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, nitakusaidia kutatua swali hili la kemia kwa kutumia Sheria ya Graham ya Usambazaji. Swali: Hamsini (50) cm³ ya gesi ya salfa IV oksidi (SO_2) husambaa kupitia plagi yenye vinyweleo kwa sekunde 80. Ni kiasi gani cha gesi ya nitrojeni IV oksidi (NO_2) kitasambaa kupitia plagi hiyo hiyo kwa sekunde 10? Tunatumia Sheria ya Graham ya Usambazaji, 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 uzito wa molekuli wa gesi zote mbili. Uzito wa molekuli wa salfa IV oksidi (SO_2): M_SO_2 = 32 + (2 × 16) = 32 + 32 = 64\, g/mol. Uzito wa molekuli wa nitrojeni IV oksidi (NO_2): M_NO_2 = 14 + (2 × 16) = 14 + 32 = 46\, g/mol. Step 2: Kokotoa kiwango cha usambazaji wa SO_2. Rate_SO_2 = 50\, cm^380\, s = 0.625\, cm^3/s Step 3: Tumia Sheria ya Graham kuhesabu kiwango cha usambazaji wa NO_2. Rate_SO_2Rate_NO_2 = sqrt(M_NO_2)M_SO_2 0.625\, cm^3/sRate_NO_2 = sqrt(46\, g/mol)64\, g/mol (0.625)/(Rate_NO_2) = sqrt(0.71875) (0.625)/(Rate_NO_2) ≈ 0.84779 Rate_NO_2 = (0.625)/(0.84779) ≈ 0.73720\, cm^3/s Step 4: Kokotoa kiasi cha NO_2 kitakachosambaa kwa sekunde 10. Kiasi_NO_2 = Rate_NO_2 × Time Kiasi_NO_2 = 0.73720\, cm^3/s × 10\, s Kiasi_NO_2 ≈ 7.372\, cm^3 Kiasi cha NO_2 kitakachosambaa ni takriban 7.4\, cm^3. Last free one today — make it count tomorrow, or type /upgrade for unlimited.