Choose the most electronegative atom from the following: K, Mg, As, CI
1 answer:
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First of all, I need to know what these five salts are. Luckily, I found a similar problem from another website which is shown in the attached picture. The Ksp is the solubility product constant. It follows the formula:
Ksp = [cation concentration]ᵃ[anion concentration]ᵇ
where a and b are the subscripts of the metal and nonmetal, respectively.
For the solutions ahead, let x be the concentration of the cation.
A. The formula is PbCr₂O₄.
2.8×10⁻¹³ = [x][x]
Solving for x, x = 5.29×10⁻⁷ M
B. The formula is Co(OH)₂.
1.3×10⁻¹⁵ = [x][x]²
Solving for x, x = 1.09×10⁻⁵ M
C. The formula is CoS.
5×10⁻²² = [x][x]
Solving for x, x = 2.24×10⁻¹¹ M
D. The formula is Cr(OH)₃.
1.6×10⁻³⁰ = [x][x]³
Solving for x, x = 3.56×10⁻⁶ M
E. The formula is Ag₂S.
6×10⁻⁵¹ = [x]²[x]
Solving for x, x = 1.82×10⁻¹⁷ M
<em>Thus,the highest concentration is letter B, Cobalt (II) Hydroxide.</em>
Ksp = [cation concentration]ᵃ[anion concentration]ᵇ
where a and b are the subscripts of the metal and nonmetal, respectively.
For the solutions ahead, let x be the concentration of the cation.
A. The formula is PbCr₂O₄.
2.8×10⁻¹³ = [x][x]
Solving for x, x = 5.29×10⁻⁷ M
B. The formula is Co(OH)₂.
1.3×10⁻¹⁵ = [x][x]²
Solving for x, x = 1.09×10⁻⁵ M
C. The formula is CoS.
5×10⁻²² = [x][x]
Solving for x, x = 2.24×10⁻¹¹ M
D. The formula is Cr(OH)₃.
1.6×10⁻³⁰ = [x][x]³
Solving for x, x = 3.56×10⁻⁶ M
E. The formula is Ag₂S.
6×10⁻⁵¹ = [x]²[x]
Solving for x, x = 1.82×10⁻¹⁷ M
<em>Thus,the highest concentration is letter B, Cobalt (II) Hydroxide.</em>
Answer:
Aluminum metal
Explanation:
In order to properly answer this or a similar question, we need to know some basic rules about galvanic cells and standard reduction potentials.
First of all, your strategy would be to find a trusted source or the table of standard reduction potentials. You would then need to find the half-equations for aluminum and gold reduction:
Since we have a galvanic cell, the overall reaction is spontaneous. A spontaneous reaction indicates that the overall cell potential should be positive.
Since one half-equation should be an oxidation reaction (oxidation is loss of electrons) and one should be a reduction reaction (reduction is gain of electrons), one of these should be reversed.
Thinking simply, if the overall cell potential would be obtained by adding the two potentials, in order to acquite a positive number in the sum of potentials, we may only reverse the half-equation of aluminum (this would change the sign of E to positive):
Notice that the overall cell potential upon summing is:
Meaning we obey the law of galvanic cells.
Since oxidation is loss of electrons, notice that the loss of electrons takes place in the half-equation of aluminum: solid aluminum electrode loses 3 electrons to become aluminum cation.