Answer:
a) [H₃O⁺] = 1.8x10⁻⁵ M
b) pH = 4.75
c) % rxn = 3.5x10⁻³ %
Explanation:
a) The dissociation reaction of HCN is:
HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)
0.5 M - x x x
The dissociation constant from the above reactions is given by:
![Ka = \frac{[H_{3}O^{+}][CN^{-}]}{[HCN]} = 6.17 \cdot 10^{-10}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH_%7B3%7DO%5E%7B%2B%7D%5D%5BCN%5E%7B-%7D%5D%7D%7B%5BHCN%5D%7D%20%3D%206.17%20%5Ccdot%2010%5E%7B-10%7D)
By solving the above quadratic equation we have:
x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]
Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.
b) The pH is equal to:
![pH = -log[H_{3}O^{+}] = -log(1.75 \cdot 10^{-5} M) = 4.75](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH_%7B3%7DO%5E%7B%2B%7D%5D%20%3D%20-log%281.75%20%5Ccdot%2010%5E%7B-5%7D%20M%29%20%3D%204.75)
Then, the pH of the HCN solution is 4.75.
c) The % reaction is the % ionization:
![\% = \frac{x}{[HCN]} \times 100](https://tex.z-dn.net/?f=%20%5C%25%20%3D%20%5Cfrac%7Bx%7D%7B%5BHCN%5D%7D%20%5Ctimes%20100%20)
Therefore, the % reaction or % ionization is 3.5x10⁻³ %.
I hope it helps you!
Thermohaline circulation that occur in the ocean part that flows across sea surface help to supply heat to polar regions.
<h3>What is Thermohaline circulation?</h3>
The thermohaline circulation refer to a section and region of the ocean circulation which is majorly driven by heat gradient and freshwater that flows across the sea surface and also from the mixture of heat and salt that occur at the inside part of the ocean.
Therefore, Thermohaline circulation help to supply heat to polar regions.
Learn more about thermohaline circulation here.
brainly.com/question/1176119
Answer:
1.26 M
Explanation:
The ion nitrate is NO₃⁻ and the Barium is from group 2 so it forms the ion Ba²⁺, so the barium nitrate has the formula: Ba(NO₃)₂. The molar masses are: Ba: 137 g/mol, N = 14 g/mol, O = 16 g/mol, so the molar mass of barium nitrate is:
137 + 2x(14 + 3x16) = 199 g/mol
The number of moles is the mass divided by the molar mass, so:
n = 25.1/199 = 0.126 mol of Ba(NO₃)₂
In 1 mol of the salt, there are 2 moles of NO₃⁻, so the number of moles of nitrate is 0.252 mol. Nitrates formed with ammonium (that can react when the solid dissolves) and with elements from group 1 and 2 are completely soluble in water. So, the moles of nitrate will remain 0.252 mol.
The molarity is the number of moles divided by the volume (0.2 L):
[NO₃⁻]= 0.252/0.2 = 1.26 M
