We determine the percent by mass of water in the compound by dividing the mass of water by the total mass. The total mass of Na2SO4.10H2O is equal to 322 g. The mass of water is 180 g.
percent by mass of water = (180 / 322)*(100 %) = 55.9%
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 10,04Have a nice day!
Answer:
This is not a balanced equation
Explanation:
Let's make it a balanced equation.
2 NH3 + H2So4 = (NH4)2So4
Glad I could help!!