<span>KE = 1/2mv^2
KE = 1/2(8)4 m/s^2
KE = 4*4
KE = 16 Joules
Kinetic energy would equal 16 J </span>
Answer:
Sn + 2H2O ==> Sn(OH)2 + 2H2
67.3 g Sn x 1 mol/119 g x 2 mol H2/mol Sn x 22.4 L/mole = answer in liters
Explanation:
Sn + 2H2O ==> Sn(OH)2 + 2H2
67.3 g Sn x 1 mol/119 g x 2 mol H2/mol Sn x 22.4 L/mole = answer in liters
C. Aluminum (Al) oxidized, zinc (Zn) reduced
<h3>Further explanation</h3>
Given
Metals that undergo oxidation and reduction
Required
A galvanic cell
Solution
The condition for voltaic cells is that they can react spontaneously, indicated by a positive cell potential.

or:
E ° cell = E ° reduction-E ° oxidation
For the reaction to occur spontaneously (so that it E cell is positive), the E° anode must be less than the E°cathode
If we look at the voltaic series:
<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au </em>
The standard potential value(E°) from left to right in the voltaic series will be greater, so that the metal undergoing an oxidation reaction (acting as an anode) must be located to the left of the reduced metal (as a cathode)
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From the available answer choices, oxidized Al (anode) and reduced Zn (cathode) are voltaic/galvanic cells.
The balanced equation for the neutralisation reaction is as follows
2H₃PO₄ + 3Mg(OH)₂ --> Mg₃(PO₄)₂ + 6H₂O
stoichiometry of H₃PO₄ to H₂O is 2:6
number of H₃PO₄ moles reacted - 0.24 mol
if 2 mol of H₃PO₄ form 6 mol of H₂O
then 0.24 mol of H₃PO₄ forms - 6/2 x 0.24 = 0.72 mol of H₂O
therefore 0.72 mol of H₂O are formed
Answer:
Hi there I think the awnser is B. or A. lol but im 90% its A.
Explanation:
xoxo hope this helps
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