The first dissociation for H2X:
H2X +H2O ↔ HX + H3O
initial 0.15 0 0
change -X +X +X
at equlibrium 0.15-X X X
because Ka1 is small we can assume neglect x in H2X concentration
Ka1 = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15)
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
HX + H2O↔ X^2 + H3O
8.2x10^-4 Y 8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2 = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m
I'm assuming that you are asking a general question because you did not include an example.
The limiting reagent is the item in the reactants (reagents) that will run out first. This is because it limits what the reaction can produce, essentially causing the leftover elements/compounds to just sit there.
Answer:
10°C
Explanation:
Heat gain by water = Heat lost by the slice of pizza
Thus,
<u>For water: </u>
Volume = 50.0 L
Density of water= 1 kg/L
So, mass of the water:
Mass of water = 50 kg
Specific heat of water = 1 kcal/kg°C
ΔT = ?
For slice of pizza:
Q = 500 kcal
So,
ΔT = 10°C
Increase in temperature = 10°C
Answer:
192.9
Explanation:
From the question,
Ke = [HCL]²/[H₂][CL₂].......................... Equation 1
Where Ke = Equilibrium constant.
Given: [HCL] = 0.0625 M, [H₂] = 0.0045 M, [CL₂] = 0.0045 M
Substitute these values into equation 1
Ke = (0.0625)²/(0.0045)(0.0045)
ke = (3.90625×10⁻³)/(2.025×10⁻⁵)
ke = 1.929×10²
ke = 192.9
Hence the equilibrium constant of the system = 192.9
Explanation:
the answer is that prokaryotic cells are not multicellular
