If we write the equation of the reaction that will take place, it is:
2HNO₃ + Na₂CO₃ → 2NaNO₃ + H₂CO3
The molar ratio of 2HNO₃ : Na₂CO₃ = 1 : 2
Therefore, we can set up the equation:
M₁V₁ = 2M₂V₂
Where the left side of the equation has the molarity and volume of HNO₃ and the right side has the molarity and concentration of Na₂CO₃. Substituting:
M₁ = (2 x 0.108 x 35.7) / 25
M₁ = 0.308 M
Answer:
Equilibrium constant of the given reaction is 
Explanation:
....
....
The given reaction can be written as summation of the following reaction-


......................................................................................

Equilibrium constant of this reaction is given as-
![\frac{[NOBr]^{2}}{[N_{2}][O_{2}][Br_{2}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNOBr%5D%5E%7B2%7D%7D%7B%5BN_%7B2%7D%5D%5BO_%7B2%7D%5D%5BBr_%7B2%7D%5D%7D)
![=(\frac{[NOBr]}{[NO][Br_{2}]^{\frac{1}{2}}})^{2}(\frac{[NO]^{2}}{[N_{2}][O_{2}]})](https://tex.z-dn.net/?f=%3D%28%5Cfrac%7B%5BNOBr%5D%7D%7B%5BNO%5D%5BBr_%7B2%7D%5D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%29%5E%7B2%7D%28%5Cfrac%7B%5BNO%5D%5E%7B2%7D%7D%7B%5BN_%7B2%7D%5D%5BO_%7B2%7D%5D%7D%29)


0.6137 g of KHP contains 1.086 × 10^21 acidic protons.
Number of moles of KHP = mass of KHP/molar mass of KHP
Molar mass of KHP = 204.22 g/mol
Mass of KHP = 0.6137 g
Number of moles of KHP = 0.6137 g/204.22 g/mol = 0.003 moles of KHP
Now, 1 each molecule of KHP contains 1 acidic proton.
For 0.003 moles of KHP there are; 0.003 × 1 × NA
Where NA is Avogadro's number.
So; 0.003 moles of KHP contains 0.003 × 1 × 6.02 × 10^23
= 1.086 × 10^21 acidic protons.
Learn more: brainly.com/question/16672114
The chemical formula for ammonia is NH3. So first, you need to find the molar mass of ammonia (how many grams in one mole).
N=14g
H3=3g
So one mole of NH3 is 17 grams, you can divide 82.9 grams by 17 grams to find the number of molecules. The answer should be 4.876 moles (molecules) of ammonia. Hope this helps!
Explanation:
When pH of the solution is 11.
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![11=-\log[H^+]](https://tex.z-dn.net/?f=11%3D-%5Clog%5BH%5E%2B%5D)
..(1)
At pH = 11, the concentration of
ions is
.
When the pH of the solution is 6.
![pH=-\log[H^+]'](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D%27)
![6=-\log[H^+]'](https://tex.z-dn.net/?f=6%3D-%5Clog%5BH%5E%2B%5D%27)
..(2)
At pH = 6, the concentration of
ions is
.
On dividing (1) by (2).
![\frac{[H^+]}{[H^+]'}=\frac{1\times 10^{-11} M}{1\times 10^{-6}}=1\times 10^{-5}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH%5E%2B%5D%7D%7B%5BH%5E%2B%5D%27%7D%3D%5Cfrac%7B1%5Ctimes%2010%5E%7B-11%7D%20M%7D%7B1%5Ctimes%2010%5E%7B-6%7D%7D%3D1%5Ctimes%2010%5E%7B-5%7D%20)
The ratio of hydrogen ions in solution of pH equal to 11 to the solution of pH equal to 6 is
.
Difference between the
ions at both pH:

This means that Hydrogen ions in a solution at pH = 7 has
ions fewer than in a solution at a pH = 6