Some Basic Concepts Of Chemistry Flashcards
(29 cards)
Significant figures
¡) Decimal no. =
¡¡) Non-Decimal no. =
¡) First non zero se sari digits
¡¡) First non zero to last non zero
¡) 0.0025
¡¡) 208
¡¡¡)5005
¡v)126000
v) 500.0
v¡)2.0034
¡)2
¡¡)3
¡¡¡)4
¡v)
v)4
v¡)5
Molecular mass
> It is the sum of atomic masses of all element resent in a compound.
Molecular mass
> It is the sum of atomic masses of all element resent in a compound.
Molecular mass of H2O
2(H=1)
2×1=2
O=16
2+16= 18u
- CO
- CO2
- HNO3
- C6H1206
- 28u
- 44u
- 63u
- 180u
Molar mass
It is molecular mass in grams.
Molar mass of H20
18g
Representation of an element
Atomic mass
ELEMENT
Atomic number
Isotopes
Atomic number same
Atomic mass different
Any two examples of isotopes
1^1H, 2^1H, 3^1H
12^6C, 14^6C
35^17Cl, 37^17Cl
Isobars
Atomic mass same
Atomic number different
Any two example of isobars
40^18Ar, 40^20Ca
59^27Co, 59^28Ni
Average atomic mass
Relative abundance=R
Atomic mass=A
Avg atomic mass= R1×A1÷100+ R2×A2÷100
Find atomic mass
Isotope>Relative abundance>Atomic mass(amu)
^12C>98.892>12
^13C>1.108>13.00335
^14C>2×10^-10>14.00317
Avg atomic= 98.892×12÷100 + 1.108×13.00335÷100 + 2×10^-10×14.00317÷100
=12.001u
(Bas carbon ka atomic number likhkr .smth daldo)
Scientific notations
<– =
–> =
+ve
-ve
- 43126
- 0.000129
- 0.009625×10^3
(Pehla digit 10 se kam 9-1)
- 4.3×10^4
- 1.29×10^-4
- 9.6×10^-6
Mole concept
Q> Define 1 mole
It is defined as the mass exactly equal to one-twelfth of the mass of one carbon-12 atom
No. Of moles?
(Using given mass and molar mass)
No. Of moles,n= Given mass/molar mass
No. Of moles?
(Using given number of atoms/molecules/particles?)
No. Of moles,n= Given no of atomes/molecules/particles ÷ 6.022×10^23
No. Of moles?
(Using given volume of gas)
No of moles,n= given vol of gas at STP÷ 22.7L
Find number of moles present in 16g O2
16g÷32g= 0.5
Find number of moles present in 12.044×10^23 atoms of sulphur.
12.044×10^23÷6.022×10^23= 2
Find number of moles present in 12.044×10^23 atoms of sulphur.
12.044×10^23÷6.022×10^23= 2