Answer : The value of 
of the weak acid is, 4.72
Explanation :
First we have to calculate the moles of KOH.
Now we have to calculate the value of of the weak acid.
The equilibrium chemical reaction is:
Initial moles 0.25 0.03 0
At eqm. (0.25-0.03) 0.03 0.03
= 0.22
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[HK]}{[HA]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BHK%5D%7D%7B%5BHA%5D%7D)
Now put all the given values in this expression, we get:
Therefore, the value of of the weak acid is, 4.72
Answer:
9.3 x 10^ -5 in standard notation is 0.000093
Explanation:
Since 10 is raised to a -5, this means that there should be 5 zeros before the non-zero digits
Answer:
Yes, it is possible. Let us consider an example of two solutions, that is, solution A having 20 percent mass RbCl (rubidium chloride) and solution B is having 15 percent by mass NaCl or sodium chloride.
It is found that solution A is having more concentration in comparison to solution B in terms of mass percent. The formula for mass percent is,
% by mass = mass of solute/mass of solution * 100
Now the formula for molality is,
Molality = weight of solute/molecular weight of solute * 1000/ weight of solvent in grams
Now molality of solution A is,
m = 20/121 * 1000/80 (molecular weight of RbCl is 121 grams per mole)
m = 2.07
Now the molality of solution B is,
m = 15/58.5 * 1000/85
m = 3.02
Therefore, in terms of molality, the solution B is having greater concentration (3.02) in comparison to solution A (2.07).
Answer:
1 mole of Al³+
Explanation:
Balanced chemical equation;
Al³⁺ + Na₃PO₄ → 3Na⁺ + AlPO₄
From the above chemical equation, it is evident that this is displacement reaction which is a reaction involving one atom or charge displacing another one in the chemical reaction.
In this reaction, Al³⁺ has a charge of +3 while Na⁺ has a charge of +1. When Al³⁺ displaces Na⁺, it meets an anion of PO₄³⁻ and since both charges are equal, i.e +3 and -3, there's is a mutual exchange of electrons and bond formation occurs.
The ratio of AlPO₄ to Al³⁺ is 1 : 1
I.e 1 mole of Al³⁺ produce 1 mole of AlPO₄
Answer:
29.75 Kg of NH₃
Solution:
In order to calculate the theoretical yield, first we will identify the limiting reactant.
According to equation,
6 g (3 moles) H₂ requires = 28 g (1 mole) N₂
So,
5250 g H₂ will require = X g of N₂
Solving for X,
X = (5250 g × 28 g) ÷ 6 g
X = 25433 g of N₂
Hence, it is found that H₂ is the limiting reactant because N₂ is provided in excess (32700 g). Therefore,
As,
6 g (3 mole) H₂ produced = 34 g ( 2 moles) of NH₃
So,
5250 g H₂ will produce = X g of NH₃
Solving for X,
X = (5250 g × 34 g) ÷ 6 g
X = 29750 g of NH₃
Or,
X = 29.75 Kg of NH₃
