A compound consists of 81.8% C and 18.2% H by mass. What is its empirical formula?
1 answer:
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Solution:
Reduction potential of metal ions are provided below. Higher the value to reduction potential, greater is the tendency of metal to remain in reduced state.
In present case, reduction potential of Au is maximum, hence it is least prone to undergo oxidation. Hence, it is least reactive.
On other hand, reduction potential of Na is minimum, hence it is most prone to undergo oxidation. Hence, it is most reactive.
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Solution:
Reduction potential of metal ions are provided below. Higher the value to reduction potential, greater is the tendency of metal to remain in reduced state.
In present case, reduction potential of Au is maximum, hence it is least prone to undergo oxidation. Hence, it is least reactive.
On other hand, reduction potential of Na is minimum, hence it is most prone to undergo oxidation. Hence, it is most reactive.
Answer:
The density of Lithium β is 0.5798 g/cm³
Explanation:
For a face centered cubic (FCC) structure, there are total number of 4 atoms in the unit cell.
we need to calculate the mass of these atoms because density is mass per unit volume.
Atomic mass of Lithium is 6.94 g/mol
Then we calculate the mass of four atoms;
⇒next, we estimate the volume of the unit cell in cubic centimeter
given the edge length or lattice constant a = 0.43nm
a = 0.43nm = 0.43 X 10⁻⁹ m = 0.43 X 10⁻⁹ X 10² cm = 4.3 X 10⁻⁸cm
Volume of the unit cell = a³ = (4.3 X 10⁻⁸cm)³ = 7.9507 X 10⁻²³ cm³
⇒Finally, we calculate the density of Lithium β
Density = mass/volume
Density = (4.6097 X 10⁻²³ g)/(7.9507 X 10⁻²³ cm³)
Density = 0.5798 g/cm³