<span> first calculate the moles of every compound used
n NH3 = (0.160 mol /L )(0 .360 L) = 0.058 moles NH3
which is equal to moles NH4OH
0.058 moles NH4OH.
n MgCl2 = (0.12 mol /L)( 0.10 liters = .012 moles MgCl2
reaction 2 NH3.H20 or 2 NH4OH + MgCl2 --> Mg(OH)2(s) + 2 NH4+ + 2Cl-
Molarity of Mg++ =(0.012 moles Mg+) / 0.46 liters = 0.026 Molar Mg++
Molarity OH= (0.058 moles OH-)/ 0.46 liters = 0.126 molar OH-
the limiting reactant is Mg ++ because it is lesser than the molarity of OH-
Now the challenge is to drive the OH-) concentration down so low that Mg(OH)2 will not precipitate out.
Ksp = 1.8 x 10^-11 = (Mg)(OH-)^2
Mg++ concentration to be .026 Molar, so let X = the (OH-) concentration
1.8 x l0^-11 = (0.026)(X)^2
(X)^2 = ( 1.8 x 10^-11) / (0.026)
X^2 = 6.92 x 10^-10
X =
2.63 x l0^-5 moles (OH-)/ L
this is the concentration where solid will form
so we need to lower the (OH-) which is
0.126 molar OH- down to 2.63 x 10^-5 molar OH- by adding NH4+ ions.
(0.126 moles/liter 0H-) - (2.63 x l0^-5 molar OH- ) = 0.123 moles per liter
OH- to be neutralized by adding NH4+
since the mole ratio is 1 : 1 then </span><span> 0.123 moles per liter NH4+ concentration to neutralize the </span><span><span> 0.123</span> moles of OH- in solution.
so to prevent the precipitation of mg(oh)2
0.123 - 0.058 = 0.065 Molar NH4+ is needed
</span>
A) Since the plot 1/[AB] vs time gives straight line, the order of the reaction with respect to A is second order:
rate constant, K = slope = 5.5 x 10⁻² M⁻¹S⁻¹
b) Rate law : Rate = k[AB]²
c) half life period of the 2nd order is inversely proportional to the initial concentration of the reactants
t 1/2 =

.

t 1/2 =

d) k = 5.5 x 10⁻² M⁻¹s⁻¹
Initial concentration of AB, [A₀] = 0.250 M
concentration of AB after 75 s = [A]
k =
![\frac{1}{t} [ \frac{1}{[A]} - \frac{1}{[Ao]} ]](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7Bt%7D%20%5B%20%5Cfrac%7B1%7D%7B%5BA%5D%7D%20-%20%20%5Cfrac%7B1%7D%7B%5BAo%5D%7D%20%5D)
[A] = 0.123 M
Equation: AB → A + B
concentration of AB after 75 s = 0.123 M
Amount of AB dissociated = 0.25 - 0.123 = 0.127 M
concentration of [A] produced = concentration of [B] produced = Amount of AB reacted = 0.127 M
Why don’t you just use google?
On the earth there is the top layer dirt then loam comes next