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)
Answer:
0.99 kg O₂
1.9 kg SO₂
Explanation:
Let's consider the reaction between sulfur and oxygen to form sulfur dioxide.
S + O₂ → SO₂
The mass ratio of S to O₂ is 32.07:32.00. The mass of oxygen required to react with 1 kg of sulfur is:
1 kg S × (32.00 kg O₂/32.07 kg S) = 0.998 kg O₂
The mass ratio of S to SO₂ is 32.07:64.07. The mass of sulfur dioxide formed when 1 kg of sulfur is burned is:
1 kg S × (64.07 kg SO₂/32.07 kg S) = 1.99 kg SO₂
we have,
wavelenght=c/f
where c= 3×10^8 m/s
f=6.3×10^12 s^-1
so wavelength=(3×10^8)/(6.3×10^12)
=0.476×10^-4 m
This is categorized as a combustion reaction.
