<h3><u>Answer; </u></h3>
=10.38 moles KOH
<h3><u>Explanation</u>;</h3>
The balanced equation.
6KOH + Al2(SO4)3 --> 3K2SO4 + 2Al(OH)3
From the equation;
1 mole of aluminum sulfate requires 6 moles of potassium hydroxide.
Moles of Aluminium sulfate; 1.73 moles
Moles of KOH;
1 mol Al2(SO4)3 : 6 mol KOH = 1.73 mol Al2(SO4)3 : x mol KOH
Thus; x = (6 × 1.73)
<u> =10.38 moles KOH </u>
Answer:
c.hg cannot be cracked for fractional distillation as there is only one of each
Explanation:
Missing table!! write the elements with the first letter of the symbol with Upper Caps letters!!!
http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/EStandardTable.htm
<span>Ni2+ +Pb(s) → Ni(s) + Pb2+
</span>The potential of the oxidation of Pb(s) --> Pb2+(aq) is 0.126 V
The potential of the reduction go Ni2+(aq) --> Ni(s) is -0.25 V
<span>Add the two together and the potential for the reaction is -0.124 V (NO SPONTANEOUS THE SIGN IS NEGATIVE)
</span><span>au3+ + al(s) → au(s) + al3+Au3+(aq) -> Au(s) +1.5 VAl -> Al3+ +1.66VV= 3.16 (SPONTANEOUS THE SIGN OF THE PONTENTIAL IS POSITIVE)</span><span>Sr2+ + Sn(s) → Sr(s) + Sn2+
</span>
Sr2+(aq) + 2 e– <span> Sr(s) V= -2.89V
</span>Sn -> Sn2+ V= 0.14 V
V= -2.75 V (no spontaneous)
<span>Fe2+ + Cu(s) → Fe(s) + Cu2+
</span>Fe2+(aq) + 2 e–<span> </span><span> Fe(s) V= -0.44 V
</span>Cu -> C2+ V = - 0.337V
V= - 0.777V (no spontaneous)
<span>D. It shows that the electrons within an atom do not have sharp boundaries.</span>
The basic units for density is
and any get of units that has those units in the proper place can be considered a density unit. The ones that has those specifically are A, B, E and F
