Module 4 Further Reading

7/24/2019 Module 4 Further Reading

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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?


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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


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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.


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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


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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.


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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


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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.


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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


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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?


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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


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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?


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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

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Module 4 Further Reading