Answer:
a)
b)
![[PCl_5]=0.0375M](https://tex.z-dn.net/?f=%5BPCl_5%5D%3D0.0375M)
Explanation:
Hello!
a) In this case, since we can see that the second reaction is equal to the half of the first reaction, we can relate the equilibrium constants as shown below:
Thus, by plugging in the the equilibrium constant of the first reaction we obtain:
b) In this case, for the described reaction we can write:
Thus, the corresponding equilibrium expression is:
![K=\frac{[PCl_3][Cl_2]}{[PCl_5]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5BPCl_5%5D%7D)
In such a way, since we know the equilibrium constant and the concentrations of PCl3 and Cl2 at equilibrium, we can compute the concentration of PCl5 at equilibrium as follows:
![[PCl_5]=\frac{[PCl_3][Cl_2]}{K}\\](https://tex.z-dn.net/?f=%5BPCl_5%5D%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7BK%7D%5C%5C)
![[PCl_5]=\frac{\frac{0.20mol}{4L} *\frac{0.12mol}{4L} }{0.04}](https://tex.z-dn.net/?f=%5BPCl_5%5D%3D%5Cfrac%7B%5Cfrac%7B0.20mol%7D%7B4L%7D%20%2A%5Cfrac%7B0.12mol%7D%7B4L%7D%20%7D%7B0.04%7D)
Best regards!
Answer:
5.90
Explanation:
Initial moles of CH3COO- = 10.0/1000 x 0.75 = 0.0075 mol
Moles of HCl added = 5.0/1000 x 0.10 = 0.0005 mol
CH3COO- + HCl => CH3COOH + Cl-
Moles of CH3COO- left = 0.0075 - 0.0005 = 0.007 mol
Moles of CH3COOH formed = moles of HCl added = 0.0005 mol
pH = pKa + log([CH3COO-]/[CH3COOH])
= -log Ka + log(moles of CH3COO-/moles of CH3COOH)
= -log(1.78 x 10^(-5)) + log(0.007/0.0005)
= 5.90
Salt lowers the freezing point, and thus the melting point of the ice
Answer:
H-BI,H-Se,H-S,H-I,H-Br
Explanation:
One thing that must be kept in mind is that atomic size increases down the group and decreases across the period. The bond lengths of species are influenced by the relative sizes of atoms or ions present in the bond.
The bonds in the answer have been arranged on basis of their decreasing atomic size because the greater the atomic size of the atoms, the greater the bond length and vice versa.
