<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
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It would be velocity over frequency look up equation
Ethanol contains an alcohol group (-OH), which includes hydrogen bonding. Hydrogen bonding is stronger than other IMFs such as Van der Waals forces and Dipole-Dipole.
Ethene only has Van der Waals forces as it is an alkene with only C-H or C-C bonds. These means that the forces are not as strong.
Since ethanol contains hydrogen bonding, it will take a higher temperature in order to break these bonds, and thus results in a higher boiling point than ethene.
The reactant in a chemical process known as the limiting reactant controls how much product can be produced. When the limiting reactant is completely used up, the reaction will come to an end.
<h3>
Find the limiting reactant ?</h3>
- As a result of 1 mol Sb4O6 reacting with 6 mol H2SO4, only 0.1 mol Sb4O6 reacts with 0.6 mol H2SO4, leaving only 0.5 mol H2SO4. This indicates that H2SO4 is the limiting reactant and Sb4O6 is present in excess.
- According to your equation, which is balanced, 0.1 mol Sb4O6 should react with 0.6 mol H2SO4, yet there is only 0.5 mol H2SO4 on hand.
- Therefore, only.083 mol of Sb4O6 are reacted.
- The reactant that is present in the limiting amount—the limiting reactant—determines the extent to which a chemical reaction occurs.
- The trick is really quite easy! We employ an augmented matrix to hold the data derived from the balancing equation Sb4O6 + 6H2SO4 --> 2Sb2(SO4)3 + 6H2O.
- Although you are provided 0.5 mol of H2SO4, the reaction requires 0.6 mol. Therefore, the limiting reactant is H2SO4.
- Only 0.0833 mol of Sb4O6 is required, but you have 0.1 mol. Sb4O6 is therefore the extra reactant.
To learn more about limiting reactant refer to:
brainly.com/question/27986321
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I am pretty sure the correct answer is B.
It makes the most sense to me.
