OBJECTIVES: DEFINE ISOTOPE AND GIVE THE MASS NUMBER AND NUMBER NEUTRONS FOR A SP

Question

OBJECTIVES: DEFINE ISOTOPE AND GIVE THE MASS NUMBER AND NUMBER NEUTRONS FOR A SPECIFIC ISOTOPE. CALCULATE THE ATOMIC WEIGHT NOF AN ELEMENT FROM ISOTOPIC ABUNDANCES EXPLAIN THE DIFFERENCE BETWEEN THE ATOMIC NUMBER AND THE ATOMIC WEIGHT OF AN ELEMENT AND FIND THIS INFORMATION FOR ANY ELEMENT Give the information...^4_2He^56_26Fe^27_13Al element: _______ element: _______ element: _______ # protons: ____ atomic #: _____ # protons: ____ # neutrons: _____ mass #: _____ # electrons: ______^40_20Ca^20_10Ne^14_6C element: _______ element: _______ element: _______ # protons: _______ atomic #:______ #protons: ______ average # neutrons: ____ # electrons: ______ # electrons: ______ atomic weight: _____ atomic weight: _______ avg. # neutrons: _______^19_9F^1_1H^2_1 element: _______ element: _______ element: _______ atomic #: _____ atomic #: ______ #protons: _______ mass #: _______ # electrons: _______ # electrons: _______ # electrons: ______ # neutrons: _______ # neutrons: ___

Explanation / Answer

1.Element(He) = Helium

Atomic no(on the subscript) = no of protons=2

Mass number(superscript) = no of protons + no of neutrons

No of neutrons = mass no - no of protons = 4-2 =2

no. of neutrons = 2

2.Element(Fe) = Iron

Atomic no = 26

Mass no = 56

3.Element(Al) = Aluminum

Atomic no = no of protons = 13

No of protons = no of electrons in a neutral element

No of electron = 13

4.Element(Ca) = Calcium

Protons = atomic no = 20

Neutrons = mass no - protons = 40-20 = 20

Atomic weight = 40u

5.Element( Ne) = Neon

Atomic no = 10

Electrons = 10

6.Element(C) = Carbon

Proton= atomic no=6

Electrons = 6

Neutrons = mass no - protons = 14-6 = 8

7.Element (F) = Fluorine

atomic no = 9

Mass no = 19

Electrons = atomic no(for neutral element) =9

8.Element(H) = Hydrogen

Atomic no = 1

Electrons = 1

Neutrons = 1-1=0

9.Element(H) = Hydrogen

Protons = atomic no = 1

Electrons = Protons = 1

Neutrons = Mass no- protons = 2-1 =1

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