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3 years ago

33 grams of NH4NO3 will produce how many grams of water

Chemistry

1 answer:

Leni [432]3 years ago

8 0

Answer:

66 grams

Explanation:

For every 1 gram of NH4NO3 equals 2 grams of water

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3 years ago

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

Drupady [299]

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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3 years ago