The isotope that is more abundant, given the data is isotope Li7
<h3>Assumption</h3>
- Let Li6 be isotope A
- Let Li7 be isotope B
<h3>How to determine whiche isotope is more abundant</h3>
- Molar mass of isotope A (Li6) = 6.02 u
- Molar mass of isotope B (Li7) = 7.02 u
- Atomic mass of lithium = 6.94 u
- Abundance of A = A%
- Abundance of B = (100 - A)%
Atomic mass = [(mass of A × A%) / 100] + [(mass of B × B%) / 100]
6.94 = [(6.02 × A%) / 100] + [(7.02 × (100 - A)) / 100]
6.94 = [6.02A% / 100] + [702 - 7.02A% / 100]
6.94 = [6.02A% + 702 - 7.02A%] / 100
Cross multiply
6.02A% + 702 - 7.02A% = 6.94 × 100
6.02A% + 702 - 7.02A% = 694
Collect like terms
6.02A% - 7.02A% = 694 - 702
-A% = -8
A% = 8%
Thus,
Abundance of B = (100 - A)%
Abundance of B = (100 - 8)%
Abundance of B = 92%
SUMMARY
- Abundance of A (Li6) = 8%
- Abundance of B (Li7) = 92%
From the above, isotope Li7 is more abundant.
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You need to find which intermolecular forces are between the molecules
dipole-dipole,h bonds, etc.
I'm not very good at explaining but this is what my prof said to help us
Identify the class of the molecule or molecules you are given. Are they nonpolar species, ions or
do they have permanent dipoles? Is there only one species or are there two?
In the case of ONE species (i.e., a pure substance), the intermolecular forces will be between
molecules of the same type. So if you are dealing with ions, the intermolecular forces will be ION-
ION or IONIC. If you are dealing with dipoles, then the intermolecular forces will be DIPOLE-
DIPOLE. If you are dealing with nonpolar species, the intermolecular forces will be DISPERSION
or VAN DER WAALS or INDUCED DIPOLE-INDUCED DIPOLE (the last three are desciptions
of the same interaction; regrettably we cannot call them nonpolar-nonpolar!).
In the case of TWO species (i.e., a mixture), the intermolecular forces will be between molecules of
one type with molecules of the second type. For example, ION-DIPOLE interactions exist between
ions dissolved in a dipolar fluid such as water.
Answer:
(a) Polarity of phase: 1. Stationary phase is polar
b) Eluent strength of solvent: 1. Increases as solvent becomes more polar
c) Nature of solutes.1. Polar
d) Nature of solute interaction:2. More soluble in mobile phase as the polarity of the mobile phase decreases
e) Polarity of phases: 1. Stationary phase is polar
f) Eluent strength of solvent: 1. Increases as solvent becomes more polar.
g) Nature of solutes: 1. Polar
h) Nature of solute interaction: 2. More soluble in mobile phase as the polarity of the mobile
Explanation:
The density of helium is 10 ^-3g.cm^-3
