Answer:
The equilibrium constant Ksp of the generic salt AB2 = 6.4777 *10^-8 M
Explanation:
Step 1: The balanced equation
AB2 ⇒ A2+ + 2B-
Step 2: Given data
Concentration of A2+ = 0.00253 M
Concentration of B- = 0.00506 M
Step 3: Calculate the equilibrium constant
Equilibrium constant Ksp of [AB2] = [A2+][B-]²
Ksp = 0.00253 * 0.00506² = 6.4777 *10^-8 M
The equilibrium constant Ksp of the generic salt AB2 = 6.4777 *10^-8 M
Answer:
I think a is the best answer. but really I can not confirm that.sorry for that.
Answer:
4.5 multiplied by 10, to the -10th power.
There is no gaseous product when NaNO3 reacts with 6M HCl.
We must note that gas evolution is one of the signs that a chemical reaction has taken place. Hence, we must consider the options and look out for an option in which reaction with HCl does not lead to a gaseous product. This is the option that does not lead to evolution of a gas.
Let us consider the reaction; NaNO3(s) + HCl(aq) ---> NaCl(s) + HNO3(aq), there is no gaseous product hence NaNO3 does not result in gas evolution upon addition of 6M HCl.
Missing parts;
Addition of 6 M HCl to which substance will NOT result in gas evolution?
(A) Al
(B) Zn
(C) K2CO3
(D) NaNO3
Learn more about evolution of gas: brainly.com/question/2186340
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
