<u>Answer:</u> The average atomic mass of lithium is 6.9241 u.
<u>Explanation:</u>
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
....(1)
- <u>For
isotope:</u>
Mass of
isotope = 6 u
Percentage abundance of
isotope = 7.59 %
Fractional abundance of
isotope = 0.0759
- <u>For
isotope:</u>
Mass of
isotope = 7 u
Percentage abundance of
isotope = 92.41%
Fractional abundance of
isotope = 0.9241
Putting values in equation 1, we get:
![\text{Average atomic mass of Lithium}=[(6\times 0.0759)+(7\times 0.9241)]](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20atomic%20mass%20of%20Lithium%7D%3D%5B%286%5Ctimes%200.0759%29%2B%287%5Ctimes%200.9241%29%5D)

Hence, the average atomic mass of lithium is 6.9241 u.
Answer:
q1 = mCpΔT
= 18.016g × 1.84J/g.K × (418.15-373.15)
= 1491.72 J
q2 = n×ΔH vap
= 1mol ×44.0kJ/mol
= 44KJ
∴ qtotal = q1+ q2
= 1.498kJ + 44.0kJ
= 45.498KJ
Explanation: The heat flow can be separated into steps.all that is being observed at a constant pressure,the heat flow is equal to the enthalpy.
Answer:
In order of decreasing miscibility
C₉H₂₀ (nonane)→C₂H₅F (fluoroethane)→C₂H₅Cl (chloroethane)→H₂O (water)
Explanation:
The solubility of a solid is a measure of its ability to dissolve in a liquid while for liquids, the miscibility is a measure of thhe liquid to mix with anoyjer liquid resulting in a soltion which can hold any amount of either liquids. Immiscible liquids are those that are not soluble or have very limited solibility with each other.
C₉H₂₀ (nonane)→C₂H₅F (fluoroethane)→C₂H₅Cl (chloroethane)→H₂O (water)
In the order of decreasing miscibility as like dissolve like, ability to dissociate and polar and organic characteristics are considered
Answer:
41.9 g
Explanation:
We can calculate the heat released by the water and the heat absorbed by the steel rod using the following expression.
Q = c × m × ΔT
where,
c: specific heat capacity
m: mass
ΔT: change in temperature
If we consider the density of water is 1.00 g/mol, the mass of water is 125 g.
According to the law of conservation of energy, the sum of the heat released by the water (Qw) and the heat absorbed by the steel (Qs) is zero.
Qw + Qs = 0
Qw = -Qs
cw × mw × ΔTw = -cs × ms × ΔTs
(4.18 J/g.°C) × 125 g × (21.30°C-22.00°C) = -(0.452J/g.°C) × ms × (21.30°C-2.00°C)
ms = 41.9 g