Explanation:
When there occurs sharing of electrons between two chemically combining atoms then it forms a covalent bond. Generally, a covalent bond is formed between two non-metals.
An ionic bond is defined as the bond formed due to transfer of one or more number of electrons from one atom to another. An ionic bond is always formed between a metal and a non-metal.
Every atom of an element will have orbitals in which electrons are found. These orbitals are known as energy level.
A molecule is defined as the smallest particle present in a substance or atom.
A metallic bond is formed due to mobile valence electrons shared by positive nuclei in a metallic crystal.
Thus, we can conclude that given statements are correctly matched as follows.
1). a chemical bond formed by the electrostatic attraction between ions - ionic bond
2). a chemical bond formed by two electrons that are shared between two atoms - covalent bond
3). the orbitals of an atom where electrons are found - energy level
4). the smallest particle of a covalently bonded substance - molecule
5). a bond characteristic of metals in which mobile valence electrons are shared among positive nuclei in the metallic crystal - metallic bond
Answer:
Chloroform= limiting reactant
0.209mol of CCl4 is formed
And 32.186g of CCl4 is formed
Explanation:
The equation of reaction
CHCl3 + Cl2= CCl4 + HCl
From the equation 1 mol of
CHCl3 reacts with 1mol Cl2 to yield 1mol of CCl4
From the question
25g of CHCl3 really with Cl2
Molar mass of CHCl3= 119.5
Molar mass of Cl2 = 71
Hence moles of CHCl3= 25/119.5 = 0.209mol
Moles of Cl2 = 25/71 = 0.352mol
Hence CHCl3 is the limiting reactant
Since 1 mole of CHCl3 gave 1mol of CCl4
It implies that 0.209moles of CHCl3 will also give 0.209mol of CCl4
Mass of CCl4 formed = moles× molar mass= 0.209×154= 32.186g
Answer:
I'm thinking cooper but not sure
Answer:
+1
Explanation:
Na₂O₂
NOTE: the oxidation number of oxygen is always –2 except in peroxides where it is –1.
Thus, we can obtain the oxidation number of sodium (Na) in Na₂O₂ as illustrated below:
Na₂O₂ = 0 (oxidation number of ground state compound is zero)
2Na + 2O = 0
O = –1
2Na + 2(–1) = 0
2Na – 2 = 0
Collect like terms
2Na = 0 + 2
2Na = 2
Divide both side by 2
Na = 2/2
Na = +1
Thus, the oxidation number of sodium (Na) in Na₂O₂ is +1
