0620_w24_qp_42
A paper of Chemistry, 0620
Questions:
6
Year:
2024
Paper:
4
Variant:
2

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Sulfur forms two chlorides, P and Q. Chloride P has the formula S2Cl 2. Chloride Q has the formula SCl 2. Both chlorides are covalently bonded and have low melting points. Suggest, in terms of attraction between particles, why these chlorides have low melting points. Chloride P, S2Cl 2, forms when sulfur reacts with chlorine. Write the symbol equation for this reaction. Complete the dot-and-cross diagram in of a molecule of chloride Q, SCl 2. Show outer electrons only. Cl S Cl Chloride P is converted to chloride Q by reaction with chlorine in a closed system. The reversible reaction reaches an equilibrium. S2Cl 2+ Cl 22SCl 2P Q The forward reaction is exothermic. Suggest two changes to the conditions which will result in a decrease in the concentration of chloride Q at equilibrium. The rate of the forward reaction in is determined by collision theory. The rate of reaction depends upon two factors: ● the frequency of collisions between particles ● the proportion of collisions which have energy greater than or equal to the activation energy. Define the term activation energy. Give the symbol for activation energy. Complete Table 3.1 to show the effect, if any, when the conditions are changed. Use only the words increases, decreases or no change. Table 3.1 change to conditions effect on the frequency of collisions between particles effect on the proportion of collisions which have energy greater than or equal to the activation energy concentration of chlorine is increased temperature is increased a catalyst is added The reaction of chloride P with chlorine is a redox reaction. S2Cl 2+ Cl 22SCl 2P Q The oxidation number of Cl in chloride P and chloride Q is –1. Use oxidation numbers to explain why: ● sulfur is oxidised in the forward reaction ● chlorine is oxidised in the reverse reaction.
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Silver bromide, AgBr, is made when aqueous silver ethanoate, CH3COOAg, is added to aqueous sodium bromide, NaBr. The equation for the reaction is shown in equation 1. equation 1 CH3COOAg + NaBr → CH3COONa + AgBr The method includes the following steps. step 1 Add 200.0 cm3 of 0.0500 mol / dm3 CH3COOAg to a beaker. This volume contains 0.0100 mol of Ag+ ions. step 2 Add 50.0 cm3 of aqueous NaBr. This volume contains 0.0100 mol of Br – ions. A precipitate forms. step 3 Filter the mixture. step 4 Dry the solid residue until all the water is removed. step 5 Record the mass of the dry residue. Complete the ionic equation for the reaction by adding the missing state symbols. Ag+( ) + Br –( ) → AgBr( ) Name a different aqueous silver salt which could be used in step 1. Use the information in step 2 to calculate the concentration of aqueous NaBr. concentration = mol / dm3 State the colour of the precipitate which forms in step 2. Use the information in step 1, step 2 and equation 1 to determine the number of moles of AgBr formed. Use this value to calculate the mass of AgBr formed. number of moles of AgBr = mass of AgBr = g Name the salt dissolved in the filtrate in step 3. The recorded mass of the dry residue in step 5 is greater than the mass calculated in because a step is missing from the procedure. Suggest the missing step. Name the substance responsible for the greater mass of the dry residue. Barium sulfate can be made by the same method but with different aqueous solutions. Suggest two aqueous solutions which can be added together to make barium sulfate. and Write the balanced symbol equation for this reaction.
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