Domashniaia Rabota Po Khimii 10 Klass Gabrielian 16 Uprazhnenie Apr 2026

(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

"Great work, Maxim. Exercise 16 just saved our lab session." ✅ Final Result The molecular formula of the substance in the exercise is . The "mystery gas" was the very fuel used in camping stoves

Since all carbon from the hydrocarbon ends up in CO2cap C cap O sub 2 " he muttered

"It's Propane!" Maxim shouted. The "mystery gas" was the very fuel used in camping stoves. He quickly printed a new label— C3H8cap C sub 3 cap H sub 8 The "mystery gas" was the very fuel used in camping stoves

M=Dair×Mair=1.52×29 g/mol≈44 g/molcap M equals cap D sub a i r end-sub cross cap M sub a i r end-sub equals 1.52 cross 29 g/mol is approximately equal to 44 g/mol The molar mass of C3H8cap C sub 3 cap H sub 8 (Propane) is:

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

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: