Answer:
helium hydrogen
Explanation:
lithium beryllium bottom carbon
Answer:
Approximately
.
Explanation:
The gallium here is likely to be produced from a
solution using electrolysis. However, the problem did not provide a chemical equation for that process. How many electrons will it take to produce one mole of gallium?
Note the Roman Numeral "
" next to
. This numeral indicates that the oxidation state of the gallium in this solution is equal to
. In other words, each gallium atom is three electrons short from being neutral. It would take three electrons to reduce one of these atoms to its neutral, metallic state in the form of
.
As a result, it would take three moles of electrons to deposit one mole of gallium atoms from this gallium
solution.
How many electrons are supplied? Start by finding the charge on all the electrons in the unit coulomb. Make sure all values are in their standard units.
.
.
Calculate the number of electrons in moles using the Faraday's constant. This constant gives the size of the charge (in coulombs) on each mole of electrons.
.
It takes three moles of electrons to deposit one mole of gallium atoms
. As a result,
of electrons would deposit
of gallium atoms
.
Taking into account the reaction stoichiometry, 340.0 moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas
<h3>Reaction stoichiometry</h3>
In first place, the balanced reaction is:
CO₂ + 4 H₄ → CH₄ + 2 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- CO₂: 1 mole
- H₄: 4 moles
- CH₄: 1 mole
- H₂O: 2 moles
<h3>Moles of CH₄ formed</h3>
The following rule of three can be applied: if by reaction stoichiometry 1 mole of CO₂ form 4 moles of CH₄, 85.1 moles of CO₂ form how many moles of CH₄?

<u><em>moles of CH₄= 340.4 moles</em></u>
Then, 340.0 moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas
Learn more about the reaction stoichiometry:
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