He Rydberg formula can be extended for use with any hydrogen-like chemical elements.
<span>1/ λ = R*Z^2 [ 1/n1^2 - 1/n2^2] </span>
<span>where </span>
<span>λ is the wavelength of the light emitted in vacuum; </span>
<span>R is the Rydberg constant for this element; R 1.09737x 10^7 m-1 </span>
<span>Z is the atomic number, for He, Z =2; </span>
<span>n1 and n2 are integers such that n1 < n2 </span>
<span>The energy of a He+ 1s orbital is the opposite to the energy needed to ionize the electron that is </span>
<span>taking it from n = 1 (1/n1^2 =1) to n2 = ∞ (1/n2^2 = 0) </span>
<span>.: 1/ λ = R*Z^2 = 1.09737x 10^7*(2)^2 </span>
<span>λ = 2.278*10^-8 m </span>
<span>E = h*c/λ </span>
<span>Planck constant h = 6.626x10^-34 J s </span>
<span>c = speed of light = 2.998 x 10^8 m s-1 </span>
<span>E = (6.626x10^-34*2.998 x 10^8)/(2.278*10^-8) = 8.72*10^-18 J ion-1 </span>
<span>Can convert this value to kJ mol-1: </span>
<span>(8.72*10^-18*6.022 x 10^23)/1*10^3 = 5251 kJ mol-1 </span>
<span>Lit value: RP’s secret book: 5240.4 kJ mol-1 (difference is due to a small change in R going from H to He+) </span>
<span>So energy of the 1s e- in He+ = -5251 kJ mol-1</span>
The correct answer is B. if im wrong please correct me if i am wrong
Answer:
I’ll give you two possible conditions:
(1) An acid is present in an aqueous solution. The acid will donate a proton to the water to form hydronium. (That’s not really what happens, but that’s how we usually think of it.)
(2) The autoionization of water: in a pure water solution (or not a pure solution, doesn’t matter), one water molecule donates a proton to another water molecule, forming equal numbers of hydronium and hydroxide ions.
Answer:
The isoelectric point is that the <u>pH </u>at which the compound is in an electronically neutral form.
For diss equations<u>, p</u>lease find them in the enclosed file.
The pIs of 2 amino acids:
- Glutamate: pI = 3,2
- Histidine: pI = 7,6
Explanation:
Formula for the pI calculation: pI = (pKa1 + pKa2)/2
Given 3 pKa :
- Acid glutamic with an acid sidechain:
Use the lower 2 pKas (corresponding with 2 -COOH groups)
pKa1 = 2,19; pKa2 = 4,25; so pI = 3,2
- Histidine with 2 amino groups:
Use the higher 2 pKas ( -COOH group and -NH= group)
pKa1 = 6; pKa2 = 9,17; so pI = 7,6
Density. water's density is always 1 g/ cm^3
