Answer:
The answer is "
"
Explanation:
We arrange oxoacids to decrease the intensity of acids in this question. Or we may conclude all this from strongest to weakest acids they order oxoacids, that's why above given order is correct.
Answer:
Yes
Explanation:
Calcium Chloride is an inorganic compound, a salt that can dissolve in water.
Answer:
Explanation:
To answer this question successfully, we need to remember that atoms are neutral species, since the number of protons, the positively charged particles, is equal to the number of electrons, the negatively charged particles. That said, we may firstly find an atom which has 3 electrons (and, as a result, 3 protons, as it should be neutral).
The number of protons is equal to the atomic number of an element. We firstly may have an atom with 3 protons and 3 electrons (atomic number of 3, this is Li).
Similarly, we may take the atomic number of 4, beryllium, and remove 1 electron from it. Upon removing an electron, it would become beryllium cation, 
.
We may use the same logic going forward and taking the atomic number of 5. This is boron. In this case, we need to remove 2 electrons to have a total of 3 electrons. Removal of 2 electrons would yield a +2-charged cation: 
.
Answer:
pH = 3.3
Explanation:
Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.
In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].
Solution using the I.C.E. table:
HNO₂ ⇄ H⁺ + KNO₂⁻
C(i) 0.55M 0M 0.75M
ΔC -x +x +x
C(eq) 0.55M - x x 0.75M + x b/c [HNO₂] / Ka > 100, the x can be
dropped giving ...
≅0.55M x ≅0.75M
Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]
=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M
pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3
Solution using the Henderson-Hasselbalch Equation:
pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]
= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]
= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3
