Answer is c
Work out the moles of 2H2O: 1.80/ (16+1+1) = 0.1
The moles of 2H2 are the same (0.1)
The Mr of H2 is 2
So the mass needed to produce 1.80g of water is
0.1 x 2 = 0.2
While metal B is being reduced from an ion to a solid metal, metal A is being oxidized.
Metal A is being displaced because, through oxidation, it transforms from a solid metal to ions in a solution.
What is Oxidation ?
Redox reactions include a change in the oxidation state of the substrate. Loss of electrons or a rise in an element's oxidation state are both considered to be oxidation.
Gaining electrons or lowering the oxidation state of an element or its constituent atoms are both examples of reduction.
In general, the term "reduction" refers to a reaction in which a chemical receives additional electrons; the compound that gets electrons is referred to as being reduced. We can think of the transformation of a ketone or an aldehyde into an alcohol as a two-electron reduction because hydride can be thought of as a proton plus two electrons.
So finally we can say that Metal A is being oxidized.
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Answer:
Solution A that will form a precipitate with Ksp = 2.3 x 10−4
Explanation:
Li₃PO₄ ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)
3S S
Where S = Solubility(mole/lit) and Ksp = Solubility product
⇒ Ksp = (3S)³ x (S)
⇒ 27S⁴ = 2.3x10−4
⇒ S = 0.05 mol/lit
Concentration of Li₃PO₄ precipitate = 0.05
<u>Solution A </u>
0.500 lit of a 0.3 molar LiNO₃ contains 0.5 x 0.3 = 0.15 mole
0.4 lit of a 0.2 molar Na₃PO₄ contains = 3 x 0.4 x 0.2 = 0.24 mole
3 LiNO₃ + Na₃PO₄ → 3 NaNO₃ + Li₃PO₄
(Mole/Stoichiometry) 
= 0.05 = 0.24
Since from (Mole/Stoichiometry) ratio we can conclude that LiNO₃ is limiting reagent.
So concentration of Li₃PO₄ is equal to 0.05.
Answer:
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