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lubasha [3.4K]
3 years ago
9

What is always true of a weak acid ?

Chemistry
2 answers:
Mariana [72]3 years ago
7 0

Answer: D. it only partially dissociates in solution.

Explanation:

Strong acid is defined as the acid which completely dissociates when dissolved in water. They have low pH. These releases H^+ ions in their aqueous states.

HNO_3(aq.)\rightarrow H^+(aq.)+NO_3^-(aq.)

Weak acid is defined as the acid which does not completely dissociates when dissolved in water. They have high pH. These releases H^+ ions in their aqueous states.

CH_3COOH\rightleftharpoons CH_3COO^-+H^+

Thus the correct option is it only partially dissociates in solution.

vichka [17]3 years ago
4 0
The answer is d per me
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Use the data given below to construct a Born-Haber cycle to determine the second ionization energy of Ca. Δ H°(kJ) Ca(s)→Ca(g) 1
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Answer :  The value of second ionization energy of Ca is 1010 kJ.

Explanation :  

The formation of calcium oxide is,

Ca(s)+\frac{1}{2}O_2(g)\overset{\Delta H_f}\rightarrow CaO(s)

\Delta H_f^o = enthalpy of formation of calcium oxide = -635 kJ

The steps involved in the born-Haber cycle for the formation of CaO:

(1) Conversion of solid calcium into gaseous calcium atoms.

Ca(s)\overset{\Delta H_s}\rightarrow Ca(g)

\Delta H_s = sublimation energy of calcium = 193 kJ

(2) Conversion of gaseous calcium atoms into gaseous calcium ions.

Ca(g)\overset{\Delta H_I_1}\rightarrow Ca^{+1}(g)

\Delta H_I_1 = first ionization energy of calcium = 590 kJ

(3) Conversion of gaseous calcium ion into gaseous calcium ions.

Ca^{+1}(g)\overset{\Delta H_I_2}\rightarrow Ca^{+2}(g)

\Delta H_I_2 = second ionization energy of calcium = ?

(4) Conversion of molecular gaseous oxygen into gaseous oxygen atoms.

O_2(g)\overset{\Delta H_D}\rightarrow OI(g)

\frac{1}{2}O_2(g)\overset{\Delta H_D}\rightarrow O(g)

\Delta H_D = dissociation energy of oxygen = \frac{498}{2}=249kJ

(5) Conversion of gaseous oxygen atoms into gaseous oxygen ions.

O(g)\overset{\Delta H_E_1}\rightarrow O^-(g)

\Delta H_E_1 = first electron affinity energy of oxygen = -141 kJ

(6) Conversion of gaseous oxygen ion into gaseous oxygen ions.

O^-(g)\overset{\Delta H_E_2}\rightarrow O^{2-}(g)

\Delta H_E_2 = second electron affinity energy of oxygen = 878 kJ

(7) Conversion of gaseous cations and gaseous anion into solid calcium oxide.

Ca^{2+}(g)+O^{2-}(g)\overset{\Delta H_L}\rightarrow CaO(s)

\Delta H_L = lattice energy of calcium oxide = -3414 kJ

To calculate the overall energy from the born-Haber cycle, the equation used will be:

\Delta H_f^o=\Delta H_s+\Delta H_I_1+\Delta H_I_2+\Delta H_D+\Delta H_E_1+\Delta H_E_2+\Delta H_L

Now put all the given values in this equation, we get:

-635kJ=193kJ+590kJ+\Delta H_I_2+249kJ+(-141kJ)+878kJ+(-3414kJ)

\Delta H_I_2=1010kJ

Therefore, the value of second ionization energy of Ca is 1010 kJ.

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