Answer:
LHS-RHS
1 C - 1 C
4 H-2 H
2 O- 3 O
So on right side 2 hydrogen are less and one oxygen is more..so
Balanced equation is
CH4+2O2==CO2 + 2H2O
Explanation:
Answer:
HCl conc.= 6.0mol/L
Explanation:
From the dissociation of HCl= 1 mole H+ and 1mol Cl-, which is equivalent stoichiometrically in concentration to that of 1 mol HCl,
Answer:
An atom with 1 valence electron and an atom with 7 valence electrons
Explanation:
Covalent bond:
It is formed by the sharing of electron pair between bonded atoms.
The atom with larger electronegativity attract the electron pair more towards it self and becomes partial negative while the other atom becomes partial positive.
For example:
In water the electronegativity of oxygen is 3.44 and hydrogen is 2.2. That's why electron pair attracted more towards oxygen, thus oxygen becomes partial negative and hydrogen becomes partial positive.
the number of valance electrons of oxygen are six and hydrogen is one that's why two hydrogen atoms are attached with one oxygen atom and complete the octet.
<u>Answer:</u> The atomic symbol of the given element is 
<u>Explanation:</u>
The general isotopic representation of an element is given as: 
where,
Z represents the atomic number of the element
A represents the mass number of the element
X represents the symbol of an element
For the given isotope: 130-iodine
Mass number = 130
Atomic number = 53
Hence, the atomic symbol of the given element is
Answer:
7. 3–ethyl–6 –methyldecane
8. 5–ethyl–2,2–dimethyl–4–propyl–4 –heptene
Explanation:
It is important to note that when naming organic compounds having two or more different substituent groups, we simply name them alphabetically.
The name of the compound given in the question above can be written as follow:
7. Obtaining the name of the compound.
Compound contains:
I. Decane.
II. 3–ethyl.
III. 6 –methyl.
Naming alphabetically, we have
3–ethyl–6 –methyldecane
8. Obtaining the name of the compound.
Compound contains:
I. 2,2–dimethyl.
II. 4–propyl.
III. 4 –heptene.
IV. 5–ethyl.
Naming alphabetically, we have
5–ethyl–2,2–dimethyl–4–propyl–4 –heptene
