I think its density is 850kg/m^3
Answer:
The
for the reaction
will be 4.69.
Explanation:
The given equation is A(B) = 2B(g)
to evaluate equilibrium constant for ![B(g) = \frac{1}{2}A](https://tex.z-dn.net/?f=B%28g%29%20%3D%20%5Cfrac%7B1%7D%7B2%7DA)
![K_c=[B]^2[A]](https://tex.z-dn.net/?f=K_c%3D%5BB%5D%5E2%5BA%5D)
= 0.045
The reverse will be ![2B\leftrightharpoons A](https://tex.z-dn.net/?f=2B%5Cleftrightharpoons%20A)
Then, ![K_c = \frac{[A]}{[B]^2}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cfrac%7B%5BA%5D%7D%7B%5BB%5D%5E2%7D)
= ![\frac{1}{0.045}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B0.045%7D)
= ![22m^{-1}](https://tex.z-dn.net/?f=22m%5E%7B-1%7D)
The equilibrium constant for
will be
![K_c = \sqrt{K_c}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Csqrt%7BK_c%7D)
![=\sqrt{22}](https://tex.z-dn.net/?f=%3D%5Csqrt%7B22%7D)
= 4.69
Therefore,
for the reaction
will be 4.69.
KE = 1/2 * m * v^2
KE = 1/2 * 130 * 23^2
KE = 34385J
Anhydrous Copper sulfate is 39.81 percent copper and 60.19 percent sulfate by mass, and in its blue, hydrous form, it is 25.47% copper, 38.47% sulfate (12.82% sulfur) and 36.06% water by mass.