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The chemical formula for sodium tetraoxosulphate (VI) is \(Na_2SO_4\). It is also known as sodium sulfate. This compound does not have any water molecules as part of its crystal structure, making it anhydrous. So, it is an example of a crystalline substance that does not possess water of crystallization.

Phenolphthalein is an indicator that is commonly used in acid-base titrations. It is colorless in acidic solutions (pH below 8.2) and turns pink or magenta in alkaline solutions (pH above 10). In the titration of ethanoic acid (acetic acid) with sodium hydroxide (a strong base), the pH will change from acidic to alkaline as the acid is neutralized by the base

Vapour density = Relative molar mass / 2 Vapour density = 30 / 2 = 15 Therefore, the vapour density of the gaseous hydrocarbon is 15.

Thermal cracking is a process used in the petroleum industry to break down long-chain alkanes (hydrocarbons) into smaller, more useful hydrocarbons. It involves the decomposition of high-molecular-weight hydrocarbons, such as heavy crude oil fractions or long-chain alkanes, into lower-molecular-weight hydrocarbons, like smaller alkanes and alkenes. During thermal cracking, high temperatures and sometimes pressure are applied to the feedstock (such as crude oil or natural gas) to initiate the decomposition reactions.

Amount (in moles) = Molarity \times Volume Amount (in moles) = 0.20 mol/dm³ \times 100.0 cm³ / 1000 cm³/dm³ = 0.020 moles Now, we can calculate the mass using the formula: Mass = Amount \times Molar Mass Mass = 0.020 moles \times 40.0 g/mol = 0.80 g So, the correct answer is:0.80 g

1 Faraday of electricity is equal to the charge of 1 mole of electrons, which is 96,485 coulombs. The reduction of copper(II) ions in solution to copper metal requires 2 electrons per ion, so 1 Faraday of electricity will deposit 0.5 moles of copper. 1 Faraday * (2 electrons/Cu ion) / (1 mole electrons/96,485 coulombs) * (1 mole Cu/2 electrons) = 0.5 moles Cu

a. A transition element is one which has incompletely filled d-orbitals. bi. B ii. C c. Across a period(from left to right) there is a gradual increase in the number of protons in the nucleus/effective nuclear-charge. This increases the force of attraction between the nucleus and the electrons hence more energy is needed to remove the outermost electron. d. -methane -ethene -ethyne -ethanol -ethanoic acid e. -Alkanols are formed by replacing one hydrogen of water by an alkyl group -Alkyl groups have a positive inductive effect/ are electron releasing/electron donating -This increases the electron density of the oxygen f. -NaCl -CaCO3 -NH3 -C g. Geometric isomerism is the existence of two compounds with the same molecular formula but differ in the arrangement of groups attached to the carbon-containing double bond h. This is because constituents of water gases(CO and H2)are both combustible and give a higher calorific/heat value than producer gas i. It is defined as the heat change/released when 1 mole of a substance is burnt in excess oxygen. OR It is defined as the heat change/released when 1 mole of a substance is completely burnt in air ji. When the same quantity of electricity is passed through different electrolytes, the relative number of moles of the elements discharged is inversely proportional to the charges on the ions of the element. ii. 96500 C liberates 1.0 mol of Ag(s) 10920 C will liberate 10920/96500 =0.113 mole

ai. \(H_2SO_4 + 2NaOH → Na_2SO_4 + 2H_2O\) ii. Concentration of NaOH(\(C_n)\) \(C_n\) = 4/40= 0.1 mol/\(dm^3\) \(\frac{C_AV_A}{C_nV_n} = \frac{1}{2}\) \(C_A \times \frac{8}{0.1} \times 20 = \frac{1}{2}\) \(C_A =0.1 \times 20 \times \frac{1}{2} \times 8 = 0.125 mol /dm^3\) bi. -The volume of a gas is negligible compared to the total volume of the container -There are no forces of attraction or repulsion between molecules of the gas/collision between molecules is elastic. ii. The volume of a gas is not negligible because at high pressure the molecules occupy appreciable volume and at low temperature the intermolecular forces of attraction and repulsion become significant. ci. Glucose. ii. -Fructose -Galactose iii. hydrolysis iv. 2(\(C_6H_{10}O_5)_n + nH_2O → nC_{12}H_{12}O_{11}\) \(C_{12}H_{12}O_{11} + H_2O → 2C_6H_12O_6\) v. -maltase -diastase d. It Is the existence of compounds with the same molecular formula but different structural formulae/arrangement of the atoms/linkage of the atoms.

ai. %0 = 100 - (52.2+13.1) =34.7 C H O 52.2/12, 13.1/1, 34.7/16 4.35, 13.1, 2.168 4.35/2.168, 13.1/2.168, 2.168/2.168 2.00, 6.04, 1 empirical formula = \(C_2H_6O\) ii. 2 x vapour density = molecular mass 2x23=46 (\(C_2H_6O)_n\) = 46 [(12 x 2) + (1 x 6) + 16]n = 46 46n = 46 n = 1 molecular formula = \(C_2H_6O\) iii1. Hydroxyl iii2. b. 2(\(C_6H_{10}O_5)_n + nH_2O --diatase→ nC_{12}H_{22}O_{11}\) \(C_{12}H_{22}O_{11} + H_2O --maltase→ C_6H_{12}O_6 + C_6H_{12}O_6\) \(C_6H_{12}O_6 --zymase→ 2C_2H_5OH + 2CO_2\) ci. The reaction starts but the metal produces a layer of an oxide/aluminum oxide layer on its surface which is impermeable(to water) and stops the reaction. ii. Add dilute hydrochloric acid to excess aluminum filter off the excess metal leave the filtrate in a warm place/evaporate the filtrate to the point of crystallization/leave it in the sun. d. The flask is connected to the condenser and the solution is heated. The water vaporizes and is converted to(liquid)water in the condenser.

a. Water is added to the \(CaCl_2\): to form a solution. \(Na_2CO\), is then added to the solution to precipitate \(CaCO_3\),which is filtered, washed, and dried. b. Electricity is passed through a solution of CuSO4 using an impure copper as the anode and pure copper as the cathode. During electrolysis,the anode loses mass as copper dissolves and the cathode gains mass as copper is deposited. Anode: \(Cu_s → Cu^{2+}_{(aq)} + 2e^-\) Cathode: \(Cu^{2+}_{(aq)} + 2e^- → Cu_{(s)}\) ci. \(Cu_{(s)} + 4HNO_{3(aq)} →Cu(NO_3)_{2(aq)} + 2NO_{2(g)} + 2H_2O_{(l)}\) ii. -blue solution -(reddish)brown gas -gas bubbles/effervescence -metal dissolves/metal disappears iii. Because copper loses electrons/oxidation number increases. iv. Moles of acid=0.025x16= 0.4 mole Mole of \(NO_2\) = moles of acid/2 =0.2 mole Volume of gas (\(NO_2)\) at s.t.p. = 0.2 x 22.4 \(dm^{-3}\) = 4.48 mol \(dm^{-3}\) di. Pure HNO3 undergoes decomposition to \(NO_2\)(brown gas). This gas dissolves in the rest of the acid leaving it with that colour. ii. -2\(KNO_3 + H_2SO_4→K_2SO_4+2HNO_3\) - \(NaNO_3 + H_2SO_4→NaHSO_4 + HNO_3\)

ai. Iron dissolves readily in dilute \(H_2SO_4\) liberating hydrogen and forming \(FeSO_4\) while with Aluminum there is no reaction. ii. With aluminum there is no reaction(1)while iron dissolves in dilute \(HNO_3\) liberating hydrogen and forming \(Fe(NO_3)_2\) bi. \(S+O_2→SO_2\) ii. Vanadium(V) oxide/ platinized asbestos. iii. To favour the formation of more product/\(SO_3\). iv. The reaction is exothermic/high temperature favours the formation of more reactants/backward reaction. v. Catalysts are easily poisoned/easily damaged by dirt. vi. -SO2 damages buildings -damages living things -causes acid rain -irritates eyes, nose and throat -causes lung and respiratory diseases eg coughing, bronchitis, etc ci. Reagents -manganese(IV)oxide/MnO2 -concentrated HCI Conditions -heating Concentrated HCI(1) Bleaching powder/CaOCl2/KMnO4crystals Room temperature/heat is not required ii. -for bleaching wood pulp and textiles -for the production of plastics -for recovering of tin from its ore/extraction of tin -as a germicide and disinfectant/insecticide/in swimming pools -to sterilize/purification of water supply -for commercial manufacture of HCI -production of dyes and drugs -manufacture of solvents eg \(CCL_4\), \(CHCI_3\) di1. Hydrogen-fused calcium chloride/conc.tetraoxosulphate(VI)acid i2. Sulphur(IV)oxide-concentrated tetraoxosulphate(VI)acid/ anhydrous Calcium chloride i3. calcium oxide/ quick lime/silica gel ii1. copper and tin ii2. Copper and zinc

Phenolphthalein is a common laboratory indicator used to detect the presence of bases in a solution. In its basic form (pH > 8.3), it turns pink or purple, indicating the presence of a base.

ai. \(H_2SO_4 + 2NaOH → Na_2SO_4 + 2H_2O\) ii. Concentration of NaOH(\(C_n)\) \(C_n\) = 4/40= 0.1 mol/\(dm^3\) \(\frac{C_AV_A}{C_nV_n} = \frac{1}{2}\) \(C_A \times \frac{8}{0.1} \times 20 = \frac{1}{2}\) \(C_A =0.1 \times 20 \times \frac{1}{2} \times 8 = 0.125 mol /dm^3\) bi. -The volume of a gas is negligible compared to the total volume of the container -There are no forces of attraction or repulsion between molecules of the gas/collision between molecules is elastic. ii. The volume of a gas is not negligible because at high pressure the molecules occupy appreciable volume and at low temperature the intermolecular forces of attraction and repulsion become significant. ci. Glucose. ii. -Fructose -Galactose iii. hydrolysis iv. 2(\(C_6H_{10}O_5)_n + nH_2O → nC_{12}H_{12}O_{11}\) \(C_{12}H_{12}O_{11} + H_2O → 2C_6H_12O_6\) v. -maltase -diastase d. It Is the existence of compounds with the same molecular formula but different structural formulae/arrangement of the atoms/linkage of the atoms.

battle - /ˈbæt.əl/ raffle - /'ræf.əl/ latter - /'læt.ə/ bitter - /ˈbɪt.ə/ kettle - /'ket.əl/ Battle and kettle have the same rhyme scheme because they both end with the syllable /təl/.

Emphatic stress is the placement of greater stress on a particular word in a phrase or sentence. In this case, the word wife is emphasized to correct a mistake that has been made in the question. To find the appropriate question to the response: I bought my WIFE a new dress, you have to find the question that has a word that is different from the emphasized word/ the word in capital letters. The correct question is A. Did I buy my husband a new dress? No. I bought my WIFE a new dress.