ch3

Question 1

DALTON’S THEORY

Answer 1

• tiny particles of matter
• chemical reactions is the rearrangement of atoms
• similar to each other for each element and different from atoms of other elements

Question 2

subatomic particles

Answer 2

-inside of an atom

Question 3

PROTONS

Answer 3

has positive charge (+)
-like charges it repel, want to be apart from each other

always the atomic #
Ex: Na has 11 atomic #
Al has 13
S has 16

Question 4

ELECTRONS

Answer 4

has negative charge (-)
-like charges it repels, wants to be apart from each other

atomic # - charge
ex: Na–> 11 - 0= 11
Al–> 13 - (+3)= 10
S–> 16-(-2)= 18

Question 5

NEUTRONS

Answer 5

have no charge
- unlike charges attracts each other

Mass # - atomic #
ex: Na –> 23-11= 12
Al –> 27-13= 14
S–> 33-16=17

Question 6

RUTHERFORD’S GOLD-FOIL EXPERIMENT

Answer 6

he came up with some positively charge particles have small and dense nucleus in the atom that deflects positive particles that comes close

Question 7

ELECTRON CONFIGURATION

Answer 7

distribution of electrons of an atom or molecule in atomic or molecular orbitals
s- 1 orbital that holds up to 2 electrons
p- 3 orbitals holds up to 3 electrons
d- 5 orbitals holds up to 10 electrons
f- 7 orbital holds up to 14electrons

Question 8

VALENCE ELECTRONS

Answer 8

are electrons that form chemical bonds with other elements

ex: Ca has 2electrons because Ca is in G2

Question 9

One atomic mass unit (amu)

Answer 9

mass exactly equal to one-twelfth the mass of one carbon-12 atom that has six protons and six neutrons

Question 10

average atomic mass

Answer 10

weighted average of the masses of the naturally occurring isotopes of the element; the mass of the atom in atomic mass units
Average Atomic Mass = ∑ (fractional abundance of isotope n) × (mass of isotope n)

Question 11

Isotopes

Answer 11

atoms with identical atomic numbers but different mass numbers (that is,
same number of protons but different numbers of neutrons)
Average atomic mass of carbon = (0.09890)(12.00000 amu) + (0.0110)(13.00335 amu)
= 12.0107 amu

Question 12

ATOMIC WEIGHT

Answer 12

weighted average of the masses of all the naturally occurring isotopes of an element
Calculation: Boron- 10 Boron-11
19.19% / 100= 0.1919 80.09%/100=0.8009
(10) (0.1919) + (11)(0.8009)=10.81

Question 13

MASS NUMBER

Answer 13

mass number = A = no. of protons + no. of electrons

ex: GA–> A= 31 + 40 =71

Question 14

NUCLEAR SYMBOL