Answer:
The jewelry is 2896.54_Kg/m^3 less dense than pure silver
Explanation:
Density of jewellery = (mass of jewellery) ÷ (volume of jewellery)
=3.25g ÷ 0.428mL = 0.00325Kg÷0.000000428m^3 = 7583.46Kg/m^3
The density of silver is 10490_Kg/m^3 which is (10490 - 7583.46) 2896.54_Kg/m^3 more dense than the jewellery
The density of Silver [Ag]
The weight of Silver per cubic centimeter is 10.49 grams or the weight of silver per cubic meter is 10490 kilograms, that is the density of silver is 10490 kg/m³; at 20°C (68°F or 293.15K) at a pressure of one atmospheres.
Mole ratio:
MgCl₂ + 2 KOH = Mg(OH)₂ + 2 KCl
2 moles KOH ---------------- 1 mole Mg(OH)₂
moles KOH ------------------- 4 moles Mg(OH₂)
moles KOH = 4 x 2 / 1
= 8 moles of KOH
molar mass KOH = 56 g/mol
mass of KOH = n x mm
mass of KOH = 8 x 56
= 448 g of KOH
hope this helps!
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.