Chemistry B Quizzes

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Chemistry B Quizzes
The number of atoms or molecules in
exactly 1.000 mole of a substance, and
is equal to 6.022 x 1023
The number of atoms or molecules in
exactly 1.000 mole of a substance, and
is equal to 6.022 x 1023
Avogrado’s Number
atoms of the same element that have
different numbers of neutrons, which
means that atoms of the same
element can have different masses
Atoms of the same element that have
different numbers of neutrons, which
means that atoms of the same
element can have different masses
1. Isotope
The SI unit for amount of a
substance.
The SI unit for amount of a
substance.
Mole
a chemical formula that shows the
simplest ratio of atoms in a compound
a chemical formula that shows the
simplest ratio of atoms in a compound
Emperical Formula
Which will have the greater number of
ions, 1 mol of nickel (II) or 1 mol of
copper (I)?
Which will have the greater number of
ions, 1 mol of nickel (II) or 1 mol of
copper (I)?
They will be the same they both are 1 mole, so
each would have 6.022 x 1023 ions
Without making a calculation,is 1.11
mol Pt more of less than 6.022 x 1023
atoms?
Without making a calculation,is 1.11
mol Pt more of less than 6.022 x 1023
atoms?
It would be more because 1 mole would =
6.022 x 1023, and there are 1.11 moles Pt,
so it would be larger than avogadro’s
number
Find the mass in grams in
4.30 x 1016 atoms of He, 4.00 g/mol
Find the mass in grams in
16
4.30 x10 atoms of He, 4.00 g/mol
4.30 x 1016 atoms He 1 mol
6.022 x 1023 atoms
4.00 g
1 mol
= 2.86 x 10 -7
grams He
How many atoms are in two moles
of mercury?
1.
2.
3.
4.
1.204 x 1022
1.204 x 1023
1.204 x 1024
6.022 x 1023
How many atoms are in two moles
of mercury?
1.204 x 1024
How many atoms are in a mole of
barium chloride, BaCl2?
1.
2.
3.
4.
6.022 x 1026
1.806 x 1024
1.204 x 1024
6.022 x 1023
How many molecules are in a mole
of barium chloride, BaCl2?
1. 6.022 x 1023
How many grams of cobalt (58.9332
g/mol) are in 4.76 mol of the element?
An element has one isotope with an atomic
mass of 27.94 amu. Another isotope has
an atomic mass of 28.96 amu. The average
atomic mass of the element is 28.02 amu.
The isotope that makes up the larger
percent of a sample of the element is the
isotope with a mass of
27.94 amu
28.96 amu
28.02 amu
An element has one isotope with an atomic
mass of 27.94 amu. Another isotope has
an atomic mass of 28.96 amu. The average
atomic mass of the element is 28.02 amu.
The isotope that makes up the larger
percent of a sample of the element is the
isotope with a mass of
27.94 amu, because it is closes to average atomic
mass of 28.02 amu
What is the molar mass of C6H12O6?
What is the molar mass of C6H12O6?
180.16 g
Simply add up the masses from the periodic
table
Boron has an isotope with a mass
of 10.013 amu that makes up
19.8% of all boron. Its other
isotope has a mass of 11.009 amu
and makes up 80.2%. What is the
average atomic mass of boron.
Boron has an isotope with a mass
of 10.013 amu that makes up
19.8% of all boron. Its other
isotope has a mass of 11.009 amu
and makes up 80.2%. What is the
average atomic mass of boron.
Boron -10 = (.198) x 10.013 = 1.983
Boron – 11 = (.802) x 11.009 = 8.83
1.983+8.83 =10.81 amu
How many moles of hydrogen are
in one mole of C2H6?
How many moles of hydrogen are
in one mole of C2H6?
6
How many hydrogen atoms are
present in one formula unit of
ammonium hydrogen phosphate,
(NH4)2HPO4?
How many hydrogen atoms are
present in one formula unit of
ammonium hydrogen phosphate,
(NH4)2HPO4?
8
A compound is 5.94% H and
94.06% O. What is its empirical
formula?
A compound is 5.94% H and
94.06% O. What is its empirical
formula?
5.94% = 5.94 g H
94.06% = 94.06 g O
5.94g H
94.06g O
1 mol
1.0079g
1 mole
15.999 g
= 5.89 mol H
5.88
=5.88 mol O
5.88
HO
The first set of calculations in determining
an empirical formula from masses of
elements produces the subscripts of 1.67
and 1.00. What do you need to do to find
the actual subscripts?
a.
b.
c.
d.
Round of both numbers
Round off 1.67, and use 1.00 as it is
Multiply both subscipts by 3
Multiply both subsripts by 6
The first set of calculations in determining
an empirical formula from masses of
elements produces the subscripts of 1.67
and 1.00. What do you need to do to find
the actual subscripts?
a. Round off 1.67, and use 1.00 as it is
A compound’s empirical formula is CH. If
the formula mass is 79.12 g/mol, what is
the molecular formula?
A compound’s empirical formula is CH. If
the molar mass is 79.12 g/mol, what is the
molecular formula?
Formula mass of compound = 79.12 g
Molar mass of CH = 13.019
Molar mass of compound = 79.12 =6.07
Molar mass of CH
13.019
Multiply each by 6 to find the molecular formula = C6H6
What is the percentage of chlorine in NaCl?
What is the percentage of chlorine in NaCl?
Molar mass of Na = 22.990g
Molar mass of Cl = 35.453 g
Molar mass of NaCl = 58.443g
% of sodium = 22.990/58.443 x 100 = 39%
% of chlorine = 35.453/583443 x 100 = 61%
Match each equation with the correct
type of reaction.
1. Ca(ClO3)2  CaCl2 +O2
2. CaO + H2O  Ca(OH)2
3. C8H18 + O2  CO2 + H2O
a. Decomposition
b. Combustion
c. Synthesis
Match each equation with the correct
type of reaction.
1. Ca(ClO3)2  CaCl2 +O2
2. CaO + H2O  Ca(OH)2
3. C8H18 + O2  CO2 + H2O
a. Decomposition (1)
b. Combustion(3)
c. Synthesis ( 2)
Balance the following equation:
ZnS + O2  ZnO + SO2
Balance the following equation:
2ZnS + 3O2  2ZnO + 2SO2
Match each equation with the correct
type of reaction.
1.
2.
3.
4.
5.
6.
AgNO3 + AlCl3  AgCl + Al(NO3)2
C8H18 + O2  CO2 + H2O
Ca(ClO3)2  CaCl2 +O2
NaOH + HCl  NaCl + HOH
CaO + H2O  Ca(OH)2
Ni(ClO3)2  NiCl2 + O2
a.
b.
c.
d.
e.
f.
Decomposition
Combustion
Synthesis
Single Replacement
Double Replacement – Precipitation
Double Replacement – Acid/Base
Match each equation with the correct
type of reaction.
1.
2.
3.
4.
5.
6.
AgNO3 + AlCl3  AgCl + Al(NO3)2
C8H18 + O2  CO2 + H2O
Ca(ClO3)2  CaCl2 +O2
NaOH + HCl  NaCl + HOH
CaO + H2O  Ca(OH)2
Ni(ClO3)2  NiCl2 + O2
a.
b.
c.
d.
e.
f.
Decomposition (6), (3)
Combustion ( 2)
Synthesis (5)
Single Replacement
Double Replacement – Precipitation (1)
Double Replacement – Acid/Base (4)

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