Answer:
N₂ = 6.022 × 10²³ molecules
H₂ = 18.066 × 10²³ molecules
NH₃ = 12.044 × 10²³ molecules
Explanation:
Chemical equation;
N₂ + 3H₂ → 2NH₃
It can be seen that there are one mole of nitrogen three mole of hydrogen and two moles of ammonia are present in this equation. The number of molecules of reactant and product would be calculated by using Avogadro number.
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
Number of molecules of nitrogen gas:
1 mol = 6.022 × 10²³ molecules
Number of molecules of hydrogen:
3 mol × 6.022 × 10²³ molecules/ 1 mol
18.066 × 10²³ molecules
Number of molecules of ammonia:
2 mol × 6.022 × 10²³ molecules/ 1 mol
12.044 × 10²³ molecules
Answer : The pH of the resulting buffer is, 5.22
Explanation : Given,
First we have to calculate the moles of 
and,
The balanced chemical reaction is:
Moles of HA left = 0.117 mol - 0.08 mol = 0.037 mol
Moles of 
= 0.08 mol
The expression used for the calculation of 
is,
Now put the value of 
in this expression, we get:
Now we have to calculate the pH of buffer.
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{[A^-]}{[HA]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
Now put all the given values in this expression, we get:
Thus, the pH of the resulting buffer is, 5.22
Answer:
H2CO3 = 2H+ + CO3-
Explanation:
It is simply what carbonic acid breaks down into when placed in water. Since carbonic acid is made up of H and CO3, these are the products.
Answer:
electrons
Reason:
neutrons and protons are inside the nucleus
Answer: Wavelength associated with the fifth line is 397 nm
Explanation:
= Wavelength of radiation
E= energy
For fifth line in the H atom spectrum in the balmer series will be from n= 2 to n=7.
Using Rydberg's Equation:
Where,
= Wavelength of radiation
= Rydberg's Constant =
= Higher energy level = 7
= Lower energy level = 2 (Balmer series)
Putting the values, in above equation, we get
Thus wavelength λ associated with the fifth line is 397 nm
