How do I get this solution, gas or heat problem right?

Use C₁V₁ = C₂V₂. Rearranged for the stock volume, V₁ = C₂V₂ ÷ C₁. To make 500 mL of 0.1 M from a 1 M stock: V₁ = 0.1 × 500 ÷ 1 = 50 mL of stock, then make it up to 500 mL total — which is roughly 450 mL of solvent added, not 500.

Make up to the final volume, in a volumetric flask, whenever accuracy matters. Volumes are not strictly additive — mixing equal volumes of ethanol and water yields about 4% less than the sum — so measuring the solvent separately introduces an error that making up to the mark avoids entirely.

Divide the mass by the molar mass: n = m ÷ M. 58.44 grams of sodium chloride is one mole because its formula mass is 58.44 g/mol. Multiply the moles by 6.02214076 × 10²³ to get the number of formula units, which since 2019 is an exact defined figure rather than a measured one.

Whichever matches the bottle you are weighing from. Copper sulfate pentahydrate weighs 249.68 g/mol against 159.61 for the anhydrous salt, so weighing out the hydrate while calculating with the anhydrous figure gives you 36% less copper than you intended. Check the label, not the formula you remember.

Take pOH = −log₁₀[OH⁻], then pH = 14 − pOH at 25 °C. A 0.001 M hydroxide solution gives pOH 3 and pH 11. The 14 comes from the ion product of water, Kw = 1.0 × 10⁻¹⁴, and it is the part that changes with temperature.

Only at 25 °C. Neutrality means [H⁺] = [OH⁻], and since Kw rises with temperature, so does the dissociation of water: at 50 °C pure water is neutral at pH 6.63, and at 100 °C at about 6.14. It is not acidic — it is neutral at a different number.

Because PV = nRT is a proportionality to absolute temperature, and the Celsius scale has an arbitrary zero. At 25 °C, using 25 instead of 298.15 inflates the answer by a factor of nearly twelve. The same applies to the Carnot efficiency and to ΔS = Q/T.

22.414 litres at 0 °C and 1 atm, or 22.711 litres at IUPAC standard conditions of 0 °C and 100 kPa. At a more practical 25 °C and 1 atm it is 24.465 litres. Because it is an ideal-gas result it holds regardless of which gas you have, to within a few percent at ordinary pressures.

It is 1 − T_cold/T_hot, in kelvin, and it is a consequence of the second law of thermodynamics: any engine that exceeded it could be run in reverse to move heat from cold to hot with no work, which is impossible. Between 800 K and 300 K the ceiling is 62.5%, and no arrangement of materials or cleverness gets past it.

The Carnot cycle is reversible, which means infinitely slow — it produces zero power. Any engine that actually delivers work has to transfer heat at a finite rate, which is inherently irreversible. Add friction, incomplete combustion and exhaust losses and a car engine ends up near 30% where its Carnot ceiling is above 60%.

ΔS = Q/T, with Q the latent heat absorbed and T the transition temperature in kelvin. Melting a kilogram of ice takes 333.55 kJ at 273.15 K, so ΔS = +1,221 J/K. Boiling the same kilogram takes 2,256 kJ at 373.15 K, giving +6,047 J/K — vaporisation disorders a substance far more than melting does.

The entropy of a system can, and does every time water freezes. What cannot decrease is the total entropy of the system plus its surroundings. Freezing releases heat into the room, and the entropy the room gains exceeds what the ice loses, so the balance still goes the right way.