Absolute zero is the lowest possible volume occupied by a gas the lowest possible temperature of a gas the lowest possible pressure exerted by a gas none of the above

At 40 âc the solution has at 40 â c the solution has blank g of k n o 3 per 100 g of water and it can contain up to blank g of k n o 3 per 100 g of water. at 0 â c the solubility is ~ blank g k n o 3 per 100 g of water, so ~ blank g k n o 3 per 100 g of water will precipitate out of solution. g of kno3 per 100 g of water and it can contain up to at 40 â c the solution has blank g of k n o 3 per 100 g of water and it can contain up to blank g of k n o 3 per 100 g of water. at 0 â c the solubility is ~ blank g k n o 3 per 100 g of water, so ~ blank g k n o 3 per 100 g of water will precipitate out of solution. g of kno3 per 100 g of water. at 0 âc the solubility is ~ at 40 â c the solution has blank g of k n o 3 per 100 g of water and it can contain up to blank g of k n o 3 per 100 g of water. at 0 â c the solubility is ~ blank g k n o 3 per 100 g of water, so ~ blank g k n o 3 per 100 g of water will precipitate out of solution. kno3 per 100 g of water, so ~ at 40 â c the solution has blank g of k n o 3 per 100 g of water and it can contain up to blank g of k n o 3 per 100 g of water. at 0 â c the solubility is ~ blank g k n o 3 per 100 g of water, so ~ blank g k n o 3 per 100 g of water will precipitate out of solution. gkno3 per 100 g of water will precipitate out of solution. a kno3 solution containing 55 g of kno3 per 100.0 g of water is cooled from 40 ∘c to 0 ∘c. what will happen during cooling?

Kp is the equilibrium constant for dissociation of the propionic acid dimer. what is the sign of the slope for a plot of the natural logarithm of kp vs. inverse temperature for this reaction?

Calculate the standard enthalpy of formation of liquid methanol, ch3oh(l), using the following information: c(graphite) + o2 latex: \longrightarrow ⟶ co2(g) latex: \delta δ h° = –393.5 kj/mol h2(g) + o2 latex: \longrightarrow ⟶ h2o(l) latex: \delta δ h° = –285.8 kj/mol ch3oh(l) + o2(g) latex: \longrightarrow ⟶ co2(g) + 2h2o(l) latex: \delta δ h° = –726.4 kj/mol