In preparing diluted solutions from concentrated solutions we can use the following formula
c1v1 = c2v2
c1 and v1 are the concentration and volume of the concentrated solution respectively
c2 and v2 are the concentrations and volume of the diluted solution respectively
Substituting these values ,
20 mL x 1.0 M = C x 60 mL
C = 0.33 M
The concentration of the resulting diluted solutions is 0.33 M
First the theoretical yield of Nabr
by use of mole ratio between FeBr3 and NaBr which is 2:6 the theoretical yield
=2.36 x6/2= 7.08 moles
the % yield = actual yield/ theoretical yield x 100
that is 6.14/7.08 x100= 86.72%
The atomic mass or relative isotopic mass refers to the mass of a single particle, and therefore is tied to a certain specific isotope of an element. The dimensionless standard atomic weight instead refers to the AVERAGE of atomic mass values of a typical naturally-occurring mixture of isotopes for a sample of an element.
You can count it by yourself using formula
m = ({first isotopic distribution%}× {first atomic.mass})+ ({second isotopic distribution%}× {second atomic.mass}) / {100}
Answer is: a. Rubidium (Rb) is more reactive than strontium (Sr) because strontium atoms must lose more electrons.
The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).
Alkaline metals (group 1), in this example rubidium, have lowest ionizations energy and easy remove valence electrons (one electron), they are most reactive metals.
Earth alkaline metals (group 2), in this example strontium, have higher ionization energy than alkaline metals, because they have two valence electrons, they are less reactive.
Rubidium electron configuration: ₃₇Rb 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s¹; one valence electron is 5s¹ orbital.
Strontium electron configuration: ₃₈Sr 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s²; two valence electrons is 5s² orbital.
