d. Fe(s) and Al(s)
<h3>Further explanation</h3>
In the redox reaction, it is also known
Reducing agents are substances that experience oxidation
Oxidizing agents are substances that experience reduction
The metal activity series is expressed in 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 more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent
The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent
So that the metal located on the left can push the metal on the right in the redox reaction
The electrodes which are easier to reduce than hydrogen (H), have E cells = +
The electrodes which are easier to oxidize than hydrogen have a sign E cell = -
So the above metals or metal ions will reduce Pb²⁺ (aq) will be located to the left of the Pb in the voltaic series or which have a more negative E cell value (greater reduction power)
The metal : d. Fe(s) and Al(s)
It's A, t<span>The figure is a molecule and an element.</span>
Molar mass:
O2 = 31.99 g/mol
C8H18 = 144.22 g/mol
<span>2 C8H18(g) + 25 O2(g) = 16 CO2(g) + 18 H2O(g)
2 x 144.22 g --------------- 25 x 31.99 g
10.0 g ----------------------?? ( mass of O2)
10.0 x 25 x 31.99 / 2 x 144.22 =
7997.5 / 288.44 => 27.72 g of O2
hope this helps!
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Idk look it up on another website
Fluorine needs one electron to fill its orbital shell. It needs 8 valence electrons
