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Pb(NO3)2 + H2S → PbS + 2HNO3 34 g H2S → 239 g pbs 5.02g pbs → (34)/(237g) χ 5.02 g = 0.714 g 34g → 22.4 dm3 =0.714 → (22.4)/34g) χ 100 = 4.70

Hygroscopic salts are compounds that readily absorb moisture from the surrounding air, and become sticky and remain solids. From the question, we were told the solid before exposure to air or placed on the table, weighed 5.0g and then after exposure to air for 8 hours, it was weighing 5.5g which suggests it has absorbed moisture. That accounts for the increase in weight. That substance is said to be a hygroscopic substance because they(hygroscopic substances) often behave in such manner.Unlike the deliquescent salts, they do not turn into solutions.

Solubility in moles per dm3 No of moles = (20)/(159.5) χ 0.125 moles in 100 g of moles = (0.125)/(100) χ 1000 = 1.25 moles

No expanded working has been attached for this question yet. The verified answer is Option B.

Na2C2O4 + CaCl2 → CaC2O4 + 2NaCl molar mass of Na2C2O4 = 134 molarity in 1000g H2O = (1.9)/(134) = 0.014 M but in 502 g mole = (0.014)/(50) χ 1000 = 0.28 M M1V1 = M2V2 V2 = (M1V1)/(M2) = (0.28 χ 50)/(0.1) = 140 = 1.40 x 10\(^2\) cm\(^3\)

Ma = (MbVb)/(Va) = (0.1 m χ 20 cm3)/(25cm) = 0.08 m M = (concentration in g/dm3)/(m) = (8)/(0.08) = 100 g

PH = -log (H+) (H+) = 10-4.348 = 4.0 χ 10-5

using \(C_1 V_1\) = \(C_2 V_2\) \(C_1\) = 11.0M, \(V_1\) = ?, \(C_2\) = 0.05M and \(V_2\) = 1\(dm^3\) \(V_1\) = \(\frac{(C_2 V_2)}{(C_2)}\) = \(\frac{(0.05 \times 1)}{(11)}\) = 0.005\(dm^3\) OPTION A is the closest value.

Ag+ + e- H = 108 g Ag = 0.11 10.8 Ag 4OH → O2 + 2H2O + 4e 4F ↔ 1mole O2 4F = 22.4 dm3 χ 0.1 = 0.56 dm3

0.1F ↔ 2.95 g Ni = 0.4F ↔ 11.8 g Ni No of moles = (11.8)/(58.2) = 0.20 moles

24mg = 1 mole of magnesium and 32g of O2 = mole of O2

No expanded working has been attached for this question yet. The verified answer is Option A.

No expanded working has been attached for this question yet. The verified answer is Option B.

The longest chain is not always the straight chain. For this compound, the longest chain is 6, which means hex... with the double bond on Carbon-3, counting from right to left, or from left to right, either way, the double bond is on C3 so the parent name will be hex-3-ene. For the substituents, we have two methyl groups (-CH3), counting from R-L, on C2 & C4. Counting from L-R will be on C3 & C5. However, since C2&C4 tends to be the shorter route, following the rule, the name of the compound is 2,4- dimethylhex-3-ene. The correct answer is option C.

When propyne (methylacetylene) reacts in the presence of sodium (Na) and liquid ammonia (NH₃), it undergoes a reduction reaction. Propyne (CH₃C≡CH) is reduced to form propene (CH₃CH=CH₂) through the trans addition of hydrogen. During the reaction, the sodium (Na) dissolves in the liquid ammonia (NH₃) to form a blue-colored solution of solvated electrons, which act as a reducing agent. The sodium is oxidized, releasing electrons that are then used to reduce the propyne. The ammonia acts as a solvent and helps to stabilize the solvated electrons. So, the sodium is essentially consumed during the reaction, and the ammonia acts as a catalyst or solvent, helping to facilitate the reduction reaction. The sodium (Na) and ammonia (NH₃) don't interfere with the product formed, which is propene (CH₃CH=CH₂), rather they play a role in facilitating the reaction, but they don't become part of the final product. The sodium is consumed during the reaction, and the ammonia can be recovered or removed, leaving behind the desired product, propene.