How do I figure that out?!
Please and thank you
2007-01-10 10:07:22 · 2 answers · asked by bretssillyjo 1 in Science & Mathematics ➔ Chemistry
20.18 gram(gram molecule=mole) occupies 22.4 litres at NTP(STP). Find out for 9 grams
22.4x9 divided by 20.18 =9.9 litres
2007-01-10 10:48:04 · answer #1 · answered by ssrvj 7 · 0⤊ 0⤋
a million. 2x 39 g ok reacts with a million mole = 22.4 L F2 at STP So 23.5g will react with ( 22.4 x 23.5 ) / (2 x 39) = 6.seventy 4 L F2 at STP 2. 2 X18 g water on decomposition supplies a million mole O2 = 22.4 L at STP So 37.4 g will supply ( 22.4 x 37.4 ) / (2 x18) L O2 at STP Convert it to vol. at a million.3 atmos and 297 ok temp. utilising ( P1V1 /T1) = ( P2V2 /T2 ) 3. Use PV =n RT to discover moles of CH4 or a million.2 x 15.8 = n x 0.0821 x (273 +27) Then multiply 'n' with sixteen to get mass of methane on account that its molar mass is sixteen g/mole 4. the quantity relation is a million vol. oxygen gasoline produces 2 vol of CO2 ( from equation) So 5 vol. CO2 would be produced with the aid of 2.5 L O2 5. value of diffusion of gasoline A Root of mol.wt. of gasoline B ------------------------------------- = ------------------------------------- value of diffusion of gasoline B Root of mol.wt. of gasoline A or 7 Root mol.wt. Unknown gasoline --------------- = -------------------------------------- 4.9 Root mol.wt. Ne =20 The mol.wt. is 40-one g/mol ( not extra room)
2016-10-30 13:53:51 · answer #2 · answered by bason 4 · 0⤊ 0⤋