I think it’s classified as a sugar. The ending -ose usually means a sugar.
1,05+ 0,69 + 1,82 = 3,56g of mixture
%CaCO₃: 1,82/3,56×100% = 0,5112×100% = 51,12%
15396 g
tell me if its correct
Answer:
V = 240.79 L
Explanation:
Given data:
Volume of butane = ?
Temperature = 293°C
Pressure = 10.934 Kpa
Mass of butane = 33.25 g
Solution:
Number of moles of butane:
Number of moles = mass/ molar mass
Number of moles = 33.25 g/ 58.12 g/mol
Number of mole s= 0.57 mol
Now we will convert the temperature and pressure units.
293 +273 = 566 K
Pressure = 10.934/101 = 0.11 atm
Volume of butane:
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
V = nRT/P
V = 0.57 mol × 0.0821 atm.L/ mol.K ×566 K / 0.11 atm
V = 26.49 L/0.11
V = 240.79 L
