Ksp - solubility product constant is equivalent to equilibrium constant, except this constant is used to determine the solubility of ions of a solid in a solution.
ksp is the product of the soluble ions in the compound. Higher the ksp value, higher the degree of solubility.
ZnCO₃ (s) ---> Zn²⁺ (aq) + CO₃²⁻ (aq)
n n
ksp = [Zn²⁺][CO₃²⁻]
In the equation equal amounts of ions Zn²⁺ and CO₃²⁻ ions are soluble.
amount of ions soluble = n
ksp is therefore equal to;
ksp = n x n
ksp = n²
ksp = 1 * 10⁻¹⁰ M
therefore
1 * 10⁻¹⁰ M = n²
n = 1 x 10⁻⁵ M
therefore concentration of CO₃²⁻ = 1 x 10⁻⁵ M
<span>The element gold has 32 isotopes, ranging from A =173 to A = 204. during alpha decay Au will loose 2 in atomic number and 4 in mass number . it will form a iridium isotope and helium . according to the above statement, the balanced equation for the alpha decay of gold 173 will be given as below.
173 169 4
79 Au-------------->77 Ir + 2He</span>
Your compound is
.
Remember that the oxidation numbers in a neutral compound must add up to zero. Cl has an oxidation number of -1 because it is a halogen K has an oxidation number of +1 because it is an alkali metal, which exhibits an oxidation state of +1 in compounds.
Since you have 6 atoms of Cl, you have -1(6) = -6 for the Cl. Since you 2 atoms of K, you have +1(2) = +2 for the K. The oxidation number of Pt must make all the oxidation numbers add up to zero:
+2 + (-6) + oxidation number of Pt = 0
-4 + oxidation number of Pt = 0
Oxidation number of Pt = 4
Answer:
Maintaining a high starting-material concentration can render this reaction favorable.
Explanation:
A reaction is <em>favorable</em> when <em>ΔG < 0</em> (<em>exergonic</em>). ΔG depends on the temperature and on the reaction of reactants and products as established in the following expression:
ΔG = ΔG° + R.T.lnQ
where,
ΔG° is the standard Gibbs free energy
R is the ideal gas constant
T is the absolute temperature
Q is the reaction quotient
To make ΔG < 0 when ΔG° > 0 we need to make the term R.T.lnQ < 0. Since T is always positive we need lnQ to be negative, what happens when Q < 1. Q < 1 implies the concentration of reactants being greater than the concentration of products, that is, maintaining a high starting-material concentration will make Q < 1.
First, we write the half equations for the reduction of the chemical species present:
Cu⁺² + 2e → Cu; E° = 0.34 V
Ni⁺² + 2e → Ni; E° = - 0.23 V
In order to determine the potential of the cell, we find the difference between the two values. For this:
E(cell) = 0.34 - (-0.23)
E(cell) = 0.57 V
The second option is correct. (The difference in values is due to different values in literature, and it is negligible)
