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Physci 3 > C8 > Flashcards

C8 Flashcards

1
Q

Explain how the law of conservation of mass is related to the balancing of chemical equations.

A

The number of atoms in reactants and in products must be equal, with mass conserved-> no atoms (mass) created or destroyed in the reaction.

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2
Q

A conical flask containing 200 mL of hydrochloric acid solution is placed on a precision balance.
0.01 mol of zinc powder is added to the flask and a loose plug of cotton wool is inserted into the top of the flask to prevent any liquid escaping.
The following chemical reaction occurs:
Zn(๐‘ ) + 2HCl(๐‘Ž๐‘ž) โŸถ ZnCl2 (aq) + H2 (๐‘”)

a) What measurement(s) must be made so that the mass of hydrogen produced can be determined? (1 mark)

A

The decrease in mass of the flask and its contents as a
result of the reaction.
OR Measure mass of flask and its contents at beginning
and at end of reaction.

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3
Q

Refer to the balanced equation, reproduced below, when answering this question:
Zn(๐‘ ) + 2HCl(๐‘Ž๐‘ž) โŸถ ZnCl2 (aq) + H2 (๐‘”)

Why is there no change in the amount of hydrogen produced when adding more than 0.12
mol of zinc?

A

From the equation n(Zn) = ยฝ n(HCl) there is a
maximum of 0.24 mol of HCl(aq) present, so there is a
maximum of 0.12 mol of Zn that can react. If more Zn
is added, there is no more acid to react
Alternate if 0.14 mol of Zn is added, it needs 0.28
mol of acid, which is not available.

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4
Q

3.49 g of XHCO3 was initially dissolved in water and made up to 250.0 mL of solution; 25.0 mL samples of the hydrogen carbonate solution were used in each titration.
c) Show that the molar mass of XHCO3 is approximately 84 g molโ€“1.

info: 4.16 x 10^-3 mol of XHCO3 was reacted in each titration.

A

n(XHCO3) in 25 mL = 4.16 x 10^-3 mol
n(XHCO3) in 250 mL = 4.16 x 10^-3 mol x 10
= 4.16 x 10-2 mol
n = m/M
โ‡’ M =m/n
M(XHCO3) = 3.49/0.0416
= 83.87 g mol^-1

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5
Q

The equations listed below produce gaseous water.
Equation 1: H2O(๐‘™) โ†’ H2O(๐‘”)
Equation 2: 2H2(๐‘”) + O2(๐‘”) โ†’ 2H2O(๐‘”)
Why is the change summarised in equation 2 classified as a chemical change, whereas the
change summarised in equation 1 is not? (2 marks)

A

Eqn 2: new chemical compound is formed in the products, therefore called a chemical change
EQN 1- no new chemicals formed, only a change of state (liquid to gas), therefore called a physical change.

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6
Q

One form of barium hydroxide is the anhydrous form, Ba(OH)2. Mr(Ba(OH)2) = 171.32
(c) Calculate the total number of ions present in this amount of dissolved barium hydroxide

A

Ba(OH)2 (aq)-> Ba^2++ 2OH^-
0.387 molโ€”โ€“> 0.387 mol + 0.774 mol
therefore total n(ions)= 0.387+0.774= 1.161 mol
actual number of ions= 1.161 x 6.02 x 10^23 in total
= 6.99 x 10^23

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7
Q

A student used titration techniques to determine the concentration of some ethanoic acid
solution, CH3COOH(๐‘Ž๐‘ž)
. The reaction occurring in the titration is:
NaOH(๐‘Ž๐‘ž) + CH3COOH(๐‘Ž๐‘ž) โ†’ CH3COONa(๐‘Ž๐‘ž) + H2O(๐‘™)
The following readings were obtained:
๏‚ท20.0 mL of sodium hydroxide solution was used in each titration.
๏‚ท The concentration of sodium hydroxide solution was 0.962 mol L-1
.
๏‚ท The average volume of ethanoic acid used for neutralisation was 23.9 mL.
(a) Show that about 0.02 mol of ethanoic acid was used in each titration.

A

n(NaOH) = c x v = (0.962 x 0.0200)
= 0.0192 mol

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8
Q

A student used titration techniques to determine the concentration of some ethanoic acid
solution, CH3COOH(๐‘Ž๐‘ž)
. The reaction occurring in the titration is:
NaOH(๐‘Ž๐‘ž) + CH3COOH(๐‘Ž๐‘ž) โ†’ CH3COONa(๐‘Ž๐‘ž) + H2O(๐‘™)
The following readings were obtained:
๏‚ท 20.0 mL of sodium hydroxide solution was used in each titration.
๏‚ท The concentration of sodium hydroxide solution was 0.962 mol L-1
.
๏‚ท The average volume of ethanoic acid used for neutralisation was 23.9 mL.
(b) Calculate the concentration of the sample of ethanoic acid.
info: n(CH3COOH)= 0.0192 mol

A

conc of CH3COOH= n/v
=0.0192/0.0239
= 0.805 mol/L

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9
Q

One hydrated form of barium hydroxide, Ba(OH)2 . xH2O, is analysed by heating a sample until all the water of crystallisation has been evaporated. The following results are obtained:
๏‚ท Mass of solid before heating = 4.57 g
๏‚ท Mass of solid after heating = 2.48 g
Use this information to calculate the value of the integer x.

A

m(Ba(OH)2)โ€”โ€“> m(Ba(OH)2) + m(XH2O)
= 4.57 g โ€”โ€”> 2.48 + 2.09
n(Ba(OH)2)โ€”โ€“> N(XH2O) (AFTER HEATING)
= 2.48/ 171.32 : 2.09/ 18.016
= 0.0145 : 0.116
= 1:8
= 1:X
X=8

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10
Q

A sample of solid sodium carbonate, Na2CO3, is contaminated with sodium chloride, NaCl. To
determine the sampleโ€™s purity, the following experimental technique is used:

The titration reaction occurring is:
Na2CO3(๐‘Ž๐‘ž) + 2HCl(๐‘Ž๐‘ž) โ†’ 2NaCl(๐‘Ž๐‘ž) + CO2(๐‘Ž๐‘ž) + H2O(๐‘™)

23.5 mL of a 0.102 mol L-1 solution of hydrochloric acid solution is neutralised by 25.0 mL of
the solution made from the impure sodium carbonate.

Show that the mass of sodium carbonate in 250.0 mL of the solution used in the titration
is about 1.3 g. (Mr (Na2CO3 = 105.99)

A

c(Na2CO3)= 0.04794 M
V= 0.25 L
n(Na2CO3)= 0.04794 X 0.25
= 0.011985 mol
m(Na2CO3)= Nm
= 0.011985 X (22.99 X 2 + 16.01 + 16 X 3)
= 1.32g

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11
Q

If 0.415 moles of A was reacted with B, how many moles of C will be produced?

A

n(C) = ยฝ x 0.415 = 0.208 mol

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12
Q

A real reaction, where aluminium is burnt in oxygen to form aluminium oxide is shown below.
4Al(s) + 3O2(g) โ†’ 2Al2O3(s)
c) What mass of Al2O3(s) will form from 89 g of Al(s)?

A

n(Al) = 89
26.98
= 3.30 mol
n(Al2O3) = ยฝ x 3.30 = 1.65 mol
M(Al2O3) = (26.98 x 2) + (3 x 16.00) = 101.96
m(Al2O3) = 1.65 x 101.96 = 168 g

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13
Q

A chemical found in the skin of green potatoes was analysed. The results were as follows:
Hydrogen = 2.24 %
Carbon = 26.7 %
Oxygen = 71.1 %.
a) Use this information to show that the compound has an empirical formula, HCO2

A

H C O

n 1.008/2.24 26.7/12.01 71.1/16
2.22 2.22 4.44

Ratio 2.22/2.22 2.22/2.22 4.44/2.22
1 1 2
โˆด molecular formula is HCO2

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14
Q

b) The compound was further analysed to find its molar mass, which is 90.03 g mol-1
. Use this
information to show that the chemical compound is oxalic acid, H2C2O4
Empirical formula= HCO2

A

Empirical mass = 1.008 + 12.01 + 2(16) = 45.02
Ratio = 90.03
45.02
= 2,
2 x HCO2 = H2C2O4

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15
Q

Oxalic acid is considered a weak acid.
i. What is meant by โ€˜a weak acidโ€™? (1 MARK)

A

Q23c i) Does not fully dissociate/ionise in solution 1

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16
Q

ii. Suggest a pH of a dilute oxalic acid solution.
(1 MARK)

A

range 3-6 accepted

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17
Q

c) Calculate the volume of water needed to be added to 1.0 L of 0.25 mol L-1 HCl to create a
0.20 mol L-1 solution.

A

C1V1=C2V2
0.25 x 1.0 = 0.20 x V
V = 0.25
0.20
= 1.25 L
โˆด add 0.25 L to the 1.0 L solution to reduce its
concentration to 0.20 mol L^-1

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18
Q

By referring to the Brรธnsted-Lowry model of acids and bases, explain the relationship that
exists between an acid like HBr when reacted with a base like NaOH. (3 marks)

A

The acid donates a proton (H+) to the base which accepts it.
HBr was the proton donor and NaOH was the proton
acceptor.

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19
Q

c) A 1.0 mol L-1 solution of HBr was also added to a strip of magnesium metal.
ii. From your equation, outline two (2) observations from the reaction.
(1 mark)

A

Bubbles of gas
Mg strip dissolved / disappears

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20
Q

CuO + HBr -> CuBr2 + H2O
If you had 10 moles of hydrogen bromide, how many moles of water would be produced?

A

5 moles of h20

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21
Q

N2 + O2 -> NO
If you had 4 moles of nitrogen, how many moles of nitrogen monoxide would be produced?

A

8 moles

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22
Q

๐‘๐‘Ž๐ถ๐‘™ + ๐ต๐‘’๐น2 โ†’ ๐ต๐‘’๐ถ๐‘™2 + ๐‘๐‘Ž๐น
If 5 moles of sodium fluoride is produced,
How many moles of BeCl2 would be produced?

A

2.5 moles

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23
Q
  1. How much sulphuric acid is required to completely react with 3.7 moles of lithium? How much
    salt would be produced?
A

3.7 mol of Li requires 3.7/2= 1.85 mol of H2SO4
produces 1.85 mol of the salt

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24
Q

Nitric acid reacts with calcium hydroxide. How many moles of each are required to produce 4.5
moles of water?

A

4.5 mol of water needs 2.23 mol of Ca(OH)2 and 4.5 mol of HNO3

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25
Q
  1. Iron(II) phosphate reacts with 2.6 mol sodium sulfate. How much of each product is formed?
A

forms 1.73 mol Na3PO4
and 2.6 mol FeSO4

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26
Q

How much aluminium nitrate is needed to react completely with 0.0123 mol of tin(IV) oxide.
What are the products and how much of each is formed?

A

needs 0.0123/3 *4= 0.0164 mol of Al(NO3)3
produces 0.0123/3 *2= 0.0082 mol Al(NO3)3
and 0.0123 mol Sn(NO3)4

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27
Q

What does a strong acid indicate? Give some examples.

A
  • dissociate almost fully (assume completely)
  • has a strong push to dissociate and a weak push to bring back tgt
  • HCL, H2SO4, HNO3
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28
Q

What does a weak acid indicate? Give some examples.

A
  • partly dissociate, majority remains undissociated
  • acetic acid
29
Q

What is the pH for an acid?

A

pH 1-7

30
Q

What is an acid?

A

H+ donors (proton donors)

31
Q

What is a base?

A

H- acceptors (proton acceptors)

32
Q

What is the pH for a base?

A

pH 7-14

33
Q

What are some examples of weak and strong base?

A

Weak: ammonia
Strong: sodium hydroxide

34
Q

What are soluble bases called?

A

Alkalis

35
Q

How do we โ€œfeelโ€ whether it is a base or an acid?

A

Bases= slippery
Acids= sour

36
Q

How do the two indicators indicate acidic or basic?

A

Litmus:
Blue litmus turns red under acidic conditions
Red litmus turns blue under basic conditions

Phenolphthalein
Colourless under acidic conditions
Pink under basic conditions

37
Q

What is the relationship between strength and concentration of an acid?

A

A strong acid will completely ionize but can still be dilute.

38
Q

When does the equivalence point occur in a titration?

A
  • when the pH is 7
  • at neutralisation
  • when the moles match
39
Q

Any acid that donates more than one H+ ion is ______

A

Any acid that donates more than one H+ ion is polyprotic

40
Q

Monoprotic

A

An acid that donates one H+ ion

41
Q

Fizzy water is described as a dilute, weak acid. Explain the meaning of dilute and weak in this context. (2 marks)

A

Dilute: small amount of dissolved CO2 per L (unit
volume) of solution
Weak: the acid only partially breaks down/ionises,
producing H+
(aq), in water.

42
Q

Zn(๐‘ ) + 2HCl(๐‘Ž๐‘ž) โŸถ ZnCl2 (aq) + H2 (๐‘”)

b) Calculate the expected mass of hydrogen produced when 0.075 mol of zinc is added.

A

n(Zn) = 0.075 mol
From equation n(H2) = n(Zn) = 0.075 mol
m(H2) = 0.075 x 2.016
= 0.151 g
= 0.151 x 1000 mg
= 151 mg

43
Q

A conical flask containing 200 mL of hydrochloric acid solution is placed on a precision balance.
0.01 mol of zinc powder is added to the flask and a loose plug of cotton wool is inserted into the top of the flask to prevent any liquid escaping.
The following chemical reaction occurs:
Zn(๐‘ ) + 2HCl(๐‘Ž๐‘ž) โŸถ ZnCl2 (aq) + H2 (๐‘”)

Refer to the balanced equation, reproduced below, when answering this question:
Zn(๐‘ ) + 2HCl(๐‘Ž๐‘ž) โŸถ ZnCl2 (aq) + H2 (๐‘”)
Why is there no change in the amount of hydrogen produced when adding more than 0.12
mol of zinc?

A

From the equation n(Zn) = ยฝ n(HCl) there is a
maximum of 0.24 mol of HCl(aq) present, so there is a
maximum of 0.12 mol of Zn that can react. If more Zn
is added, there is no more acid to react
Alternate if 0.14 mol of Zn is added, it needs 0.28
mol of acid, which is not available.

44
Q

Zn(๐‘ ) + 2HCl(๐‘Ž๐‘ž) โŸถ ZnCl2 (aq) + H2 (๐‘”)

As the reaction proceeds, will the pH of the contents of the flask increase, stay the same
or decrease? Justify your choice.

A

pH will increase
The solution is becoming less acidic as acidic H2 ions are released as gas/ closer to neutral
(pH 7) as the acid is reacting.

45
Q

A laboratory technician diluted a 1.42 mol Lโ€“1 solution of sulfuric acid solution to prepare
2.00 L of dilute acid. The concentration of the diluted acid was found to be 0.103 mol Lโ€“1
.
a) Determine the volume of the more concentrated acid used for the preparation.

A

n(dilute H2SO4) prepared = 2 x 0.103 = 0.206 mol
๏œn(conc H2SO4) needed = 0.206
c =n/V
โ‡’ V = n/c
V = 0.206/1.42 = 145 mL
Using dilution: n(dilute H2SO4) = n(conc H2SO4)
c1V1 = c2V2
2 x 0.103 = 1.42 x V2
V2 = 0.206/1.42
= 145 mL

46
Q

2XHCO3 (๐‘Ž๐‘ž) + H2SO4 (๐‘Ž๐‘ž) โŸถ X2SO4 (๐‘Ž๐‘ž) + 2H2O(๐‘™) + 2CO2 (๐‘”)
M(XHCO3)= 83.87
Identify the identity of X.

A

M(HCO3 ) = 1.008 + 12 + 48 = 61.008 g mol 1
M(X) = 83.87 61.008 = 22.9 g mol 1
Hence X is sodium

47
Q

Inorganic sources of dissolved carbon in seawater include the hydrogen carbonate ion and other
carbon compounds.
(a) Calculate the concentration, in mol L-1
, of the hydrogen carbonate ion, HCO3 (๐‘Ž๐‘ž) in a sample
of seawater if the concentration of this ion is 140 mg/L.

A

M(HCO3)= (1.008+12.01+48)
= 61.02 gmol^-1
concentration= 140 mgL^-1
= 0.140 gL^-1
= 0.140/61.02 mol L^-1
= 0.00229 molL^-1

48
Q

Carbonic acid, H2CO3(aq), is one of the carbon compounds present in seawater.
(b) Carbonic acid is a weak and dilute acid. Explain what is meant by these terms: weak acid and dilute acid

A

weak acid= low % dissociation in solution, e.g. HX(aq)โ€“5%โ€“> H^+(g)+X^-(aq)
dilute acid= small number of mole of acid dissolved per litre, e.g. c(HX)= 0.050 molL^-1

49
Q

450 mL of hydrochloric acid solution of concentration 6.25 mol L-1 was spilt on the floor of a
laboratory. Powdered limestone powder, which contains calcium carbonate, CaCO3 was sprinkled
on the acid to neutralise it.
The following reaction occurs:
CaCO3(๐‘ ) + 2HCl(๐‘Ž๐‘ž) โ†’ CaCl2(๐‘Ž๐‘ž) + H2O(๐‘™)+CO2(๐‘”)
(Mr(CaCO3) = 100.1)
Info: around 3 mol of HCL was spilt
The limestone powder contains 92.5% calcium carbonate. Other chemicals in the powder do not
react with the acid.
(b) Calculate the mass of limestone powder needed to neutralise the acid spilt.

A

n(CaCO3)= 2.81 * 1/2 = 1.41 mol
mass of pure CaCo3= n * M
= 1.41 * 100.1
= 141 g
if 92.5% pure, mass of limestone= 141 * 100/92.5
= 152 g

50
Q

Solid samples of three white compounds are labelled as A, B, and C.
The following tests were performed on the compounds to identify them.
๏‚ท The solubility of each compound was tested and pH measured.
๏‚ท A small amount of each solid was placed in separate test tubes and a dilute solution of hydrochloric acid, HCl(aq) was added to each.
The observations are summarised in the table below:
A B C
Solubility Does not dissolve Does not dissolve Dissolves to give a
solution of pH 14
Reaction with acid Solid disappears;
gas evolved
Solid disappears Solid disappears
Use the following list of compounds to identify A, B, and C, giving reasons for your choice:
๏‚ท aluminium oxide, Al2O3,
๏‚ท sodium chloride, NaCl,
๏‚ท sodium hydroxide, NaOH,
๏‚ท calcium carbonate, CaCO3.
(One chemical from the list will not be included in your answer.)
The inclusion of chemical equations is not necessary, but may help in your explanation.
(4 marks)

A

A= CaCO3 (s) which is insoluble in water but reacts with dilute acid to release CO2, ie CaCO3(s) + 2HCL(aq)-> CaCl2(aq) + CO2(g) + H20(l)
B= Al2O3(s) which is insoluble in water but dissolves in acid with no gas formed.
ie Al2O3(s) + 6HCl(aq)-> 2AlCl3(aq) + 3H2O(l)
C= NaOH(s) dissolves readily in water forming an alkaline (basic solution due to OH^-(aq) therefore pH=14
NaCl not one of the solids listed because it would dissolve in water to form a neutral solution (pH=7)

51
Q
  1. Acid rain is caused by gaseous pollutants such as nitrogen oxides and sulphur dioxide being released into the atmosphere. These gases react with water vapour to form very dilute solutions that may contain sulphur- and nitrogen-based acids. Over many years acid rain slowly damages buildings, monuments and unprotected metal, as well as having other very undesirable environmental effects.

Assume sulphuric acid is the key active ingredient in acid rain and that calcium carbonate is a major constituent of marble.

Show the chemical equation for the reaction of sulphuric acid and calcium carbonate and use this to explain why a marble monument is eroded by acid rain (3 marks)

A

H2SO4(aq) + CaCO3(s) -> CaSO4 + CO2+ H2O (l)
- acid rain reacts with solid CaCO3 statue
- none of the products are solid
- statue erodes

52
Q
  1. Acetic (ethanoic) acid, CH3COOH(aq), is present in vinegar as the principal component other than water. The balanced equation for the reaction between acetic acid in solution and sodium hydroxide solution is:

CH3COOH(aq) + NaOH(aq) ๏ƒ  CH3COONa(aq) + H2O(l)

In preparation for finding the concentration of acetic acid in a particular brand of vinegar, a solution of sodium hydroxide was made by dissolving 5.00g of sodium hydroxide in 250.0mL of water.

b. What steps would you take to turn the solution prepared in part a. into a solution with concentration of 0.0500 mol L-1? Express the relevant mathematical equation that allows you to perform this process. (2 marks)

A

c1v1=c2v2
0.500 * 0.250 = 0.0500 v2
-> v2= 2.5 total volume
add 2.25 L water to the solution

53
Q

a. What is pH a measure of? How can we use a phenolphthalein indicator to distinguish between an acidic or a basic solution? (2 marks)

A

pH is a measure of concentration of H+ ions
Colorless in acid, turns pink in base

54
Q

b. Explain the difference between concentration and strength of an acid.
(2 marks)

A

Strength is ability do dissociate into ions:
- strong= full dissociate
- weak= partial dissociation
Concentration is amount of acid dissolved in solution

55
Q

c. Use your answers in part a and b to explain how a strong acid can under certain conditions have the same pH as a weak acid. (2 marks)

A

Strong and concentrated acid has high H+ concentration, low pH
If diluted, the concentration of H+ ions will lower and pH rise to around 5

56
Q
  1. Whilst stirring, potassium hydroxide solution is slowly added to dilute sulphuric acid. The pH of the resulting solution is continuously measured.
    i. Use the Lowry-Bronsted theory of acids and bases to explain what is happening in this case. (2 marks
A

H2SO4 donates proton (acid) while KOH accepts protons (base)

57
Q
  1. A 7.532 g of hydrated magnesium chloride, MgCl2ยทxH2O, was heated until its mass no longer changed. The (now anhydrous) solid was then weighed and found to be 3.527 g. Determine the degree of hydration of MgCl2ยทxH2O. (3 marks)
A

n(MgCol2)= m(MgCl2)/ M(MgCl2)= 3.527/ (24.31 + 2*35.5)= 0.037 mol
n(H2O)= m(H2O)/ M(H2O)= 7.532-3.527/ 2.016+16= 0.223 mol
0.037:0.223
1:6 ratio
therefore MgCl2.6H2O ie 6

58
Q

Relative atomic mass

A

The weighted average of the mass of an atom relative to 1/12 of a c-12 atom

59
Q

Oven cleaner

A

pH 14
basic

60
Q

Hair remover

A

pH 13
basic

61
Q

Household ammonia

A

pH 11
basic

62
Q

Auto battery acid

A

pH 1
acidic

63
Q

Lemon juice

A

pH 2.3
acidic

64
Q

black coffee

A

pH 5.5
acidic

65
Q

milk

A

pH 6.5
acidic

66
Q

human blood

A

pH 7.3
basic

67
Q

bleach

A

pH 9.5
basic

68
Q

Write a balanced chemical equation to represent the formation of solid aluminium chloride, AlCl3, from solid aluminium and gaseous chlorine.

A

3Cl2(g) + 2Al(s) โ†’ 2AlCl3(s)

69
Q

Determine the concentration of K+ and SO4^2- ions in a 0.13 mol L^-1 solution of K2SO4.

A

k+: 0.26
so4: 0.13

Physci 3 (6 decks)

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