Atomic structure
Electron configuration
Elements, compounds & mixtures
Mol principles
Reacting masses
100

Define an isotope and state one way isotopes of the same element are similar and one way they differ.

Same element (same protons); different mass numbers (different neutrons).

100

Write the electron configurations for Na and Cl atoms, then for Na⁺ and Cl⁻ ions.

Na 2,8,1 → Na⁺ 2,8; Cl 2,8,7 → Cl⁻ 2,8,8.

100

Define element, compound, and mixture; give one example of each.

Element—one type of atom (e.g., O₂); Compound—chemically bonded atoms (e.g., H₂O); Mixture—physically combined (e.g., air).

100

Calculate Mr of CaCO₃. (Ar Ca = 40, C = 12, O = 16)

40 + 12 + (3×16) = 100.

100

How many moles are in 18 g of water? (Ar H = 1, O = 16)

 n = m/Mr = 18/18 = 1.00 mol.

200

Element X has mass number 35 and atomic number 17. For neutral X and for X⁻, state the number of protons, neutrons, and electrons.

Neutral X: p=17, n=18, e=17. 

X⁻: p=17, n=18, e=18.

200

An atom has configuration 2,8,3. State its period, (main) group, and whether it is a metal or non-metal.

Period 3, Group 13 (III), metal (Al).

200

State the separation method to obtain:
(a) salt from salty water,
(b) water from salty water,
(c) sand from a sand–salt–water mixture.

(a) Evaporation/crystallisation, (b) Simple distillation, (c) Filtration.

200

Balance: Al + O₂ → Al₂O₃

4Al + 3O₂ → 2Al₂O₃.

200

Magnesium burns in oxygen: 2Mg + O₂ → 2MgO. If 6.0 g of Mg react completely, calculate the mass of MgO formed. (Ar Mg = 24, O = 16)

n(Mg)=6.0/24=0.25 mol → n(MgO)=0.25 mol → mass=0.25×40=10.0 g.

300

Chlorine exists as Cl-35 (75%) and Cl-37 (25%). Calculate the relative atomic mass, Aᵣ(Cl), to 1 d.p.

(35×0.75)+(37×0.25)=35.5.

300

Which is larger: Mg or Mg²⁺? Explain briefly in terms of shells/attraction.

Mg is larger; Mg²⁺ has lost its outer shell and has stronger attraction nucleus→electrons.

300

In paper chromatography, a dye travels 4.0 cm while the solvent front travels 8.0 cm. Calculate Rf.

Rf = 4.0/8.0 = 0.50.

300

Calculate % by mass of calcium in CaCO₃. (Ar Ca = 40, C = 12, O = 16)

(40/100)×100% = 40%.

300

Heptane burns completely: C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O. What minimum mass of O₂ is needed to burn 30.0 g of heptane? (Ar C = 12, H = 1, O = 16)

n(C₇H₁₆)=30/100=0.30 mol → n(O₂)=11×0.30=3.3 mol → mass=3.3×32=105.6 g.

400

For the ion ²⁷₁₃Al³⁺, give: protons, neutrons, electrons.

p=13, n=14, e=10.

400

Explain why reactivity increases down Group 1 but decreases down Group 7.

G1: more shells → more shielding → outer e⁻ lost more easily (↑reactivity). G7: more shells → harder to gain e⁻ (↓reactivity).

400

You have sand, salt, and water mixed. Outline the sequence of methods to separate and obtain dry samples of both sand and salt.

Filter (sand residue); Evaporate/Crystallise filtrate to get salt; Dry both solids.

400

A compound contains C 40.0 g, H 6.7 g, O 53.3 g. Find the empirical formula. (Ar C = 12, H = 1, O = 16)

Moles: C 40/12=3.33, H 6.7/1=6.7, O 53.3/16=3.33 → ratio 1:2:1 → CH₂O.

400

2H₂ + O₂ → 2H₂O. If 20.0 g of O₂ is used up, what mass of H₂O is formed? (Ar H = 1, O = 16)

n(O₂)=20.0/32=0.625 mol 

n(H₂O)=2×0.625=1.25 mol → mass = 1.25×18 = 22.5 g.

500

An ion Y²⁻ has electron configuration 2,8,8. If the nucleus contains 18 neutrons, identify Y and give its mass number.

Y atom has 16 e⁻ → S; mass no. = p+n = 16+18=34 → ³⁴S.

500

Atoms A and B have electron configurations A: 2,8,2 and B: 2,8,7. State the ion each is most likely to form.

A → A²⁺, B → B⁻

500

You must separate and collect pure liquids from the miscible solution:
Ethanol (bp 78 °C) + water (bp 100 °C)
State the best separation method and explain how it works.

Fractional distillation.

Substances with different boiling points travel up the fractionating column and boil off at different temperatures. The vapor is collected and then condensed in the condenser.

500

A compound has empirical formula C₃H₈O and Mr = 120. Determine its molecular formula. (Ar C = 12, H = 1, O = 16)

Empirical mass = 60 → factor 120/60=2 → C₆H₁₆O₂.

500

4Al + 3O₂ → 2Al₂O₃. What is the minimum mass of O₂ needed to react with 25.0 g of Al? (Ar Al = 27, O = 16)

n(Al)=25.0/27=0.9259 mol → n(O₂)=(3/4)×0.9259=0.694 mol → mass = 0.694×32 = 22.2 g (3 s.f.).

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