Answer:
Oxidation: Cr → Cr⁴⁺ + 4 e⁻
Reduction: 4 e⁻ + O₂ → 2 O²⁻
Explanation:
Let's consider the following redox reaction.
Cr + O₂ → CrO₂
Cr is oxidized. Its oxidation number increases from 0 to +4. The corresponding half-reaction is:
Cr → Cr⁴⁺ + 4 e⁻
O is reduced. Its oxidation number decreases from 0 to -2. The corresponding half-reaction is:
4 e⁻ + O₂ → 2 O²⁻
<span>The two populations will show cospeciation.</span>
Answer:
The answer to your question is 2.92 x 10⁻⁶ g
Explanation:
Data
mass of K = ?
atoms of K = 4.50 x 10¹⁶
Process
To answer this question just remember that the atomic mass of any element is equivalent to Avogadro's number.
1.- Look for the atomic number of Potassium
Atomic number = 39.10 g
2.- Use proportions and cross multiplication
39.10 g of K --------------- 6.023 x 10²³ atoms
x ----------------4.50 x 10¹⁶ atoms
x = (4.50 x 10¹⁶ x 39.10) / 6.023 x 10²³
-Simplify
x = 1.7595 x 10¹⁸ / 6.023 x 10²³
-Result
x = 2.92 x 10⁻⁶ g
Answer is: cobalt(II) hydroxide.
Ksp(lead(II) chromate) = [Pb²⁺][CrO₄²⁻]. [Pb²⁺] = √2,8·10⁻¹³ = 5,3·10⁻⁷ M.<span>
Ksp(cobalt(II) hydroxide) = </span>[Co²⁺][OH⁻]². [Co²⁺] = ∛1,3·10⁻¹⁵ = 1,1·10⁻⁵ M.
Ksp(cobalt(II) sulfide) = [Co²⁺][S²⁻]. [Co²⁺] = √5·10⁻²² = 2,23·10⁻¹¹ M.
Ksp(chromium(III) hydroxide) = [Cr³⁺][OH⁻]³. [Cr³⁺] = √(√1,6·10⁻³⁰) = 3,6·10⁻⁸.
Ksp(silver sulfide) = [Ag⁺]²[S²⁻].
C) coal !
Have a nice day!\
:D