Answer:
Kb = 6.22x10⁻⁷
Explanation:
Triethanolamine, C₆H₁₅O₃N, is in equilibrium with water:
C₆H₁₅O₃N(aq) + H₂O(l) ⇄ C₆H₁₅O₃NH⁺(aq) + OH⁻(aq)
Kb is defined from concentrations in equilibrium, thus:
Kb = [C₆H₁₅O₃NH⁺] [OH⁻] / [C₆H₁₅O₃N]
The equilibrium concentration of these compounds could be written as:
[C₆H₁₅O₃N] = 0.486M - X
[C₆H₁₅O₃NH⁺] = X
[OH⁻] = X
pH is -log [H⁺], thus, [H⁺] = 10^-pH = 1.820x10⁻¹¹M
Also, Kw = [OH⁻] ₓ [H⁺];
1x10⁻¹⁴ = [OH⁻] ₓ [H⁺]
1x10⁻¹⁴ = [OH⁻] ₓ [1.820x10⁻¹¹M]
5.495x10⁻⁴M = [OH⁻], that means <em>X = 5.495x10⁻⁴M</em>
Replacing in Kb formula:
Kb = [5.495x10⁻⁴M] [5.495x10⁻⁴M] / [0.486M-5.495x10⁻⁴M]
<em>Kb = 6.22x10⁻⁷</em>
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36.0 g of glucose divided by 180 g/mol = 0.200 moles of glucose
find molarity
0.200 moles of glucose / 2 liters = 0.100 molar solution
(hope this helps)
Kelvin is a temperature scale designed so that zero degrees K is defined as absolute zero (at absolute zero, a hypothetical temperature, all molecular movement stops - all actual temperatures are above absolute zero) and the size of one unit is the same as the size of one degree Celsius.
Equation of decomposition of ammonia:
N2+3H2->2NH3
Euilibrium constant:
Kc=(NH3)^2/((N2)((H2)^3))
As concentration of N2=0.000105, H2=0.0000542
so equation will become:
3.7=(NH3)^2/(0.000105)*(0.0000542)^3
NH3=√(3.7*0.000105*(0.0000542)^3)
NH3=7.8×10⁻⁹
So concentration of ammonia will be 7.8×10⁻⁹.
Answer:
12 moles of cesium xenon heptafluoride
Explanation:
The reaction of cesium fluoride with xenon hexafluoride is CeF + XeF6 -> CeXeF7 and the reaction is balanced as written. So the mole ratio is 1:1:1. We are given 12 moles of CeF and 14 moles of XeF6 are reacting, but after the 12 moles of CeF react completely, the reaction will stop as we have run out of one of our reactants. So only 12 moles of CeXeF7 will be produced.
