Answer:
[H2] = 0.012 M
[N2] = 0.019 M
[H2O] = 0.057 M
Explanation:
The strategy here is to account for the species at equilibrium given that the concentration of [NO]=0.062M at equilibrium is known and the quantities initially present and its stoichiometry.
2NO(g) + 2H2(g) ⇒ N2(g) + 2H2O(g)
i mol 0.10 0.050 0.10
c mol -0.038 -0.038 +0019 +0.038
e mol 0.062 0.012 00.019 0.057
Since the volume of the vessel is 1.0 L, the concentrations in molarity are:
[NO] = 0.062 M
[H2] = 0.012 M
[N2] = 0.019 M
[H2O] = 0.057 M
Main function of a restriction enzyme is to cleve (Cut) DNA bases.
Hope this helps!
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The answer is: 1.5 moles of oxygen are present.
V(O₂) = 33.6 L; volume of oxygen.
p(O₂) = 1.0 atm; pressure of oxygen.
T = 0°C; temperature.
Vm = 22.4 L/mol; molar volume at STP (Standard Temperature and Pressure).
At STP one mole of gas occupies 22.4 liters of volume.
n(O₂) = V(O₂) ÷ Vm.
n(O₂) = 33.6 L ÷ 22.4 L/mol.
n(O₂) = 1.50 mol; amount of oxygen.
