Hey there!
RbOH
Rb: 1 x 85.468 = 85.468
O: 1 x 16 = 16
H: 1 x 1.008 = 1.008
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102.476
The molar mass of RbOH is 102.476 g/mol.
Hope this helps!
Answer:
15 mL of the solute
Explanation:
From the question given above, the following data were obtained:
Solution = 50 mL
Solvent = 35 mL
Solute =?
Solution is simply defined as:
Solution = solute + solvent
With the above formula, we can easily obtain the solute in the solution as follow:
Solution = 50 mL
Solvent = 35 mL
Solute =?
Solution = solute + solvent.
50 = solute + 35
Collect like terms
50 – 35 = solute
15 = solute
Solute = 15 mL
Therefore, 15 mL of the solute is required.
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 transfer of electrons is an ionic bond, therefore, we need to determine which substance includes an ionic bond.
CO2, is a covalent bond, which is the sharing of electrons, not transfer.
KBr, is an ionic bond, because (k) loses an electron to bromine.
So, our final answer is:- <span>(3) KBr</span>
