Lithium's atomic weight is 6.94 g/mol.
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The answer you are looking for is the coefficient.
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Answer:
Mass fraction: 73,6% n-hexane; 26,4% dichloromethane
Mole fraction: 73,0% n-hexane; 27,0% dichloromethane
Explanation:
With a basis of 100 mL:
Mass of n-hexane:
85 mL × = 55,7 g
Mass of dichloromethane
15 mL × = 20,0 g
Total mass = 20,0 g + 55,7 g = 75,7 g
Mass fraction of n-hexane:
=73,6%
Mass fraction of dichloromethane:
= 26,4%
Moles of n-hexane:
55,7 g × = 0,65 moles
Mass of dichloromethane
20,0g × = 0,24 moles
Total moles: 0,65 moles + 0,24 moles = <em>0,89 moles</em>
Molar fraction of n-hexane:
=73,0%
Molar fraction of dichloromethane:
= 27,0%
I hope it helps!
The type of chemical reaction is "Double displacement"
If we abbreviate the formula for nicotine as Nic, then the equations for two different equilibria of Nic in water are
Nic + H2O ---> NicH+ + OH-
NicH+ + H2O ---> NicH2 2+ + OH-
We can write the Kb1 expression for the first equation as
Kb1 = 1.0×10^-6 = [NicH+][OH-] / [Nic]
1.0×10^-6 = x^2 / 1.85×10^-3 - x
Approximating that x is negligible compared to 1.85×10^-3 simplifies the equation to
1.0×10^-6 = x^2 / 1.85×10^-3
x = 0.0000430
x = [OH-] = 4.30×10^-5 M
From the Kb2 expression
Kb2 = 1.3×10-11 = [NicH2 2+][OH-] / [NicH+]
1.1×10^-10 = x^2 / 4.30×10^-5 - x
Approximating that x is negligible compared to 4.30×10^-5 simplifies the equation to
1.1×10^-10 = x^2 / 4.30×10^-5
x = [OH-] = 6.88×10^-8
The concentration [OH-] can be computed as
[OH-] = 4.30×10^-5 M + 6.88×10^-8 M = 4.30×10^-5 M
This shows that the second equilibrium has a negligible effect on the pH.
We can now calculate for pH:
pOH = -log [OH-] = -log (4.30×10^-5 M) = 4.37
pH = 14 - pOH = 14 - 4.37 = 9.63