<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:
Explanation:
scheelite is CaWO₄
Mol weight = 288
80 g of scheelite = 80 / 288 = 27.77 x 10⁻² moles
27.77 x 10⁻² moles of scheelite = 27.77 x 10⁻² x 6.02 x 10²³ molecules of scheelite
= 167.17 x 10²¹ molecules of scheelite
1 molecule of scheelite contains 4 atoms of oxygen
167.77 x 10²¹ molecules of scheelite contains 4 x 167.77 x 10²¹ atoms of oxygen .
= 671.08 x 10²¹ atoms of oxygen .
= 671 x 10²¹ atoms .
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The current required to accumulate the 1.22 grams of nickel in 0.5 hours is 2.23 A.
<h3>What is current?</h3>
The current is given as the product of the charge with time. In the electrochemical analysis of the nickel, there will be a reduction of the nickel ion to nickel. The formation is given as:

There is the deposition of 1 mole of Ni with 2 electrons transfer. The transfer of charge for 1 mole that is 58.7 grams Nickel is:

The mass of Ni to be deposited is 1.22 grams. The charge required is given as:

The current required to transfer 4010.7 C of charge in 1800 seconds is given as:

Thus, the current required to accumulate the 1.22 grams of nickel in 0.5 hours is 2.23 A.
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