First, calculate the number of moles of $\textO_2$ produced: $n = \fracPVRT = \frac(1.00 \text atm)(0.120 \text L)(0.0821 \text L atm/mol K)(298 \text K) = 0.00491 \text mol$ The molar mass of $\textKClO_3$ is 122.55 g/mol. The theoretical yield of $\textO_2$ from 0.500 g of $\textKClO_3$ is: $0.500 \text g \times \frac1 \text mol122.55 \text g \times \frac3 \text mol O_22 \text mol KClO_3 \times \frac32.00 \text g1 \text mol O_2 = 0.195 \text g O_2$ Percent yield $= \frac0.00491 \text mol \times 32.00 \text g/mol0.195 \text g \times 100% \approx 80.5%$
Add HNO₃ to the complex solution → brown precipitate of AgCl reforms. 1972 ap chemistry free response answers
One of the most cited problems from 1972 involved a complex dry mixture of potassium hydroxide ( cap K cap O cap H ), potassium carbonate ( cap K sub 2 cap C cap O sub 3 ), and potassium chloride ( cap K cap C l The Challenge: A 5.00g sample is reacted with 0.100L of . Students had to: Calculate the % of cap K sub 2 cap C cap O sub 3 based on 249mL of cap C cap O sub 2 gas produced. Use back-titration data with cap N a cap O cap H to find the percentages of the remaining components. The Solution Path: Use the Ideal Gas Law ( ) to find the moles of cap C cap O sub 2 . At 740 torr and 22°C, Relate moles of cap C cap O sub 2 cap K sub 2 cap C cap O sub 3 (1:1 ratio). Calculation: cap K sub 2 cap C cap O sub 3 2. Transition Metal Coordination First, calculate the number of moles of $\textO_2$
In 1972, the Free Response (now known as Section II) required students to demonstrate deep conceptual knowledge without the aid of modern graphing calculators. The focus was on "first principles"—the ability to derive relationships and explain why a reaction occurs, rather than just plugging numbers into a formula. Key Questions and Conceptual Answers 1. Equilibrium and Solubility Product ( Kspcap K sub s p end-sub Students had to: Calculate the % of cap