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Anna007 [38]
3 years ago
9

Why can you only change 1 variable in an experiment??? #Science

Chemistry
1 answer:
Montano1993 [528]3 years ago
7 0
<span>It is important to test only one variable when conducting an experiment because that one variable will only have one change. You want to know whether or not that one change is what caused the outcome of the experiment. If you use too many you don't know what really caused the outcome.</span>
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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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