Answer:
a. pka = 3,73.
b. pkb = 10,27.
Explanation:
a. Supposing the chemical formula of X-281 is HX, the dissociation in water is:
HX + H₂O ⇄ H₃O⁺ + X⁻
Where ka is defined as:
![ka = \frac{[H_3O^+][X^-]}{[HX]}](https://tex.z-dn.net/?f=ka%20%3D%20%5Cfrac%7B%5BH_3O%5E%2B%5D%5BX%5E-%5D%7D%7B%5BHX%5D%7D)
In equilibrium, molar concentrations are:
[HX] = 0,089M - x
[H₃O⁺] = x
[X⁻] = x
pH is defined as -log[H₃O⁺]], thus, [H₃O⁺] is:
![[H_3O^+]} = 10^{-2,40}](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%7D%20%3D%2010%5E%7B-2%2C40%7D)
[H₃O⁺] = <em>0,004M</em>
Thus:
[X⁻] = 0,004M
And:
[HX] = 0,089M - 0,004M = <em>0,085M</em>
![ka = \frac{[0,004][0,004]}{[0,085]}](https://tex.z-dn.net/?f=ka%20%3D%20%5Cfrac%7B%5B0%2C004%5D%5B0%2C004%5D%7D%7B%5B0%2C085%5D%7D)
ka = 1,88x10⁻⁴
And <em>pka = 3,73</em>
b. As pka + pkb = 14,00
pkb = 14,00 - 3,73
<em>pkb = 10,27</em>
I hope it helps!
The answer is <span>D.when the aim is to show electron distributions in shells. This is because there are some instances when elements don't possess a regular or normal electron configuration. There are those who have special electron configurations wherein a lower subshell isn't completely filled before occupying a higher subshell. It is best to visualize such cases using the orbital notation.</span>
Answer:
The estimated feed rate of logs is 14.3 logs/min.
Explanation:
The product of the process is 2000 tons/day of dry wood pulp, of 85 wt% of cellulose. That represents (2000*0.85)=1700 tons/day of cellulose.
That cellulose has to be feed by the wood chips, which had 47 wt% of cellulose in its composition. That means you need (1700/0.47)=3617 tons/day of wood chips to provide all that cellulose.
Th entering flow is wood chips with 45 wt% of water. This solution has an specific gravity of 0.640.
To know the specific gravity of the wood chips we have to write a volume balance. We also know that Mw=0.45*M and Mc=0.55*M.

The specific gravity of the wood chips is 0.494.
The average volume of a log is

The weight of one log is

To provide 3617 ton/day of wood chips, we need


The feed rate of logs is 14.3 logs/min.
The concentration of a solution is the number of moles of solute per fixed volume of solution.
Concentration (C) = number of moles of solute (n) / volume of the solution (v)
we have to find the volume of the solution when 36.0 g of Ca(OH)₂ is added to water to make a solution of concentration 0.530 M
mass of Ca(OH)₂ added - 36.0 g
number of moles of Ca(OH)₂ - 36.0 g / 74.1 g/mol = 0.486 mol
we know the concentration of the solution prepared and the number of moles of Ca(OH)₂ added, substituting these values in the above equation, we can find the volume of the solution
C = n/v
0.530 mol/L = 0.486 mol / V
V = 0.917 L
answer is 0.917 L
Answer:
See explanation
Explanation:
Now , we have the equation of the reaction as;
2H2S(g) + 302(g)------->2SO2(g) + 2H2O(g)
This equation shows that SO2 gas is produced in the process. Let us recall that this same SO2 gas is the anhydride of H2SO4. This means that it can dissolve in water to form H2SO4
So, when SO2 dissolve in rain droplets, then H2SO4 is formed thereby lowering the pH of rain water. This is acid rain.