State hess's law.how is it used
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For example, to check the enthalpy change that occurs when benzene undergoes incomplete combustion to water and carbon monoxide is not an easy task, because the products invariably contain CO2. However, by combining the reactions of the complete combustion of benzene and the combustion of CO, you can get the reaction you want.
Reaction wanted: 2C6H6 + 9O2 → 12CO + 6H2O
Reactions provided: 2C6H6 + 15O2 → 12CO2 + 6H2O and 2CO + O2 → 2CO2, and their associated ΔH.
Rearrange the reactions so that, when they add up, they result in the wanted reaction.
2C6H6 + 15O2 → 12CO2 + 6H2O (leave as is; no changes to ΔH)
12CO2 → 12CO + 6O2 (reverse and multiply by 6; this changes the sign of ΔH and multiplies it by 6)
Added up, it will result in 2C6H6 + 9O2 → 12CO + 6H2O. Add up the ΔH values for the rearranged reactions to find ΔH for this particular reaction.
Answer: Hess's Law says the total enthalpy change does not rely on the path taken from beginning to end. Enthalpy can be calculated in one grand step or multiple smaller steps.
To solve this type of problem, organize the given chemical reactions where the total effect yields the reaction needed. There are a few rules that you must follow when manipulating a reaction.
The reaction can be reversed. This will change the sign of ΔHf.
The reaction can be multiplied by a constant. The value of ΔHf must be multiplied by the same constant.
Any combination of the first two rules may be used.
Finding a correct path is different for each Hess's Law problem and may require some trial and error.
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For example,
"Which of the following molecules is expected to have one or more unpaired electrons? Check all that apply.
O2-
F2+
N22-
O22-"
Any help would be appreciated. ...?
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Final answer:
To determine unpaired electrons, look at the valence electrons in an ion's molecule, accounting for extra or absent electrons due to the ion's charge. For instance, O2- and F2+ have one unpaired electron, but N22- is not a valid ion, and O22- electrons are all paired.
Explanation:
To determine if a molecular ion will have one or more unpaired electrons, we need to look at the number of valence electrons in the molecule and take into account any extra or absent electrons due to the charge of the ion.
For example, O2- has 12 + 1 = 13 valence electrons (6 from each oxygen atom and 1 extra due to the negative charge). It requires 14 for all the electrons to be paired (2 in each oxygen's inner shell and 4 bonds or lone pairs in the outer shell), thus there is one unpaired electron in O2-.
However, in F2+, there are only 13 electrons because one electron is lost due to the positive charge. Like oxygen, fluorine also prefers to have 7 electrons in its outer shell, so there are also unpaired electrons in F2+.
N22- doesn’t exist because nitrogen normally forms triple bond and doesn’t need or want to take extra electrons, so this molecular ion is not valid.
O22- has 14 + 2 = 16 valence electrons and this total number of valence electrons is an even number, which indicates all the electrons are paired.
Learn more about Unpaired Electrons here:
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Boron consists of two isotopes, boron-10 and boron-11.
Given that its relative atomic mass is 10.2
Find the abundance of each isotope.
Let y/100 = the abundance of copper-10
and (100 - y)/100 = the abundance of copper-11
10.2 = (y/100 x 10) + [(100 - y)/100 x 11]
10.2 = 10y/100 + 1100/100 - 11y/100
1020 = 10y + 1100 - 11y
-80 = -y
y = 80
Abundance of boron-10 = 10/100 = 10%
Abundance of boron-11 = 100 - 10 = 90%
Answers
To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant temperature and number of moles of the gas the product of PV is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:
P1V1 =P2V2
V2 = P1 x V1 / P2
V2 = 42.0 x 12.5 / 75.0
V2 = 7.0 L
Answer:
7.00 L
Explanation:
The only thing that varies between the two situations is pressure and volume.
we have the ideal gas equation
We know that n = moles of substance remain constant, also the temperature and n corresponding to the ideal gas constant
Situation N1
Situation N2
As nrT are equal both times, therefore we can match this term in both equations
We equate both equations
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Answer:
Explanation:
A pure substance is a substance that consists of only one type of particle, either atoms, molecules, or ions, and has a consistent and uniform composition throughout. Pure substances cannot be separated into other substances by physical means. They have well-defined chemical properties and distinct physical properties, such as melting point and boiling point.
Two examples of pure substances are:
Elemental Hydrogen (H2): Elemental hydrogen consists of diatomic molecules, each containing two hydrogen atoms bonded together. It is a pure substance because it consists only of hydrogen molecules and cannot be separated into other substances without chemical reactions.
Water (H2O): Water is a pure substance composed of water molecules, each consisting of two hydrogen atoms and one oxygen atom bonded together. It is also a pure substance because it has a consistent and uniform composition throughout and cannot be separated into other substances without breaking its molecular bonds.
These examples illustrate the concept of pure substances, where the composition is homogeneous and consistent throughout the entire sample.