Using the Rydberg formula, the spectral line of H - atom is suitable for this purpose is Paschen, ∞ → 3.
- Using the Rydberg formula;
1/λ = RH(1/nf^2 - 1/ni^2)
Given that;
λ = wavelength
RH = Rydberg constant
nf = final state
ni = initial state
- When final state = 3 and initial state = ∞
Then;
1/λ = 1 × 10^7 m-1 (1/3^2 - 1/ ∞^2)
1/λ = 1 × 10^7 m-1 (1/3^2 )
λ = 900 nm
Hence, the correct answer is Paschen, ∞ → 3
Learn more about the Rydberg formula; brainly.com/question/17753747
Answer:
The solution is given below
Explanation:
Heat, q= mc∆T
q= 125g x 4.18 J/g∙°C x (21.18x- 24.28) °C
q= -1619.75J
NEGATIVE SIGN INDICATES THAT HEAT IS ABSORBED.
Enthalpy Change, ∆H = 1619.75 7/ 10.5 g
= 154.26 J/g
No. of moles of KBr = Mass of KBr/ Molecular Weight of KBr
=10.5g/119gmol-1
=0.088 mol
∆H= 1619.75 J/ 0.088 mol
= 18.41 kJ/mol
The neutron has no charge because it is a neutral particle.
Answer:
4.96E-8 moles of Cu(OH)2
Explanation:
Kps es the constant referring to how much a substance can be dissolved in water. Using Kps, it is possible to know the concentration of weak electrolytes. Then, pKps is the minus logarithm of Kps.
Now, we know that sodium hydroxide (NaOH) is a strong electrolyte, who is completely dissolved in water. Therefore the pH depends only on OH concentration originating from NaOH. Let us to figure out how much is that OH concentration.
![pH= -log[H]\\pH= -log (\frac{kw}{[OH]})](https://tex.z-dn.net/?f=pH%3D%20-log%5BH%5D%5C%5CpH%3D%20-log%20%28%5Cfrac%7Bkw%7D%7B%5BOH%5D%7D%29)
![8.23 = - log(\frac{Kw}{[OH]} \\10^{-8.23} = Kw/[OH]\\ [OH] = Kw/10^{-8.23}](https://tex.z-dn.net/?f=8.23%20%3D%20-%20log%28%5Cfrac%7BKw%7D%7B%5BOH%5D%7D%20%5C%5C10%5E%7B-8.23%7D%20%3D%20Kw%2F%5BOH%5D%5C%5C%20%5BOH%5D%20%3D%20Kw%2F10%5E%7B-8.23%7D)
![[OH]=1.69E-6](https://tex.z-dn.net/?f=%5BOH%5D%3D1.69E-6)
This concentration of OH affects the disociation of Cu(OH)2. Let us see the dissociation reaction:

In the equilibrum, exist a concentration of OH already, that we knew, and it will be added that from dissociation, called "s":
The expression for Kps is:
![Kps= [Cu^{2+}] [OH]^2](https://tex.z-dn.net/?f=Kps%3D%20%5BCu%5E%7B2%2B%7D%5D%20%5BOH%5D%5E2)
The moles of (CuOH)2 soluble are limitated for the concentration of OH present, according to the next equation.

"s" is the soluble quantity of Cu(OH)2.
The solution for this third grade equation is 
Now, let us calculate the moles in 1 L:
