Answer:
C₅ H₁₂ O
Explanation:
44 g of CO₂ contains 12 g of C
30.2 g of CO₂ will contain 12 x 30.2 / 44 = 8.236 g of C .
18 g of H₂O contains 2 g of hydrogen
14.8 g of H₂0 will contain 1.644 g of H .
total compound = 12.1 out of which 8.236 g is C and 1.644 g is H , rest will be O
gram of O = 2.22
moles of C, O, H in the given compound = 8.236 / 12 , 2.22 / 16 , 1.644 / 1
= .6863 , .13875 , 1.644
ratio of their moles = 4.946 : 1 : 11.84
rounding off to digits
ratio = 5 : 1 : 12
empirical formula = C₅ H₁₂ O
Your answer is going to be 200g
Answer:
2.47L
Explanation:
Using the combined gas law equation as follows:
P1V1/T1= P2V2/T2
Where;
P1 = initial pressure (mmHg)
P2 = final pressure (mmHg)
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question;
P1 = 705mmHg
P2 = 760mmHg (STP)
V1 = 3.00L
V2 = ?
T1 = 35°C = 35 + 273 = 308K
T2 = 273K (STP)
Using P1V1/T1= P2V2/T2
705 × 3/308 = 760 × V2/273
2115/308 = 760V2/273
Cross multiply
308 × 760V2 = 2115 × 273
234,080V2 = 577,395
V2 = 577,395 ÷ 234,080
V2 = 2.47L
