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ElenaW [278]
3 years ago
14

A useful single-displacement reaction involves thermite, which is a

Chemistry
1 answer:
Licemer1 [7]3 years ago
7 0
The reaction of iron (III) oxide and aluminum is initiated by heat released from a small amount "starter mixture". This reaction is an oxidation-reduction reaction, a single replacement reaction, producing great quantities of heat (flame and sparks) and a stream of molten iron and aluminum oxide which pours out of a hole in the bottom of the pot into sand.

The balanced chemical equation for this reaction is:

2 Al(s) + Fe2O3(s) --> 2Fe(s) + Al2O3(s) + 850 kJ/mol

Curriculum Notes
This chemical reaction can be used to demonstrate an exothermic reaction, a single replacement or oxidation-reduction reaction, and the connection between ∆H calculated for this reaction using heats of formation and Hess' Law and calculating ∆H for this reaction using qrxn = mc∆T and the moles of limiting reactant. This reaction also illustrates the role of activation energy in a chemical reaction. The thermite mixture must be raised to a high temperature before it will react.

To determine how much thermal energy is released in this reaction, heats of formation values and Hess' Law can be used.

By definition, the deltaHfo of an element in its standard state is zero.

2 Al(s) + Fe2O3(s) --> 2Fe (s) + Al2O3 (s)

The deltaH for this reaction is the sum of the deltaHfo's of the products - the sum of the deltaHfo's of the reactants (multiplying each by their stoichiometric coefficient in the balanced reaction equation), i.e.:

deltaHorxn = (1 mol)(deltaHfoAl2O3) + (2 mol)(deltaHfoFe) - (1 mol)(deltaHfoFe2O3) - (2 mol)(deltaHfoAl)

deltaHorxn = (1 mol)(-1,669.8 kJ/mol) + (2 mol)(0) - (1 mol)(-822.2 kJ/mol) - (2mol)(0 kJ/mol)

deltaHorxn = -847.6 kJ

The melting point of iron is 1530°C (or 2790°F).
MARK ME BRAINLIEST

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Answer:

(a) AL

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(c)Al

Explanation:

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The higher the electron is on energy level the farther its from nucleus and more loosely bonded thus need lesser energy.

By looking at electron configuration we can figure out which electron will need more energy.

<h3>(a)Na, Mg, Al</h3>

1s², 2s², 2p⁶, 3s², 3p⁶,4s², 3d¹

Na₁₁ ⇒ 1s², 2s², 2p⁶, 3s¹

Mg₁₂ ⇒ 1s², 2s², 2p⁶, 3s²

Al₁₃ ⇒ 1s², 2s², 2p⁶, 3s², 2p¹

Al will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.

<h3>(b) K, Ca, Sc</h3>

K₁₉⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s²

Ca₂₀⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s²

Sc₂₁⇒1s², 2s², 2p⁶, 3s², 3p⁶,4s², 3d¹

Sc will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.

<h3>(c) Li, Al, B</h3>

Li₃ ⇒ 1s², 2s¹

Al₁₃ ⇒ 1s², 2s², 2p⁶, 3s², 2p¹

B₅ ⇒ 1s², 2s², 2p¹

Al will have lowest IE₃ as its third electron will be in highest energy level and more loosely bonded to nucleus.

3 0
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A chemical change affects on the molecular level of matter, which makes it irreversible. Combustion is a pretty good exmple. Physical changes are reversible and dont alter the formula. Hope this helped!
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As you move across a period, the atomic mass increases because the atomic number also increases. ... The atomic mass for any given atom mainly comes from the mass of the protons and neutrons.

Explanation:

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