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Stella [2.4K]
2 years ago
5

If the specific heat of iron is 0.449 J/g·K, what is the enthalpy change of this process for 100 g of Fe (s)? Is the process end

othermic or exothermic? Explain.
Chemistry
1 answer:
katrin2010 [14]2 years ago
8 0

Full Question:

Although it is not really a chemical reaction, one could consider a cooling process to be analogous to the enthalpy change that occurs in a reaction. Consider the transformation

Fe (s, T = 300 K) → Fe (s, T = 250 K)

If the specific heat of iron is 0.449 J/g·K, what is the enthalpy change of this process for 100 g of Fe (s)? Is the process endothermic or exothermic? Explain.

Answer:

ΔH = - 2245J

Explanation:

Specific heat capacity, Cp = 0.449 J/g·K

ΔH=MCpΔT

ΔT = 250K - 300K = - 50K

ΔH = 100 * 0.449 * (-50)

ΔH = - 2245J

The iron will lose (give off) heat energy to cool down. Hence, it is an exothermic process. The negative sign of the enthalpy confirms this.

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3 years ago
1. Using the balanced equation, answer the following questions:
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Answer:

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                     b)  31.90 g of KCl

Explanation:

                  The balance chemical equation for given decomposition reaction is as follow;

                                   2 KClO₃ → 2 KCl + 3 O₂

<h3>Part 1:</h3>

Step 1: <u>Calculate Moles for given amount of KClO₃;</u>

                    Moles  =  Mass / M.Mass

                    Moles  =  34.35 g / 122.55 g/mol

                    Moles  =  0.280 moles of KClO₃

Step 2: <u>Find out  moles of O₂ produced;</u>

According to eq,

                  2 moles of KClO₃ produces  =  3 moles of O₂

So,

            0.280 moles of KClO₃ will produce  =  X moles of O₂

Solving for X,

                    X  =  0.280 mol × 3 mol / 2 mol

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Step 3: <u>Calculate No. of Molecules of O₂ as,</u>

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No. of Molecules  =  0.42 mol × 6.022 × 10²³ molecules/mol

No. of Molecules  =  2.53 × 10²³ molecules of O₂

<h3>Part 2:</h3>

Step 1: <u>Calculate Moles for given amount of KClO₃;</u>

                    Moles  =  Mass / M.Mass

                    Moles  =  52.53 g / 122.55 g/mol

                    Moles  =  0.428 moles of KClO₃

Step 2: <u>Find out  moles of KCl produced;</u>

According to eq,

                  2 moles of KClO₃ produces  =  2 moles of KCl

So,

            0.428 moles of KClO₃ will produce  =  X moles of KCl

Solving for X,

                    X  =  0.428 mol × 2 mol / 2 mol

                     X =  0.428 moles of KCl

Step 3: <u>Calculate Mass of KCl as;</u>

                         Mass  =  Moles × M.Mass

                         Mass  =  0.428 mol × 74.55 g/mol

                         Mass  =  31.90 g of KCl

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