Pearson Edexcel International GCSE (9–1) · CHEMISTRY · Unit
Principles of Chemistry
991 practice questions
Topics covered
- States of Matter102 practice questions
- Elements, Compounds and Mixtures69 practice questions
- Atomic Structure108 practice questions
- The Periodic Table103 practice questions
- Chemical Formulae, Equations and Calculations Part 119 practice questions
- Chemical Formulae, Equations and Calculations Part 2144 practice questions
- Ionic Bonding143 practice questions
- Covalent Bonding68 practice questions
- Metallic Bonding19 practice questions
- Electrolysis198 practice questions
Practice questions
211–220 of 407 · Easy
Question 211
1 mark · Easy
Answer and see the solution →Write the balanced symbol equation for the reaction between potassium hydroxide and nitric acid.
Question 212
3 marks · Easy
Answer and see the solution →A student uses 28.0 cm³ of hydrochloric acid to neutralise 25.0 cm³ of sodium hydroxide solution with a concentration of 0.120 mol/dm³. The equation for the reaction is:
(a)[1 mark]Calculate the moles of sodium hydroxide used.
(b)[1 mark]State how many moles of hydrochloric acid react with this amount, using the equation.
(c)[1 mark]Calculate the concentration of hydrochloric acid in mol/dm³.
Question 213
2 marks · Easy
Answer and see the solution →A titration uses 20.0 cm³ of 0.175 mol/dm³ nitric acid ( [3m[33mHNO[0m[1;31m[0m[0;32m3[0m ) solution for neutralisation.
Calculate the number of moles of nitric acid used. Show your steps.
Question 214
1 mark · Easy
Answer and see the solution →What is the formula used to calculate the volume (in dm³) of a solution, given the number of moles and concentration in mol/dm³?
Question 215
3 marks · Easy
Answer and see the solution →Potassium hydroxide reacts with nitric acid according to the equation:
25.0 cm³ of 0.200 mol/dm³ potassium hydroxide solution is mixed with 25.0 cm³ of 0.200 mol/dm³ nitric acid.
(a)[1 mark]Calculate the number of moles of potassium hydroxide in the mixture.
(b)[1 mark]Calculate the number of moles of nitric acid in the mixture.
(c)[1 mark]State the simplest whole-number ratio of reactant volumes needed for complete neutralisation according to the equation.
Question 216
2 marks · Easy
Answer and see the solution →Calculate the volume in cm³ of 0.150 mol/dm³ sulfuric acid solution that contains 0.0300 mol of sulfuric acid.
Question 217
2 marks · Easy
Answer and see the solution →Explain why, when rearranging the concentration equation to solve for volume, the number of moles must be divided by concentration, and not the other way around.
Question 218
3 marks · Easy
Answer and see the solution →The following table shows the reaction between sodium sulfate and barium chloride:
Reactant Volume (cm³) Concentration (mol/dm³) Sodium sulfate 50.0 0.100 Barium chloride excess — Use the data given to answer the following:
(a)[1 mark]Calculate the number of moles of sodium sulfate used.
(b)[1 mark]Use the balanced equation to state how many moles of barium sulfate will be formed.
(c)[1 mark]State the minimum volume of sodium sulfate solution (in cm³) needed to ensure 0.0100 mol of barium sulfate forms, using the same concentration.
Question 219
1 mark · Easy
Answer and see the solution →State the information you must know to calculate the volume of a solution required to react with a known amount of another reactant.
Question 220
3 marks · Easy
Answer and see the solution →20.0 cm³ of a solution contains 0.00400 mol of a substance.
(a)[1 mark]Calculate the volume of the solution in dm³.
(b)[1 mark]Calculate the concentration of the solution in mol/dm³.
(c)[1 mark]State the formula used for this calculation.