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Hitman42 [59]
3 years ago
6

Standard Thermodynamic Quantities for Selected Substances at 25 ∘C Reactant or product ΔHf∘(kJ/mol) Al(s) 0.0 MnO2(s) −520.0 Al2

O3(s) −1675.7 Mn(s) 0.0 Part A The thermite reaction, in which powdered aluminum reacts with manganese oxide, is highly exothermic. 4Al(s)+3MnO2(s)→2Al2O3(s)+3Mn(s) Use standard enthalpies of formation to find ΔH∘rxn for the thermite reaction.
Chemistry
1 answer:
Svet_ta [14]3 years ago
4 0

Answer:

-1815.4 kJ/mol

Explanation:

Starting with standard enthalpies of formation you can calculate the standard enthalpy for the reaction doing this simple calculation:

∑ n *ΔH formation (products) - ∑ n *ΔH formation (reagents)

This is possible because enthalpy is state function meaning it only deppends on the initial and final state of the system (That's why is also possible to "mix" reactions with Hess Law to determine the enthalpy of a new reaction). Also the enthalpy of formation is the heat required to form the compound from pure elements, then products are just atoms of reagents organized in a different form.

In this case:

ΔH rxn = [(2 * -1675.7) - (3 * -520.0)] kJ/mol = -1815.4 kJ/mol

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The correct option is: CH4 + O2 → CO2 + H2O.

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Arrange the molecules H2O, NH3, Ar, NaCl in order of expected increasing boiling points. 1. None of these 2. NaCl, H2O, NH3, Ar
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Answer:

Option (4) is correct

Explanation:

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The radioisotope potassium-40 decays to argon-40 by positron emission with a half-life of 1.3 Ã 109 yr. A sample of moon rock wa
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Explanation:

The age of the rock can be calculated using the equation:

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Now from the data , we have 78 argon-40  atoms for every 22 potassium-40 atoms, we can deduce that originally we had 22 + 78 = 100 atoms of potassium-40 so this is our N₀.

When we look at the equation, we see that k is unknown, but we can calculate it from the half-life which is given by the equation:

k =  0.693/ t half-life = 0.693/ 1.3 x 10⁹ yr = 5.33 x 10⁻¹⁰ yr⁻¹

Now we are in position to answer the question.

ln ( 78/100 ) =  - (5.33 x 10⁻¹⁰ yr⁻¹ ) t

- 0.249 = - 5.33 x 10⁻¹⁰ yr⁻¹  t

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