domashniaia rabota po khimii 10 klass gabrielian 16 uprazhnenie

Domashniaia Rabota Po Khimii 10 Klass Gabrielian 16 Uprazhnenie -

He checked the density. The problem mentioned the relative density of the gas compared to air is approximately

n(H)=2×n(H2O)=2×(7.2 g18 g/mol)=0.8 moln open paren cap H close paren equals 2 cross n open paren cap H sub 2 cap O close paren equals 2 cross open paren the fraction with numerator 7.2 g and denominator 18 g/mol end-fraction close paren equals 0.8 mol Step 2: Finding the Ratio Maxim looked at his notes. The ratio of ."So the simplest formula is C3H8cap C sub 3 cap H sub 8 ," he realized. "But is it the real one?" Step 3: The Final Verification He checked the density

(3×12)+(8×1)=36+8=44 g/molopen paren 3 cross 12 close paren plus open paren 8 cross 1 close paren equals 36 plus 8 equals 44 g/mol The Discovery "But is it the real one

n(C)=n(CO2)=13.2 g44 g/mol=0.3 moln open paren cap C close paren equals n open paren cap C cap O sub 2 close paren equals the fraction with numerator 13.2 g and denominator 44 g/mol end-fraction equals 0.3 mol Since each H2Ocap H sub 2 cap O molecule has two hydrogen atoms: 7 was quiet, except for the low hum of the ventilation hood

"Great work, Maxim. Exercise 16 just saved our lab session." ✅ Final Result The molecular formula of the substance in the exercise is .

The lab at School No. 7 was quiet, except for the low hum of the ventilation hood. Maxim sat hunched over his desk, staring at a silver gas cylinder that had lost its label years ago. His teacher, Elena Petrovna, had given him a challenge: "Identify this gas, or we can't use it for the demonstration tomorrow."

"First," he muttered, "I need to find the amount of substance for Carbon and Hydrogen."