<span>2Li⁺(aq) + Zn⁰(s) → 2Li⁰(s) + Zn²⁺(aq)
</span>2Li⁺(aq) + 2e⁻ → 2Li⁰(s)
Zn⁰(s) → Zn²⁺(aq) +2e⁻
2 electrons are transferred from atom of Zn⁰ to 2 ions of Li⁺.
72g
Explanation:
Given parameters:
Number of moles of LiOH = 3moles
Unknown:
Mass of LiOH = ?
Solution:
A mole of substance is a unit used to make quantitative measures in chemistry.
It is the amount of substance that contains the avogadro's number of particles.
The mole is related to mass using the expression below;
Mass of a substance = number of moles x molar mass
Molar mass of LiOH = 7 + 16 + 1 = 24g/mol
Mass of LiOH = 3 x 24 = 72g
learn more:
Number of moles brainly.com/question/1841136
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Answer: Region 3
Explanation: The temperature and time graph suggests that region 3 is the region in which the substance can co exist in both the phases that is solid phase and liquid phase.
Region 1 explains that the the solid has just started melting and there occurs a break point and then region 2 again explains that the solid is taking more time with temperature to get converted into the liquid and thus region 3 explains the equilibrium between the two phases.
<u>Answer:</u> The ionic compound formed is
(barium fluoride)
<u>Explanation:</u>
Ionic compound is formed by the complete transfer of electrons from 1 atom to another atom. The cation is formed by the loss of electrons by metals and anions are formed by gain of electrons by non metals.
Taking the metal as barium and non-metal as fluorine.
Barium is the 56th element of the periodic table having electronic configuration of ![[Xe]6s^2](https://tex.z-dn.net/?f=%5BXe%5D6s%5E2)
This element will loose 2 electrons and will form
ion
Fluorine is the 9th element of the periodic table having electronic configuration of ![[He]2s^22p^5](https://tex.z-dn.net/?f=%5BHe%5D2s%5E22p%5E5)
This element will gain 1 electron and will form
ion
By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.
Hence, the ionic compound formed is
(barium fluoride)
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