A)2014
B)Cl2
C)Mg
D)Mg2+
The metal loses electrons, and when it becomes Mg2+, it loses two electrons, becoming oxidized. So, the correct option is D.
Initially, the word "oxidation" was used to refer to chemical processes in which a component reacts with oxygen. An illustration of this is the oxidation of magnesium in the formation of magnesium oxide when magnesium and oxygen react.
Antoine Lavoisier used the term "oxidation" to describe the reaction of a material with oxygen. The meaning was later expanded to cover additional reactions in which electron are lost, irrespective of whether oxygen was present, after it was realised that the substance loses electrons when it is oxidised.
In the aforementioned example, the metal loses electrons; specifically, it loses two electrons as it changes from Mg+ to Mg2+, getting oxidized.
Therefore, the correct option is D
Learn more about Oxidation reaction, here:
#SPJ7
Answer:
Mg ²⁺
Explanation:
Τhe metal loses electrons and in forming Mg²⁺ ,it loses 2 electrons and hence oxidized.
Mg(s) ⇒ Mg²⁺ + 2e⁻
Answer:
27·92 %
Explanation:
Given chemical equation is
2C4H10 + 13O2 -----> 8CO2 + 10H2O
So according to the above balanced chemical equation for 2 moles of C4H10, 10 moles of water is produced
Molecular weight of C4H10 is 58 g
Molecular weight of water is 18 g
So for 116 g of C4H10, 180 g of water is produced and therefore for 1 g of C4H10 (180 ÷ 116) g of water is produced
∴ For 1 g of C4H10 1·552 g of water is produced
For 3000 g of C4H10, (3000 × 1·552) g of water is produced
∴ Number of grams of water produced for 3000 g of C4H10 is 4656 g
Percent yield =((experimental amount) ÷ (theoretical amount)) × 100
Here experimental amount is 4656 g
theoretical amount is 1300 g
∴ Percent yield = (1300 ÷ 4656) × 100 = 27·92 %
∴ Percent yield = 27·92 %