Answer:
Solubility of O₂(g) in 4L water = 3.42 x 10⁻² grams O₂(g)
Explanation:
Graham's Law => Solubility(S) ∝ Applied Pressure(P) => S =k·P
Given P = 0.209Atm & k = 1.28 x 10⁻³mol/L·Atm
=> S = k·P = (1.28 x 10⁻³ mole/L·Atm)0.209Atm = 2.68 x 10⁻³ mol O₂/L water.
∴Solubility of O₂(g) in 4L water at 0.209Atm = (2.68 x 10⁻³mole O₂(g)/L)(4L)(32 g O₂(g)/mol O₂(g)) = <u>3.45 x 10⁻² grams O₂(g) in 4L water. </u>
The answer is 16.75 it rounds to 17 Hope this helps! ;D
CaBr₃ = 40 + 80 * 3 = 280 g/mol
hope this helps!
Hello!
The property that is being demonstrated when Janice was given a mixture of alcohol and water and her teacher told her that she could use temperature to separate these compounds is boiling.
The boiling point is the temperature when a substance goes from liquid state to gas state. If two liquid substances with different boiling points are in a mixture, they can be separated by heating one of them to its boiling point and condensing the gas in another container in a process called distillation. The difference in boiling points of alcohols (usually lower than water) and water allows to separate them in a mixture.
Have a nice day!
Used the Ideal gas law to get the grams of carbon dioxide
PV=nRT Where p is pressure, v is volume, n is a number of moles, t is temperature and r is the ideal gas constant. The value of R is 0.0821.
Solution
First, find the mole of carbon dioxide
n= PV/RT
= 1 x 1.0 / 0.081 x (2.73 x 102)
= 1 / 0.081 x 278.46
= 1 / 22.56
= 0.045 mol
Convert the moles into grams
Grams of CO2 = 0.045 mol x 44 g/mol = 1.98 g
So the grams of CO2 is 1.98g
