7/24/2019 Module 4 Further Reading
1/13
7/24/2019 Module 4 Further Reading
2/13
Chemistry: Building Blocks of the World
Module 4, Topic 2 Practice Questions
Calculate the number of O atoms in 3.50 moles of O atoms.
How many moles of Ag atoms are present if you have 4.63 x 1022Ag atoms?
How many individual Cs atoms are in 0.250 moles of Cs atoms?
7/24/2019 Module 4 Further Reading
3/13
Chemistry: Building Blocks of the World
Module 4, Topic 2 Practice Questions (Answers)
Calculate the number of O atoms in 3.50 moles of O atoms.
Number of atoms = n x NA
Number of atoms = 3.50 x 6.022 x 1023
Number of atoms = 2.11 x 1024O atoms
How many moles of Ag atoms are present if you have 4.63 x 1022Ag atoms?
n =
n =
n = 0.0769 mol
How many individual Cs atoms are in 0.250 moles of Cs atoms?
Number of atoms = n x NA
Number of atoms = 0.250 x 6.022 x 1023
Number of atoms = 1.51 x 1023Cs atoms
7/24/2019 Module 4 Further Reading
4/13
Chemistry: Building Blocks of the World
Module 4, Topic 4 Practice Questions
Calculate the number of moles of iron (Fe) in 7.24 grams of pure iron metal.
Calculate the mass of neon (Ne) present in 3.25 moles of neon atoms.
Calculate the number of moles of carbon (C) C in 11.3 grams of pure carbon.
Calculate the mass of gold (Au) present in 0.0955 moles of gold atoms.
7/24/2019 Module 4 Further Reading
5/13
Chemistry: Building Blocks of the World
Module 4, Topic 4 Practice Questions (Answers)
Calculate the number of moles of iron (Fe) in 7.24 grams of pure iron metal.
n =
n =
n = 0.130 mol
Calculate the mass of neon (Ne) present in 3.25 moles of neon atoms.
m = n x M
m = 3.25 x 20.18
m = 65.6 grams
Calculate the number of moles of carbon (C) C in 11.3 grams of pure carbon.
n =
n =
n = 0.941 mol
Calculate the mass of gold (Au) present in 0.0955 moles of gold atoms.
m = n x M
m = 0.0955 x 197.0
m = 18.8 grams
7/24/2019 Module 4 Further Reading
6/13
Chemistry: Building Blocks of the World
Module 4, Topic 5 Practice Questions
Calculate the number of moles of NaCl in 18.5 grams of solid NaCl.
Calculate the mass of methane (CH4) present in 2.27 moles of methane gas.
Calculate the number of moles of N2O4in 50.0 grams of N2O4gas.
Calculate the mass of glucose (C6H12O6) present in 0.0324 moles of glucoseatoms.
7/24/2019 Module 4 Further Reading
7/13
Chemistry: Building Blocks of the World
Module 4, Topic 5 Practice Questions (Answers)
Calculate the number of moles of NaCl in 18.5 grams of solid NaCl.
n =
M = 22.99 + 35.45 = 58.44 g/mol
n =
n = 0.317 mol
Calculate the mass of methane (CH4) present in 2.27 moles of methane gas.
m = n x M M = 12.01 + (4 x 1.008) = 16.042 g/mol
m = 2.27 x 16.042
m = 36.4 grams
Calculate the number of moles of N2O4in 50.0 grams of N2O4gas.
n =
M = (2 x 14.01) + (4 x 16.00) = 92.02 g/mol
n =
n = 0.543 mol
Calculate the mass of glucose (C6H12O6) present in 0.0324 moles of glucoseatoms.Answer: Mass = 5.84 gramsm = n x M M = (12.01 x 6) + (1.008 x 12) + (16.00 x 6)
= 180.156 g/mol
m = 0.0324 x 180.156
m = 5.84 grams
7/24/2019 Module 4 Further Reading
8/13
Chemistry: Building Blocks of the World
Module 4, Topic 6 Practice Questions
Calculate the number of moles of NaCl in 2.20 litres of 1.85 M NaCl solution.
Calculate the concentration of a NaOH solution prepared by dissolving 0.0253moles of NaOH in 75.0 mL of water.
Calculate the volume of 0.357 M KF required to provide 0.0500 moles of KF.
7/24/2019 Module 4 Further Reading
9/13
Chemistry: Building Blocks of the World
Module 4, Topic 6 Practice Questions (Answers)
Calculate the number of moles of NaCl in 2.20 litres of 1.85 M NaCl solution.
n = c x v
n = 1.85 x 2.20
n = 4.07 mol
Calculate the concentration of a NaOH solution prepared by dissolving 0.0253moles of NaOH in 75.0 mL of water.
c =
c =
c = 0.337 M or mol/L
Calculate the volume of 0.357 M KF required to provide 0.0500 moles of KF.
v =
v =
n = 0.140 L or 140 mL
7/24/2019 Module 4 Further Reading
10/13
Chemistry: Building Blocks of the World
Module 4, Topic 7 Practice Questions
What is the concentration of a 250 mL solution containing 0.970 grams ofNaNO3?
Calculate the mass of potassium chloride (KCl) in 22.47 mL of 0.124 M KClsolution.
What volume of water is required to make a 0.120 M Li2CO3solution from 2.40grams of Li2CO3?
What mass of NaCl is required to produce 50.0 mL of 0.0500 M NaCl solution?
7/24/2019 Module 4 Further Reading
11/13
Chemistry: Building Blocks of the World
Module 4, Topic 7 Practice Questions (Answers)
What is the concentration of a 250 mL solution containing 0.970 grams ofNaNO3?
M = 22.99 + 14.01 + (3 x 16.00) = 85.00 g/mol
n =
=
= 0.0114 mol
c =
=
= 0.0456 M or mol/L
Calculate the mass of potassium chloride (KCl) in 22.47 mL of 0.124 M KClsolution.
M = 39.10 + 35.45 = 74.55 g/mol
n = c x v = 0.124 x 0.02247 = 0.00279 mol
m = n x M = 0.00279 x 74.55 = 0.208 grams
What volume of water is required to make a 0.120 M Li2CO3solution from 2.40grams of Li2CO3?M = (2 x 6.939) + 12.01 + (3 x 16.00) = 73.888 g/mol
n =
=
= 0.0325 mol
v =
=
= 0.271 L or 271 mL
What mass of NaCl is required to produce 50.0 mL of 0.0500 M NaCl solution?
M = 22.99 + 35.45 = 58.44 g/mol
n = c x v = 0.0500 x 0.0500 = 0.00250 mol
m = n x M = 0.00250 x 58.44 = 0.146 grams
7/24/2019 Module 4 Further Reading
12/13
Chemistry: Building Blocks of the World
Module 4, Topic 9 Practice Questions
Question 1:
N2(g)+ 3F2(g)2NF3(g)
What mass of fluorine gas (F2) is required to produce 850 grams of NF3?
Question 2:
2HCl(aq)+ Na2CO3(aq)CO2(g)+ H2O(l)+ 2NaCl(aq)
Calculate the mass of CO2gas that could be produced when 7.24 mL of 1.22 Mhydrochloric acid (HCl) is mixed with sodium carbonate (Na2CO3).
Question 3:
H2SO4(aq)+ NaOH(aq)H2O(l)+ Na2SO4(aq)
Balance the above equation and then determine the concentration of thesulphuric acid solution (H2SO4) if 20.00 mL of sulphuric acid solution reactedcompletely with 19.83 mL of 0.212 M NaOH?
7/24/2019 Module 4 Further Reading
13/13
Chemistry: Building Blocks of the World
Module 4, Topic 9 Practice Questions (Answers)
Question 1:
N2(g)+ 3F2(g)2NF3(g)
What mass of fluorine gas (F2) is required to produce 850 grams of NF3?M(NF3) = 14.01 + (3 x 19.00) = 71.01 g/mol, M(F2) = (2 x 19.00) = 38.00 g/mol
n(NF3) =
=
= 11.97 mol
n(F2) = 11.97 x
= 17.96 mol
m(F2) = n x M = 17.96 x 38.00 = 682 grams
Question 2:
2HCl(aq)+ Na2CO3(aq)CO2(g)+ H2O(l)+ 2NaCl(aq)
Calculate the mass of CO2gas that could be produced when 7.24 mL of 1.22 Mhydrochloric acid (HCl) is mixed with sodium carbonate (Na2CO3).M(CO2) = 12.01 + (2 x 16.00) = 44.01 g/mol
n(HCl) = c x v = 1.22 x 0.00724 = 0.00883 mol
n(CO2) = 0.00883 x
= 0.00442 mol
m(CO2) = n x M = 0.00442 x 44.01 = 0.194 grams
Question 3:
H2SO4(aq)+ 2NaOH(aq)2H2O(l)+ Na2SO4(aq)
Balance the above equation and then determine the concentration of thesulphuric acid solution (H2SO4) if 20.00 mL of sulphuric acid solution reactedcompletely with 19.83 mL of 0.212 M NaOH?
n(NaOH) = c x v = 0.212 x 0.01983 = 0.00420 mol
n(H2SO4) = 0.00420 x
= 0.00210 mol
c(H2SO4) =
=
= 0.105 M or mol/L