Answer:
I think its b
Explanation:
but I wouldn't depend on this answer
Answer:
Explanation:
Hello!
In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:
Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:
Best regards!
Answer:
Mole fraction of solute is 0.0462
Explanation:
To solve this we use the colligative property of lowering vapor pressure.
First of all, we search for vapor pressure of pure water at 25°C = 23.8 Torr
Now, we convert the Torr to mmHg. Ratio is 1:1, so 23.8 Torr is 23.8 mmHg.
Formula for lowering vapor pressure is:
ΔP = P° . Xm
Where ΔP = P' (Vapor pressure of solution) - P° (Vapor pressure of pure solvent)
Xm = mole fraction
24.9 mmHg - 23.8 mmHg = 23mmHg . Xm
Xm = (24.9 mmHg - 23.8 mmHg) / 23mmHg
Xm = 0.0462
FLOOR. Most of the molecules that dont sick to something fall on the floor so most would be on the FLOOR.
%(NaHCO3)= ((mass NaHCO3)/(mass NaHCO3 + mass water))*100%
m=Volume*Density
Density of water =1 g/ml
m(water) = Volume(water)*Density(water) = 600.0 ml * 1g/ml=600g water
%(NaHCO3)= ((20.0 g)/(20.0 g + 600 g))*100%=0.0323*100%=32.3%
