Year 11 Chemistry Exam Questions Part 2: Metals
Posted on May 4, 2017 by DJ Kim
Year 11 Chemistry Exam Questions: Metals
Assess your depth of knowledge and understanding for the Year 11 Chemistry ‘Metals’ module. Work through these Metals exam questions below.
You can refer to the Glossary of Key Words for the meaning of any verbs in the questions.
Detailed answers are provided at the bottom of this blog post.
How did Thomson’s “plum pudding” model of the atom differ from Rutherford’s “planetary” model?
Why was Newland’s periodic table not widely accepted by society?
Explain why the date of widespread use of aluminium is much later than that of copper.
What is an alloy? Give an example.
What is the definition of ionisation energy?
Explain the trend in first ionisation energies of elements going down a group in the periodic table.
Explain why elements with atomic numbers 3, 11, 19, 37 and 55 have lower first ionisation energies than the elements immediately preceding them (atomic numbers 2, 10, 18, 36, 54).
Write the formulae of the following acids:
(i) Sulfuric acid
(ii) Nitric acid
(iii) Hydrochloric acid
(iv) Phosphoric acid
Write balanced chemical equations for the following reactions:
(i) Sodium metal in water
(ii) Magnesium metal in nitric acid
(iii) Zn metal with oxygen gas
(iv) Copper with silver nitrate
Which of the following metals (Ag, Zn, K, Ca, Mg, Au, and Cu) will react with cold water and which is the most reactive?
Write dissociation equations for the following compounds:
(i) HNO3(aq) →
(ii) H3PO4(aq) →
(iii) CaCl2(aq) →
Consider the following reaction:
2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)
(i) Write the full ionic equation
(ii) Write the net ionic equation
(iii) Write the half equations
Write a balanced net ionic equation for the reaction that occurs when iron(III) chloride solution mixes with sodium hydroxide solution to form a reddish precipitate.
For the following reaction, write the oxidation half equation and the reduction half equation.
Mg(s) + 2AgNO3(aq) → Mg(NO3)2(aq) + 2Ag(s)
Find the molar mass of the following compounds:
(i) Magnesium chloride
(ii) Lithium bromide
(iii) Potassium sulfate
(iv) Ammonium nitrate
Aluminium chloride can be formed from the reaction of aluminium and chlorine gas.
(i) Write a balanced chemical equation for this reaction.
(ii) What volume of chlorine gas is required to react with 2.4 x 1022 atoms of aluminium at STP?
(iii) What mass of aluminium chloride will be produced?
Consider the following reaction:
2NO(g) + O2(g) → 2NO2(g)
If 2.15 g of nitrogen monoxide reacts with 4.23 g of oxygen, how many grams of nitrogen dioxide will form and how many grams of excess reactant will remain?
An unknown compound is found to consist of carbon, hydrogen, and oxygen. The analysis revealed that the compound contains 64.8% carbon and 13.6% hydrogen by mass. The approximate molar mass was 200 g mol-1. What is its molecular formula?
- Thomson’s plum pudding model depicted electrons embedded randomly in a positively charged sphere. Rutherford’s planetary model portrays the atom as mostly empty space with a tiny, dense, positively charged core called the nucleus, which is surrounded by the electrons at a very far distance away from the nucleus.
- Newland’s periodic table placed elements such as oxygen and iron into the same chemical family. Oxygen is a non-metal and iron is a metal, hence they do not share the similar properties and cannot be grouped together.
- Aluminium is more reactive than copper, thus aluminium forms more stable compounds compared to copper compounds. As a consequence, more energy is required for the extraction of aluminium to allow it to be used. Aluminium could not be extracted by means of heating (smelting) like copper, it required the advent of electrolysis which provided greater energy. Hence, aluminium was discovered and used at a later date than copper.
- An alloy is a mixture containing a metal and one or more elements. An example of an alloy is tungsten steel, which consists of iron, chromium and tungsten.
- The energy needed to remove one electron from one mole of gaseous atoms of the element.
- As you go down a group, first ionisation energy decreases as there is decreasing effective nuclear charge due to outer electrons being farther from the nucleus. Thus the outermost electrons are less attracted to the nucleus and less energy would be required to remove an electron.
- Elements with atomic numbers 3, 11, 19, 37 and 55 belong in Group 1 and elements with atomic numbers 2, 10, 18, 36, 54 belong in Group 18 (VIII). Going across the periodic table left to right, first ionisation energy increases. This is due to an increased effective nuclear charge, resulting in greater attraction of the outer electron to the nucleus that would require more energy to remove it.
- (i) Sulfuric acid H2SO4
(ii) Nitric acid HNO3
(iii) Hydrochloric acid HCl
(iv) Phosphoric acid H3PO4
- (i) 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
(ii) Mg(s) + 2HNO3(aq) → Mg(NO3)2(aq) + H2(g)
(iii) 2Zn(s) + O2(g) → 2ZnO(s)
(iv) Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s)
- K and Ca. K is most reactive
- (i) HNO3(aq) → H+(aq) + NO3–(aq)
(ii) H3PO4(aq) → 3H+(aq) + PO43-(aq)
(iii) CaCl2(aq) → Ca2+(aq) + 2Cl–(aq)
- (i) 2Al(s) + 6H+(aq) + 3SO42-(aq) → 2Al3+(aq) + 3SO42-(aq) + 3H2(g)
(ii) 2Al(s) + 6H+(aq) → 2Al3+(aq) + 3H2(g)
(iii) 2Al(s) → 2Al3+(aq) + 6e–
6H+(aq) + 6e– → 3H2(g)
- FeCl3(aq) + 3NaOH(aq) → Fe(OH)3(s) + 3NaCl(aq)
- 2Mg(s) → Mg2+(aq) + 2e–
2Ag+(aq) + 2e– → 2Ag(s)
- (i) MgCl2: 24.31 + 35.45 x 2 = 93.21 g mol-1
(ii) LiBr: 6.941 + 79.90 = 86.84 g mol-1
(iii) K2SO4: 39.10 x 2 + 32.07 + 16.00 x 4 = 174.3 g mol-1
(iv) NH4NO3: 14.01 x 2 + 1.008 x 4 + 16.00 x 3 = 80.05 g mol-1
- (i) 2Al(s) + 3Cl2(g) → 2AlCl3(s)
(ii) n(Al) = particles/Avogadro’s number
= 2.4 x 1022/6.022 x 1023 mol
= 0.03985…moln(Cl2) = 3/2 x n(Al)
= 0.05978… molV(Cl2) = n(Cl2) x MV
= 0.05978… x 22.71
= 1.4 L (2 sig. fig.)(iii) n(AlCl3) = n(Al) = 0.03985…mol
m(AlCl3) = n(AlCl3) x mm(AlCl3)
= 0.03985… x (26.98 + 35.45 x 3) g
= 5.3 g (2 sig. fig.)
- n(NO) = m(NO)/mm(NO)
= 2.15/(14.01 + 16.00) mol
= 0.07164…moln(O2) = m(O2)/mm(O2)
= 4.23/(16.00 x 2) mol
= 0.1321875 molNO is the limiting reactantn(NO2) = n(NO) = 0.07164…mol (1:1 ratio)
m(NO2) = n (NO2) x mm(NO2)
= 0.07164… x (14.01 + 16.00 x 2) g
= 3.30 g (3 sig. fig.)
O2 is the excess reactant
n(O2 needed to react) = n(NO)/2
= 0.07164…/ 2 mol
= 0.035821… mol
n(O2 remaining) = n(O2 total) – n(O2 needed to react)
= 0.1321875 – 0.035821… mol
= 0.096366… mol
m(O2 remaining) = n (O2 remaining) x mm(O2)
= 0.096366… mol x (16.00 x 2)
= 3.08 g (3 sig. fig.)
- Use the table below to calculate the ratio between C, H and O atoms.
C H O Mass (assuming 100 g) 64.8 g 13.6 g 21.6 g Moles (mol) 5.3955… 13.49206… 1.35 Ratio 4 10 1
Empirical formula: C4H10O
No. of empirical units = molar mass / empirical mass
= 2.69 (round up to 3)
Molecular formula: 3 x C4H10O = C12H30O3
Read more: If you missed part 1 of this guide you can visit Year 11 Chemistry Exam Questions Part 1 Chemical Earth.
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