Answer:
Option D. 13.44
Explanation:
We'll begin by calculating the number of mole in 47.7g of copper(II) oxide, CuO.
This can be obtained as follow:
Mass of CuO = 47.7 g
Molar mass of CuO = 63.5 + 16 = 79.5 g/mol
Mole of CuO =.?
Mole = mass /Molar mass
Mole of CuO = 47.7/79.5
Mole of CuO = 0.6 mole
Next, we shall write the balanced equation for the reaction. This is given below:
CuO + H2 —> Cu + H2O
From the balanced equation above,
1 mole of CuO reacted with 1 mole of H2 to produce 1 mole of Cu and 1 mole of H2O.
Next, we shall determine the number of mole of H2 needed to react completely with 0.6 mole of CuO.
This can be obtained as follow:
From the balanced equation above,
1 mole of CuO reacted with 1 mole of H2.
Therefore, 0.6 mole of CuO will also react with 0.6 mole of H2.
Finally, we shall determine the volume occupied by 0.6 mole of H2 at STP.
This can be obtained as follow:
1 mole of H2 occupied 22.4 dm³ at STP.
Therefore, 0.6 mole of H2 will occupy = 0.6 × 22.4 = 13.44 dm³.
Therefore, 13.44 dm³ of H2 is needed for the reaction.
The activity series goes top to bottom, most active to least active elements, going: Li, K, Ba, Sr, Ca, Na, Mg, Mn, Zn, Fe, Cd, Co, Ni, Sn, Pb, H, Cu, Ag, Hg, Au.
Thus, your list of metals would go from most reactive to least reactive: Li, K, Mg, Zn, Fe, Cu, Au
The given 2.6 µC of charge is due to a buildup of electrons, each of which has a charge of 1.6 x 10^-19 C. The 2.6 <span>µC is equivalent to 2.6 x 10^-6 C, so we can divide this by the individual charge of an electron:
</span>2.6 x 10^-6 C / 1.6 x 10^-19 (C/electron) = 1.625 x 10^13 electrons
