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3 years ago

Study the reactions.

Chemistry

1 answer:

REY [17]3 years ago

8 0

Answer:

B) by first reversing Reaction 2 and then adding the enthalpy of Reaction 1 and the enthalpy of Reaction 2

Explanation:

C2H2 (g)+2H2 (g)→C2H6 (g) ΔH=−75 kcal

C2H4 (g)+H2 (g)→C2H6 (g) ΔH=−33 kcal

C2H2 (g)+H2 (g)→C2H4 (g) ΔH= ?

First step - Reverse reaction 2.

C2H6 (g) → C2H4 (g) + H2 (g) ΔH= 33 kcal

Second step - Add reaction 1 and the reversed reaction 2. This gives us:

C2H2 (g) + 2H2 (g) + C2H6 (g) → C2H6 (g) + C2H4 (g) + H2 (g)

Simplifying gives:

C2H2 (g)+H2 (g)→C2H4 (g)

The correct option is:

B) by first reversing Reaction 2 and then adding the enthalpy of Reaction 1 and the enthalpy of Reaction 2

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Answer:

The specific heat capacity of the metal is 1.307 J/g °C

Explanation:

<u>Step 1:</u> Data given

mass of the metal = 312 grams

initial temperature of the metal ( before plunged in the water) = 277.845 °C

initial temperature water = 11.945 °C

Final temperature of water (and aslo the metal) = 99.062 °C

Specific heat capacity of wayer = 4.184 J/g °C

<u>Step 2:</u> Calculate the specifi heat capacity of the metal

Loss of Heat of the Metal = Gain of Heat by the Water

Qmetal = -Qwater

m(metal) * C(metal) * ΔT(metal) = - m(water) * C(water) * ΔT(water)

with mass of metal = 312 grams

with C(metal) = TO BE DETERMINED

with ΔT (metal) = 99.062 - 277.845 = -178.783 °C

with mass of water = 200 grams

with C(water) = 4.184 J/°C * g

with ΔT (water) = 99.062 - 11.945 = 87.117 °C

312 * C(metal) * -178.783 = - 200* 4.184 * 87.117

C(metal) = -72899.51 / 55780.296 = 1.307

The specific heat capacity of the metal is 1.307 J/g °C

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3 years ago