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3 years ago

Calculate the maximum amount of useful work that can be obtained and comment on the spontaneity for the reaction at 25C :

Chemistry

1 answer:

harkovskaia [24]3 years ago

3 0

Answer:

$1.41 *10^{3} kJ/mol$

Explanation:

First, we find in the tables the ΔH of formation of each compound. As you can see in the (image 1)

Then we solve the ecuation for ΔH°reaction

ΔH°reaction=∑ΔH°f(products)−∑ΔH°f(Reactants)

ΔH°reaction= (-2* 393.5 - 2*285.8) - (52.4 + 0) kJ/mol

ΔH°reaction = -1.41 *10^3 kJ/mol

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How many moles of iron is 6.022 x 10^22 atoms of iron? (Report answer as a number rounded to one place past the decimal.) *

yuradex [85]

<h3>Answer:</h3>

$\displaystyle 0.1 \ mol \ Fe$

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

6.022 × 10²² atoms Fe (iron)

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

Set up: $\displaystyle 6.022 \cdot 10^{22} \ atoms \ Fe(\frac{1 \ mol \ Fe}{6.022 \cdot 10^{23} \ atoms \ Fe})$
Divide: $\displaystyle 0.1 \ mol \ Fe$

7 0

3 years ago

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what i

Ostrovityanka [42]

Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :

[butane]=1.0 M , [isobutane]=2.5 M

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?

Answer:

The equilibrium concentration of each gas:

[Butane] = 1.14 M

[isobutane] = 2.86 M

Explanation:

Butane ⇄ Isobutane

At equilibrium

1.0 M 2.5 M

After addition of 0.50 M of butane:

(1.0 + 0.50) M -

After equilibrium reestablishes:

(1.50-x)M (2.5+x)

The equilibrium expression will wriiten as:

$K_c=\frac{[Isobutane]}{[Butane]}$

$2.5=\frac{2.5+x}{(1.50-x)}$

x = 0.36 M

The equilibrium concentration of each gas:

[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M

[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M

3 0

3 years ago

Atomic radius of bromine

Keith_Richards [23]

Answer: The answer is 185 Pm

4 0

3 years ago

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Compared to winds on the earth the winds on Saturn are much stronger. <br><br> True or False?

Yanka [14]

This is most definitly true

3 0

3 years ago

What is the parent isotope when Cl-33 is formed during a beta decay?

zzz [600]

Answer:

33/16 S

Explanation:

In beta decay, the atomic number of the daughter nucleus increases by one unit while the mass of the daughter nucleus remains the same as that of the parent nucleus.

Hence, if we know that a beta decay has occurred, then the parent nucleus must have the same mass as its daughter nucleus but have an atomic number that is less than that of the daughter nucleus by only one unit, hence the answer above.

7 0

3 years ago