation in the following way:
![pH=pKa+log( \frac{[CH_3NH_2]}{[CH_3NH_3Cl]} )=10,65+log( \frac{0,18M}{0,73M} )=10,04](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%20%5Cfrac%7B%5BCH_3NH_2%5D%7D%7B%5BCH_3NH_3Cl%5D%7D%20%29%3D10%2C65%2Blog%28%20%5Cfrac%7B0%2C18M%7D%7B0%2C73M%7D%20%29%3D10%2C04)
So, the pH of this buffer solution is 10,04
Have a nice day!
First, we need to determine the pKa of the base. It can be found applying the following equation:
Now, we can apply the Henderson-Hasselbach's equation in the following way:
So, the pH of this buffer solution is 10,04
Have a nice day!