Use a copy of the Periodic Table where you need one. One mole of any gas occupies 24 dm3 at room temperature and pressure.
Dimethylglyoxime has the chemical formula (CH3CNOH)2. How many atoms does one molecule of dimethylglyoxime contain?
Source: 2023 GESS Sec 3 End-of-Year P1 Q19
The formula of iridium sulfate is Ir2(SO4)3 and that of zinc hypochlorite is Zn(ClO)2. What is the formula of iridium hypochlorite?
Source: 2023 GESS Sec 3 End-of-Year P1 Q20
A piece of zinc strip is burnt in oxygen to form zinc oxide. Which of the following chemical equations is correct?
Source: 2023 XMS Sec 3 WA2 Q3
Which chemical equation is not balanced?
Source: 2023 XMS Sec 3 WA2 Q5
Mercury hydroxide reacts with phosphoric acid in the reaction below. w Hg(OH)2 + x H3PO4 → y Hg3(PO4)2 + z H2O. Which values of w, x, y and z correctly balance the equation?
| w | x | y | z | |
|---|---|---|---|---|
| A | 2 | 3 | 1 | 4 |
| B | 3 | 3 | 2 | 7 |
| C | 3 | 2 | 1 | 6 |
| D | 6 | 4 | 2 | 11 |
Source: 2023 GESS Sec 3 End-of-Year P1 Q23
Which compound contains the highest percentage of nitrogen by mass?
Source: 2023 GESS Sec 3 End-of-Year P1 Q24
A new element was discovered. The relative abundance of its isotopes is shown below. What is its relative atomic mass?
| nucleon number of isotope | 63 | 65 | 67 |
|---|---|---|---|
| relative abundance / % | 68.54 | 29.51 | 1.95 |
Source: 2023 GESS Sec 3 End-of-Year P1 Q26
What is the empirical formula of a compound consisting of 4.8 g of carbon and 0.8 g of hydrogen?
Source: 2023 GESS Sec 3 End-of-Year P1 Q27
A sample of nitrogen of mass 5.00 g is completely converted into an oxide of nitrogen. The mass of the oxide formed is 19.30 g. What is the empirical formula of the oxide?
Source: SJI 2020 Year 3 End-of-Year P1 Q15
Balance the following chemical equations.
Source: 2023 XMS Sec 3 WA2 Q6
Write the chemical equations for the following reactions.
Source: 2022 XMS Sec 3 WA2 B3(a)
A student prepares barium sulfate by mixing two solutions, barium nitrate and potassium sulfate. Barium sulfate is collected as the residue.
Source: SJI 2020 Year 3 End-of-Year P2 A5(a)
Carbon can be reacted with boron to form compound R. A sample of R was found to contain 78.6% boron by mass.
Source: 2024 CHS Sec 3 End-of-Year Section B Q3(b)
25.0 cm3 of 0.40 mol/dm3 sulfuric acid was reacted with exactly 35 cm3 of sodium hydroxide. H2SO4 + 2NaOH → Na2SO4 + 2H2O
Source: 2023 GESS Sec 3 End-of-Year P2 B11(a) and (b)
Wollastonite, CaSiO3, can be produced from the reaction between calcium phosphate and silicon dioxide. 2Ca3(PO4)2 + 6SiO2 → P4O10 + 6CaSiO3. 100 g of calcium phosphate, Ca3(PO4)2, was used in this reaction. The actual mass of wollastonite produced was 87 g.
Source: 2023 GESS Sec 3 End-of-Year P2 A9
D16[1]
The formula has two units of CH3CNOH: 4 carbon, 8 hydrogen, 2 nitrogen and 2 oxygen atoms, 16 atoms in total.
DIr(ClO)3[1]
Ir2(SO4)3 shows the ions are Ir3+ and SO42−. Zn(ClO)2 shows the ions are Zn2+ and ClO−. So Ir3+ and ClO− give Ir(ClO)3.
A2Zn + O2 → 2ZnO[1]
Oxygen is diatomic, O2, and zinc oxide is ZnO (Zn2+ and O2−).
AC6H12O6 + 3O2 → 6CO2 + 6H2O[1]
In A there are 12 oxygen atoms on the left (6 + 6) but 18 on the right (12 + 6). It needs 6O2.
Cw = 3, x = 2, y = 1, z = 6[1]
There are 3 Hg on the right, so w = 3 and y = 1. The values in D fit for Hg, P and y, but z is wrong, as H and O are not balanced.
CNO[1]
Al(NO3)3: 19.7%. Mg3N2: 28%. NO: 1430 × 100 = 46.7%. NO2: 30.4%.
B63.67[1]
Ar = 68.54100 × 63 + 29.51100 × 65 + 1.95100 × 67 = 63.67
BCH2[1]
Moles: C = 4.812 = 0.4, H = 0.81 = 0.8. The ratio is 1 : 2, so CH2.
DN2O5[1]
Mass of oxygen = 19.30 − 5.00 = 14.30 g. Moles: N = 5.0014 = 0.357, O = 14.3016 = 0.894. The ratio is 1 : 2.5 = 2 : 5.
(1), 12, 8 [1]
(1), 5, 3, 4 [1]
2, 3, 6, (1) [1]
2Na + 2H2O → H2 + 2NaOH [1]
2HNO3 + CaO → Ca(NO3)2 + H2O [1]
2C3H6 + 9O2 → 6CO2 + 6H2O [1]
Ba(NO3)2 + K2SO4 → BaSO4 + 2KNO3 [1]
Ba2+ + SO42− → BaSO4 [1]
Masses in 100 g: B = 78.6 g and C = 21.4 g. [1]
Moles: B = 78.611 = 7.15 and C = 21.412 = 1.78. [1]
Ratio 7.15 ÷ 1.78 = 4.01, about 4 : 1. The empirical formula is B4C. [1]
Moles of sulfuric acid = 0.025 × 0.40 = 0.01 mol [1]
Moles of NaOH = 2 × 0.01 = 0.02 mol, so concentration = 0.020.035 = 0.571 mol/dm3 [1]
H+(aq) + OH−(aq) → H2O(l) [1]
Moles of Ca3(PO4)2 = 100310 = 0.32258 mol [1]
Ca3(PO4)2 : CaSiO3 = 2 : 6 = 1 : 3, so moles of CaSiO3 = 0.96774 mol [1]
Theoretical mass of CaSiO3 = 0.96774 × 116 = 112.26 g [1]
Percentage yield = 87112.26 × 100% = 77.5% [1]
A common mistake is to include the number in front of a formula when working out its Mr.