Answer:
The correct answer is 5.447 × 10⁻⁵ vacancies per atom.
Explanation:
Based on the given question, the at 750 degree C the number of vacancies or Nv is 2.8 × 10²⁴ m⁻³. The density of the metal is 5.60 g/cm³ or 5.60 × 10⁶ g/m³. The atomic weight of the metal given is 65.6 gram per mole. In order to determine the fraction of vacancies, the formula to be used is,
Fv = Nv/N------ (i)
Here Nv is the number of vacancies and N is the number of atomic sites per unit volume. To find N, the formula to be used is,
N = NA×P/A, here NA is the Avogadro's number, which is equivalent to 6.022 × 10²³ atoms per mol, P is the density and A is the atomic weight. Now putting the values we get,
N = 6.022 × 10²³ atoms/mol × 5.60 × 10⁶ g/m³ / 65.6 g/mol
N = 5.14073 × 10²⁸ atoms/m³
Now putting the values of Nv and N in the equation (i) we get,
Fv = 2.8 × 10²⁴ m⁻³ / 5.14073 × 10²⁸ atoms/m^3
Fv = 5.44669 × 10⁻⁵ vacancies per atom or 5.447 × 10⁻⁵ vacancies/atom.
Answer:
(B). it's metallic bonding
<span>The process of removing or adding electrons is called ionization </span>
Answer:
a) 88.48%
b) 0.05625 mol
Explanation:
2CH₃CH₂OH(l) → CH₃CH₂OCH₂CH₃(l) + H₂O(g) Reaction 1
CH₃CH₂OH(l) → CH₂═CH₂(g) + H₂O(g) Reaction 2
a) CH₃CH₂OH = 46.0684 g/mol
CH₃CH₂OCH₂CH₃ = 74.12 g/mol
1 mol CH₃CH₂OH ______ 46.0684 g
x ______ 50.0 g
x = 1.085 mol CH₃CH₂OH
1 mol CH₃CH₂OCH₂CH₃ ______ 74.12 g g
y ______ 35.9 g
y = 0.48 mol CH₃CH₂OCH₂CH₃
100% yield _____ 0.5425 mol CH₃CH₂OCH₂CH₃
w _____ 0.48 mol CH₃CH₂OCH₂CH₃
w = 88.48%
b) Only 0.96 mol of ethanol reacted to form diethyl ether. This means that 0.125 mol of ethanol did not react. 45% of 0.125 mol reacted to form ethylene. Therefore, 0.05625 mol of ethanol reacted by the side reaction (reaction 2). Since 1 mol of ethanol leads to 1 mol of ethylene, 0.05625 mol of ethanol produces 0.05625 mol of ethylene.
