<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>
Answer:
pH = 2.
Explanation:
A weak acid is in equilibrium with its ions in a solution, so it must have an equilibrium constant (Ka). And, pKa = -logKa
Ka = 10⁻⁴
So, for CH₃COOH the equilibrium must be:
CH₃COOH(aq) ⇄ H⁺(aq) + CH₃COO⁻(aq)
1 M 0 0 Initial
-x +x +x Reacted
1-x x x Equilibrium
And the equilibrium constant:
![Ka = \frac{[H+]x[CH3COO-]}{[CH3COOH]}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH%2B%5Dx%5BCH3COO-%5D%7D%7B%5BCH3COOH%5D%7D)
Supposing x << 1:
10⁻⁴ = x²
x = √10⁻⁴
x = 10⁻² M, so the supposing is correct.
So,
pH = -log[H⁺]
pH = -log10⁻²
pH = 2
The answer is going to be C. Although producing at a fast rate, fossil fuel cannot be re-used or recycled. Wind can be re-used, geothermal energy can be re-used, and hydro energy can be re-used.
Hope this Helps! :)
<span>This question asksyou to apply Hess's law.
You have to look for how to add up all the reaction so that you get the net equation as the combustion for benzene. The net reaction should look something like C6H6(l)+ O2 (g)-->CO2(g) +H2O(l). So, you need to add up the reaction in a way so that you can cancel H2 and C.
multiply 2 H2(g) + O2 (g) --> 2H2O(l) delta H= -572 kJ by 3
multiply C(s) + O2(g) --> CO2(g) delta H= -394 kJ by 12
multiply 6C(s) + 3 H2(g) --> C6H6(l) delta H= +49 kJ by 2 after reversing the equation.
Then,
6 H2(g) + 3O2 (g) --> 6H2O(l) delta H= -1716 kJ
12C(s) + 12O2(g) --> 12CO2(g) delta H= -4728 kJ
2C6H6(l) --> 12 C(s) + 6 H2(g) delta H= - 98 kJ
______________________________________...
2C6H6(l) + 16O2 (g)-->12CO2(g) + 6H2O(l) delta H= - 6542 kJ
I hope this helps and my answer is right.</span>
