This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Haya, nitatatua maswali haya tena kwa kutumia Sheria ya Graham ya Usambazaji (Graham's Law of Diffusion). Swali la 6: Tunatumia Sheria ya Graham, ambayo inasema kwamba kiwango cha usambazaji wa gesi kinapingana na mzizi wa mraba wa uzito wake wa molekuli. (Rate_1)/(Rate_2) = sqrt((M_2)/(M_1)) Ambapo Rate = (Volume)/(Time). Step 1: Kokotoa viwango vya usambazaji kwa gesi ya hidrojeni na gesi P. Kiwango cha usambazaji wa hidrojeni (Rate_H_2): Rate_H_2 = 500\, cm^350\, s = 10\, cm^3/s Kiwango cha usambazaji wa gesi P (Rate_P): Rate_P = 853\, cm^3400\, s = 2.1325\, cm^3/s Uzito wa molekuli wa hidrojeni (M_H_2) ni 2 × 1 = 2\, g/mol. Step 2: Tumia Sheria ya Graham kuhesabu uzito wa molekuli wa gesi P (M_P). Rate_H_2Rate_P = sqrt((M_P)/(M_H_2)) 10\, cm^3/s2.1325\, cm^3/s = sqrt((M_P)/(2)) 4.6894 = sqrt((M_P)/(2)) Step 3: Piga mraba pande zote mbili na utatue kwa M_P. (4.6894)^2 = (M_P)/(2) 21.9805 = (M_P)/(2) M_P = 21.9805 × 2 M_P = 43.961 Uzito wa molekuli wa gesi P ni takriban 44. Swali la 7: Kiwango cha usambazaji wa hidrojeni (Rate_H_2) ni 10\, cm^3/s. Uzito wa molekuli wa hidrojeni (M_H_2) ni 2\, g/mol. a) Kiasi cha oksijeni (O_2) kinachosambaa kwa sekunde 20. Uzito wa molekuli wa oksijeni (M_O_2) ni 2 × 16 = 32\, g/mol. Step 1: Tumia Sheria ya Graham. Rate_H_2Rate_O_2 = sqrt(M_O_2)M_H_2 (10)/(Rate_O_2) = sqrt((32)/(2)) (10)/(Rate_O_2) = sqrt(16) (10)/(Rate_O_2) = 4 Step 2: Kokotoa Rate_O_2 na kisha kiasi cha O_2. Rate_O_2 = (10)/(4) = 2.5\, cm^3/s Kiasi cha O_2 = Rate_O_2 × Time Kiasi cha O_2 = 2.5\, cm^3/s × 20\, s Kiasi cha O_2 = 50\, cm^3 b) Kiasi cha pentane (C_5H_12) kinachosambaa kwa sekunde 30. Uzito wa molekuli wa pentane (M_C_5H_12) ni (5 × 12) + (12 × 1) = 60 + 12 = 72\, g/mol. Step 1: Tumia Sheria ya Graham. Rate_H_2Rate_C_5H_12 = sqrt(M_C_5H_12)M_H_2 (10)/(Rate_C_5H_12) = sqrt((72)/(2)) (10)/(Rate_C_5H_12) = sqrt(36) (10)/(Rate_C_5H_12) = 6 Step 2: Kokotoa Rate_C_5H_12 na kisha kiasi cha C_5H_12. Rate_C_5H_12 = (10)/(6) = 1.6667\, cm^3/s Kiasi cha C_5H_12 = Rate_C_5H_12 × Time Kiasi cha C_5H_12 = 1.6667\, cm^3/s × 30\, s Kiasi cha C_5H_12 = 50\, cm^3 c) Kiasi cha heptane (C_7H_16) kinachosambaa kwa sekunde 28. Uzito wa molekuli wa heptane (M_C_7H_16) ni (7 × 12) + (16 × 1) = 84 + 16 = 100\, g/mol. Step 1: Tumia Sheria ya Graham. Rate_H_2Rate_C_7H_16 = sqrt(M_C_7H_16)M_H_2 (10)/(Rate_C_7H_16) = sqrt((100)/(2)) (10)/(Rate_C_7H_16) = sqrt(50) (10)/(Rate_C_7H_16) ≈ 7.071 Step 2: Kokotoa Rate_C_7H_16 na kisha kiasi cha C_7H_16. Rate_C_7H_16 = (10)/(7.071) ≈ 1.4142\, cm^3/s Kiasi cha C_7H_16 = Rate_C_7H_16 × Time Kiasi cha C_7H_16 = 1.4142\, cm^3/s × 28\, s Kiasi cha C_7H_16 = 39.6\, cm^3 Swali la 8: Gesi X: Kiasi = 200\, cm^3, Muda = 20\, s Hidrojeni (H_2): Kiasi = 200\, cm^3, Muda = 5\, s Uzito wa molekuli wa hidrojeni (M_H_2) ni 2\, g/mol. Step 1: Kokotoa viwango vya usambazaji kwa gesi X na hidrojeni. Kiwango cha usambazaji wa gesi X (Rate_X): Rate_X = 200\, cm^320\, s = 10\, cm^3/s Kiwango cha usambazaji wa hidrojeni (Rate_H_2): Rate_H_2 = 200\, cm^35\, s = 40\, cm^3/s Step 2: Tumia Sheria ya Graham kuhesabu uzito wa molekuli wa gesi X (M_X). (Rate_X)/(Rate_H_2) = sqrt(M_H_2)M_X 10\, cm^3/s40\, cm^3/s = sqrt((2)/(M_X)) (1)/(4) = sqrt((2)/(M_X)) Step 3: Piga mraba pande zote mbili na utatue kwa M_X. ((1)/(4))^2 = (2)/(M_X) (1)/(16) = (2)/(M_X) M_X = 2 × 16 M_X = 32 Uzito wa molekuli wa gesi X ni 32. That's 2 down. 3 left today — send the next one.