Fluorine has the highest. Fluorine's electronegativity is 4.0
The energy that is
essential to break one C-H bond is 414 kJ/mol. Since, there are four C-H bonds
in CH4, the energy Δ HCH4 for
breaking all the bonds is calculated as Δ HCH4 = 4 x bond energy of C-H bond. By
multiplying the 4 with the 414 kJ/mol you can get the answer of 1656 kJ/mol CH4
molecules.
Answer:
chlorine is therefore an anion
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
Answer:
The correct answer is - 29.45 / 100 x 25.6 = 7.5392 grams
Explanation:
It is given in the question that in 100 gms of CaSO4 there are 29.45 grams of Ca present and there is 25.6 gram of total CaSO4 sample present, So, to calculate the exact value of calcium in this given sample is:
mass of Ca = total amount of sample*percentage of calcium in sample /100
M of Ca =25.6*29.45/100
M of Ca = 7.5392 grams
Thus, the correct procedure is given by 29.45 / 100 x 25.6 = 7.5392 grams
