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 4.30 x1016 atoms of He, 4.00 g/mol

4.30 x 1016 atoms He 1 mol 4.00 g = 2.86 x 10 -7

6.022 x 1023 atoms 1 molgrams He

How many atoms are in two moles of mercury?

1. 1.204 x 1022

2. 1.204 x 1023

3. 1.204 x 1024

4. 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. 6.022 x 1026

2. 1.806 x 1024

3. 1.204 x 1024

4. 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 = 1.983Boron – 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 H94.06% = 94.06 g O

5.94g H 1 mol = 5.89 mol H1.0079g 5.88

94.06g O 1 mole =5.88 mol O 15.999 g 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. Round of both numbersb. Round off 1.67, and use 1.00 as it isc. Multiply both subscipts by 3d. 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 =6.07Molar 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 = 22.990gMolar mass of Cl = 35.453 gMolar 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. Decompositionb. Combustionc. 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. AgNO3 + AlCl3 AgCl + Al(NO3)2

2. C8H18 + O2 CO2 + H2O3. Ca(ClO3)2 CaCl2 +O2

4. NaOH + HCl NaCl + HOH5. CaO + H2O Ca(OH)2

6. Ni(ClO3)2 NiCl2 + O2

a. Decompositionb. Combustionc. Synthesisd. Single Replacemente. Double Replacement – Precipitationf. Double Replacement – Acid/Base

Match each equation with the correct type of reaction.

1. AgNO3 + AlCl3 AgCl + Al(NO3)2

2. C8H18 + O2 CO2 + H2O3. Ca(ClO3)2 CaCl2 +O2

4. NaOH + HCl NaCl + HOH5. CaO + H2O Ca(OH)2

6. Ni(ClO3)2 NiCl2 + O2

a. Decomposition (6), (3)b. Combustion ( 2)c. Synthesis (5)d. Single Replacemente. Double Replacement – Precipitation (1)f. Double Replacement – Acid/Base (4)