Dependent variable: amount of dandelions killed by werd
Independent variable: health of dandelions/where they grew (if one is malnourished/older/dry it might be killed by the weed killer with more ease)
The value of Keq = 110
Let's know how to get it :
According to the equilibrium reaction equation :
I2(g) + Br2(g) → 2IBr(g)
now we need to get the value of Keq
when the Keq expression is:
Keq = [IBr]^2/ [I] [Br]
when we have [IBr] at equilibrium = 0.84 so we need to get [I] &[Br] at equilibrium .
when [Br2] = [I2] = 0.5 - 0.84 / 2
= 0.08 M
so, by substitution
∴ Keq = (0.84)^2/(0.08)^2
= 110
The hormone that promotes active tubular secretion of potassium ions into, along with sodium absorption from filtrate in the distal convoluted tubule and collecting duct is Aldosterone. This hormone is produced by the adrenal gland and acts mainly in the functional unit of the kidney to aid in the conservation of sodium, secretion of potassium, water retention and to stabilize blood pressure. It indirectly regulates blood levels of electrolytes and helps to maintain the blood pH.
Answer:
a) MZ₂
b) They have the same concentration
c) 4x10⁻⁴ mol/L
Explanation:
a) The solubility (S) is the concentration of the salt that will be dissociated and form the ions in the solution, the solubility product constant (Kps) is the multiplication of the concentration of the ions elevated at their coefficients. The concentration of the ions depends on the stoichiometry and will be equivalent to S.
The salts solubilization reactions and their Kps values are:
MA(s) ⇄ M⁺²(aq) + A⁻²(aq) Kps = S*S = S²
MZ₂(aq) ⇄ M⁺²(aq) + 2Z⁻(aq) Kps = S*S² = S³
Thus, the Kps of MZ₂ has a larger value.
b) A saturated solution is a solution that has the maximum amount of salt dissolved, so, the concentration dissolved is solubility. As we can notice from the reactions, the concentration of M⁺² is the same for both salts.
c) The equilibrium will be not modified because the salts have the same solubility. So, let's suppose that the volume of each one is 1 L, so the number of moles of the cation in each one is 4x10⁻⁴ mol. The total number of moles is 8x10⁻⁴ mol, and the concentration is:
8x10⁻⁴ mol/2 L = 4x10⁻⁴ mol/L.
