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Likurg_2 [28]
3 years ago
8

Find the empirical formula of the following compound: 0.77 mol of iron atoms combined with 1.0 mol of oxygen atoms.

Chemistry
1 answer:
eimsori [14]3 years ago
7 0

The empirical formula is Fe₃O₄.

The empirical formula is the <em>simplest whole-number ratio of atoms</em> in a compound.

The ratio of atoms is the same as the ratio of moles, so our job is to calculate the molar ratio of Fe to O.

I like to summarize the calculations in a table.

<u>Element</u>   <u>Moles</u>    <u>Ratio</u>¹   <u>×3</u>²   <u>Integers</u>³

    Fe         0.77       1         3             3

    O           1.0         1.3      3.9          4

¹ To get the molar ratio, you divide each number of moles by the smallest number (0.77).

² If the ratio is not close to an integer, multiply by a number (in this case, 3) to get numbers that are close to integers.

³ Round off these numbers to integers (3 and 4).

The empirical formula is Fe₃O₄.

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Calculate the standard enthalpy change of formation of CHA given that the standard enthalpies change of combustion of methane, g
Nostrana [21]

Answer : The standard enthalpy of formation of methane is, -74.8 kJ/mole

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of CH_4 will be,

C(s)+2H_2(g)\rightarrow CH_4(g)    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

(1) CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)     \Delta H_1=-890kJ/mole

(2) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-393.4kJ/mole

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-285.7kJ/mole

Now we will reverse the reaction 1, multiply reaction 3 by 2 then adding all the equations, we get :

(1) CO_2(g)+2H_2O(l)\rightarrow CH_4(g)+2O_2(g)     \Delta H_1=890kJ/mole

(2) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-393.4kJ/mole

(3) 2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)    \Delta H_3=2\times (-285.7kJ)=-571.4kJ/mole

The expression for enthalpy of formation of CH_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(+890kJ/mole)+(-393.4kJ/mole)+(-571.4kJ/mole)

\Delta H=-74.8kJ/mole

Therefore, the standard enthalpy of formation of methane is, -74.8 kJ/mole

3 0
3 years ago
Define ionization energy and explain why the second aisle and ionization energy of the element is higher than its ionization ene
seraphim [82]

Define ionization energy and explain why the second aisle and ionization energy of the element is higher than its ionization energy

The energy needed to remove one or more electrons from a neutral atom to form a positively charged ion is a physical property that influences the chemical behavior of the atom. By definition, the first ionization energy of an element is the energy needed to remove the outermost, or highest energy, electron from a neutral atom in the gas phase.

The process by which the first ionization energy of hydrogen is measured would be represented by the following equation.

H(g) -----> H+(g) + e-     deltaHo = -1312.0 kJ/mol


6 0
3 years ago
Assuming the volume is constant, if a gas has an
anyanavicka [17]

Answer:

672kP

Explanation:

Equate the initial pressure with respect the first temperature. The unknown pressure to the second temperature and you cross multiply

4 0
3 years ago
Read 2 more answers
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frez [133]

Answer:

I would say B but i am not sure hope it helps

Explanation:

5 0
3 years ago
There are two naturally occurring isotopes of copper. 63cu has a mass of 62.9296 amu. 65cu has a mass of 64.9278 amu. determine
SSSSS [86.1K]
1) You need to use the atomic mass of copper.


You can find it in a periodic table. It is 63.546 amu.


2) The atomic mass is the weigthed mass of the different isotopes.


This is, the atomic mass of one element is the atomic mass of each isotope times its corresponding abundance:


=> atomic mass of the element = abundance isotope 1 * atomic mass isotope 1 + abundance isotope 2 *  atomic mass isotope 2 + ....+abundance isotope n * atomic mass isotope n.


3) The statement tells there are two isotopes so the abundance of one is x and the abundance of the other is 1 - x


=> 63.546 amu = x * 62.9296 amu + (1-x)*64.9278


=> 63.546 = 62.9296x + 64.9278 - 64.9278x


=> 64.9278x - 62.9296 = 64.9278 - 63.546


=> 1.9982x = 1.3818


=> x = 1.3818 / 1.9982 = 0.6915 = 69.15%


=> 1 - x = 1 - 0.6915 = 0.3085 = 30.85%


Answer:


Cu-63 69.15%;


Cu-65 : 30.85%
3 0
3 years ago
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