To find the temperature at which the volume of the gas would be 0.550 L, given that it is 0.432 L at -20.0 °C, apply Charles’s Law.
The formula is v1/T1 = v2/T2
Known values:
V1 = 0.550 L
T1 = ?
T2 = -20°C + 273 = 253 K
V2 = 0.432 L
Rearranging for T1:
T1 = (V1 × T2) / V2
Calculating:
T1 = (0.55 L × 253) / 0.432 L = 322.11 K or 49.11°C
Response:

Clarification:
Hello,
In this scenario, since a single drop equates to 0.05 mL of the solution provided, with a concentration of 0.02 g/mL, the mass of oleic acid in one drop calculates to:

Best wishes.
(c) Cu + S → CuS is classified as a redox reaction
Explanation:
The following reactions are presented:
(a) K₂CrO₄ + BaCl₂ → BaCrO₄ + 2 KCl
(b) Pb²⁺ + 2 Br⁻ → PbBr₂
(c) Cu + S → CuS
Reaction (c) represents a redox reaction, as the oxidation states of the elements are changing. In this case:
Cu + S → CuS
In its elemental form, Cu has an oxidation state of 0, while in CuS (copper sulfide), its oxidation state changes to +2.
Similarly, S in its elemental form has an oxidation state of 0 and is -2 in CuS (copper sulfide).
Learn more about:
redox reactions
Explanation:
Initial moles of ethanoic acid = 0.020 mol
At equilibrium, half of the ethanoic acid molecules have reacted.
Thus, moles of ethanoic acid reacted = 0.020 mol * (50% / 100%)
= 0.010 mol
Moles of ethanoic acid remaining = 0.020 mol - 0.010 mol = 0.010 mol
The moles of product
gas formed are determined as follows:
0.010 mol CH3COOH * (1 mol
/ 2 mol CH3COOH)
= 0.005 mol 
Consequently, the total moles of gas present in the vessel at equilibrium are 0.010 mol CH3COOH and 0.005 mol 
Total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol
Next, let’s determine the pressure:
0.020 mol of gas has a pressure of 0.74 atm; so under the same conditions, we find the pressure exerted by 0.015 mol of gas:
P1/n1 = P2/n2
P2 = P1*(n2 / n1)
= 0.74 atm * (0.015 mol / 0.020 mol)
= 0.555 atm