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natima [27]
2 years ago
15

The final overall chemical equation is Upper Ca upper O (s) plus upper C upper O subscript 2 (g) right arrow upper C a upper C u

pper O subscript 3 (s).. When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and has its sign changed. is halved. has its sign changed. is unchanged.
Chemistry
1 answer:
GenaCL600 [577]2 years ago
7 0

Answer:

the enthalpy of the second intermediate equation is halved and has its sign changed.

Explanation:

Let us take a look at the first and second intermediate reactions as well as the overall reaction equation for the process under review;

First reaction;

Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ

Second reaction;

2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ

Hence the overall equation is now;

CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH = ?

According to the Hess law of constant heat summation, the enthalpy of the overall reaction is supposed to be obtained as a sum of the enthalpy of both reactions but this will not give the enthalpy of the overall reaction in this case. The enthalpy of the overall reaction is rather obtained by halving the enthalpy of the second intermediate reaction and reversing its sign before taking the sum as shown below;

Enthalpy of Intermediate reaction 1 + ½(- Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction

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Ann [662]

Solution:

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2 years ago
An unknown compound melting at 131 - 133 C. It is thought to be one of the following compounds: trans-cinnamic acid (133-134); b
Vesnalui [34]

Answer:

benzamide

Explanation:

Compound            melting Point ,ºC          Melting Pont Mixture, ºC

       X                          131 - 133

trans-cinnamic            133 - 134                      110 - 120

acid

benzamide                 128 - 130                       130-132

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The reason there is not an exact match is not due due to the presence of impurities. The presence of impurities always lower the melting point ( it is a coligative property such as the melting point depresion of salt and water )

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5 0
2 years ago
Please help!!
Sati [7]

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Since the reaction is already balanced, we can now identify which is the limiting and excess reagent.

First, we need to determine the number of moles of each chemical in the equation. This is crucial for determining the limiting and excess reagent.

<span>Assuming that there is the same amount  of solution X for each reactant</span>

1.0 M NaOH ( X ) = 1.0 moles NaOH

1.00 x 10-5 M C25N3H30Cl ( X ) = 1.00 x 10-5 moles C25N3H30Cl

<span>The result showed that the crystal violet has lesser amount than NaOH. Thus, the limiting reactant in this chemical reaction is crystal violet and the excess reactant is NaOH.</span>

3 0
2 years ago
When a 1.00 L sample of water from the surface of the Dead Sea (which is more than 400 meters below sea level and much saltier t
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Now put all the given values in the formula of molarity, we get

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Sindrei [870]

Answer:

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Explanation:

Hope this helped :)

5 0
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