demonstrates the formation of a solid precipitate
Answer:
The number of OH group found in found in a 6 carbon alditol is 6
Explanation:
The structure of a 6 carbon alditol is shown on the first uploaded image
The answer I will give is an approximate number, but it will be close. The problem is that every periodic table is slightly different. Some round as I will, and some carry the mass of the elements out to 3 decimal places. You should go back and put the correct numbers in to the mass that I use.
Formula
KOH
Givens
Solution
Molar Mass = 39 + 16 + 1
Molar Mass = 56 grams / mole
Explanation:
he concluded that the gas is comming out
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)
