Answer:
Explanation:
1 Al = 26.98 AMU
3 Br = 239.7 AMU
AlBr3 = 266.69 AMU
Al = (26.98)/(266.69) = 10.1%
Br3 = (239.7)/(266.69) = 89.9%
Answer:
I thing Calcium Perchlorate
Answer:
Kc = [CO][Cl₂]/[COCl].
Kp = P(CO)P(Cl₂)/P(COCl).
Explanation:
For the balanced reaction:
COCl₂(g) ⇄ CO(g) + Cl₂(g).
The equilibrium constant can be expressed as concentration equilibrium constant (Kc) or pressure equilibrium constant (Kp).
The equilibrium constant is the ratio of the product of products concentrations to the product of the reactants concentrations.
Kc = [CO][Cl₂]/[COCl].
Kp = P(CO)P(Cl₂)/P(COCl).
Answer:
1.62 g
Explanation:
Given that:
Concentration of HCN = 0.119 M
Assuming the ka 4.00 × 10⁻¹⁰
The pKa of HCN (hydrocyanic acid) = -log (Ka)
= - log ( 4.00 × 10⁻¹⁰)
= 9.398
pH of buffer = 8.809
Using Henderson Hasselbach equation:
![pH = pKa + log \dfrac{[conjugate\ base ]}{acid}](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2B%20log%20%5Cdfrac%7B%5Bconjugate%5C%20%20base%20%5D%7D%7Bacid%7D)
![pH = pKa + log \dfrac{[CN^-]}{[HCN]}](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2B%20log%20%5Cdfrac%7B%5BCN%5E-%5D%7D%7B%5BHCN%5D%7D)
![8.809 = 9.398 +log \dfrac{[CN^-]}{[HCN]}](https://tex.z-dn.net/?f=8.809%20%3D%209.398%20%2Blog%20%5Cdfrac%7B%5BCN%5E-%5D%7D%7B%5BHCN%5D%7D)
![log \dfrac{[CN^-]}{[HCN]}= 8.809 - 9.398](https://tex.z-dn.net/?f=log%20%5Cdfrac%7B%5BCN%5E-%5D%7D%7B%5BHCN%5D%7D%3D%208.809%20-%209.398)
![log \dfrac{[CN^-]}{[HCN]}= -0.589](https://tex.z-dn.net/?f=log%20%5Cdfrac%7B%5BCN%5E-%5D%7D%7B%5BHCN%5D%7D%3D%20-0.589)
![\dfrac{[CN^-]}{[HCN]}= 0.2576](https://tex.z-dn.net/?f=%5Cdfrac%7B%5BCN%5E-%5D%7D%7B%5BHCN%5D%7D%3D%200.2576)
[CN^-] = 0.2576[HCN]
[CN^-] = 0.2756 (0.119) L
[CN^-] = 0.033 M
∴
The amount of NaCN (sodium cyanide) is calculated as follows:
= 1.62 g
