X:5.8g=16:(23+1+12+3*16)
x:5.=16:84
x:=5.8* 16/84
this is approximately 1.1
Air is mainly composed of N2 (78%), O2 (21%) and other trace gases. Now, the total pressure of air is the sum of the partial pressures of the constituent gases. The partial pressure of each gas, for example say O2, can be expressed as:
p(O2) = mole fraction of O2 * P(total, air) ----(1)
Thus, the partial pressure is directly proportional to the total pressure. If we consider a sealed container then, as the temperature of air increases so will its pressure. Based on equation (1) an increase in the pressure of air should also increase the partial pressure of oxygen.
Answer: 0.4 moles
Explanation:
Given that:
Volume of gas V = 11L
(since 1 liter = 1dm3
11L = 11dm3)
Temperature T = 25°C
Convert Celsius to Kelvin
(25°C + 273 = 298K)
Pressure P = 0.868 atm
Number of moles N = ?
Note that Molar gas constant R is a constant with a value of 0.00821 atm dm3 K-1 mol-1
Then, apply ideal gas equation
pV = nRT
0.868atm x 11dm3 = n x (0.00821 atm dm3 K-1 mol-1 x 298K)
9.548 atm dm3 = n x 24.47atm dm3mol-1
n = (9.548 atm dm3 / 24.47atm dm3 mol-1)
n = 0.4 moles
Thus, there are 0.4 moles of the gas.
The cnidarias life cycle has 2 life cycles polyp and medusa
