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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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Question 13: Consider the strength of the Hβ absorption line in the spectra of stars of various surface temperatures. This is th
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The absorption and strength of the H-beta lines change with the temperature of the stellar surface, and because of this, one can find the temperature of the star from their absorption lines and strength. To better comprehend, let us look into the concept of the atom's atomic structure.  

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Therefore, after captivating the photons the electrons jump from level 2 to level 4, which shows that there is an increase in the stellar surface temperature and at the same time one can witness a decline in the strength of the H-beta lines. In case, if the temperature of the surface increases too much, then one will witness no attachment of electron with the hydrogen atom and thus no H lines, and if the temperature of the surface becomes too low, then the electrons will stay in the ground state and no formation of H lines will take place in that condition too.  

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5 0
3 years ago
What is the empirical formula of a substance that contains 12.0 g of c, 2.00 g of h, and 5.33 g of o?
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Empirical formula is the simplest ratio of whole numbers of components making up a compound. 
empirical formula can be calculated as follows
                      C                             H                          O
mass          12.0 g                      2.00 g                   5.33 g 
number of moles  
                  12.0 g / 12 g/mol    2.00 g / 1 g/mol       5.33 g / 16 g/mol 
                    = 1.00 mol                = 2.00 mol           = 0.333 mol
divide by the least number of moles
                   1.00 / 0.333              2.00 / 0.333          0.333/ 0.333 
                    = 3.00                     = 6.01                    = 1.00
the number of atoms 
C - 3
H - 6
O - 1
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8 0
3 years ago
Please help!! A compound with a molar mass of 544.0 g/mol is made up of 26.5 grams Carbon, 2.94 grams
lukranit [14]

Answer:

1. Empirical formula = C₃H₄O₆

2. Molecular formula = C₁₂H₁₆O₂₄

Explanation:

From the question given above, the following data were obtained:

Molar mass of compound = 544 g/mol

Mass of Carbon (C) = 26.5 g

Mass of Hydrogen (H) = 2.94 g

Mass of oxygen (O) = 70.6 g

Empirical formula =?

Molecular formula =?

1. Determination of the empirical formula of the compound.

C = 26.5 g

H = 2.94 g

O = 70.6 g

Divide by their molar mass

C = 26.5 / 12 = 2.208

H = 2.94 / 1 = 2.94

O = 70.6 / 16 = 4.4125

Divide by the smallest

C = 2.208 / 2.208 = 1

H = 2.94 / 2.208 = 1.33

O = 4.4125 / 2.208 = 2

Muitiply through by 3 to express in whole number.

C = 1 × 3 = 3

H = 1.33 × 3 = 4

O = 2 × 3 = 6

Empirical formula = C₃H₄O₆

2. Determination of the molecular formula of the compound.

Molar mass of compound = 544 g/mol

Empirical formula = C₃H₄O₆

Molecular formula = [C₃H₄O₆]ₙ

[C₃H₄O₆]ₙ = 544

[(12×3) + (4×1) + (16×6)]n = 544

[36 + 4 + 96]n = 544

136n = 544

Divide both side by 136

n = 544 / 136

n = 4

Molecular formula = [C₃H₄O₆]ₙ

Molecular formula = [C₃H₄O₆]₄

Molecular formula = C₁₂H₁₆O₂₄

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