A. Zn²⁺
<h3>Further explanation</h3>
Given
Cations of several elements
Required
The least to be reduced
Solution
If we look at the voltaic series:
<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe²⁺-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Fe³⁺-Ag-Pt-Au </em>
The electrode which is easier to reduce than the hydrogen (H2) electrode has a positive sign (E red= +) and is located to the right of the voltaic series (right of H)
The electrode which is easier to oxidize than the hydrogen (H2) electrode and is difficult to experience reduction has a negative sign (E red= -) and is located to the left of the voltaic series (left of H)
Or you can look at the standard reduction potential value of the metals in the answer options, and the most negative reduction E° value which will be difficult to reduce.
The Zn metal is located far left of the other metals in the answer choices, so it is the most difficult to reduce
Answer:

Explanation:
They gave us the masses of two reactants and asked us to determine the mass of the product.
This looks like a limiting reactant problem.
1. Assemble the information
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
Mᵣ: 239.27 32.00 207.2
2PbS + 3O₂ ⟶ 2Pb + 2SO₃
m/g: 2.54 1.88
2. Calculate the moles of each reactant

3. Calculate the moles of Pb from each reactant

4. Calculate the mass of Pb

The mass of NaN3 needed to produce 17.2 L nitrogen at STP is calculated as follows
find the moles of N2 produced at STP
At STP 1mole of gas = 22.4 L , what about 17.2 L of nitrogen
by cross multiplication
= (1 mole x17.2 L)/ 22.4 L= 0.768 moles
2NaN3 =2Na +3 N2
by use of mole ratio between NaN3 to N2 (2:3) the moles of NaN3 = 0.768 x2/3 = 0.512 moles of NaN3
mass of NaN3 is therefore =moles of NaN3 xmolar mass of NaN3
=0.512moles x 65 g/mol =33.28 grams of NaN3
88226Ra→ 86222Rn + 24He is the reaction that describes an alpha emission because radiations are released.
<h3>Describes an alpha emission?</h3>
Alpha radiation occurs when the nucleus of an atom becomes unstable and alpha particles are released in order to restore stability. Alpha decay occurs in elements that have high atomic numbers, such as uranium, radium, and thorium etc.
So we can conclude that 88226Ra→ 86222Rn + 24He is the reaction that describes an alpha emission because radiations are released.
Learn more about radiation here: brainly.com/question/893656
#SJ1