The mass of 165.0 g sample that remains after 90.0 minutes is 5.16 grams
calculation
lambda㏑2/18= 0.0385
m(t)= 165 x e( 0.0385 x90) =5.16g
Answer:
If you continue to cool water past 4 degrees Celsius, its density starts to plummet (you can see this in the graph). At zero degrees, i.e., the temperature at which water turns into ice, the density of water is actually quite low. It turns out that ice has a lower density than water, and any object that has a lower density than the liquid form on which it’s kept (in this case, water) will be able to float!
Explanation:
Answer:
In oxidation reduction reactions, one species gets reduced by taking on electron(s) and another species gets oxidized by losing electrons. The movement of electrons can be used to do work. ... The electron flow can be run through a wire and these electrons can be used to do work (like run a battery). Hope this helps.
Answer:
130.4 grams of sucrose, would be needed to dissolve in 500 g of water.
Explanation:
Colligative property of boiling point elevation:
ΔT = Kb . m . i
In this case, i = 1 (sucrose is non electrolytic)
ΔT = Kb . m
0.39°C = 0.512°C/m . m
0.39°C /0.512 m/°C = m
0.762 m (molality means that this moles, are in 1kg of solvent)
If in 1kg of solvent, we have 0.712 moles of sucrose, in 500 g, which is the half, we should have, the hallf of moles, 0.381 moles
Molar mass sucrose = 342.30 g/m
Molar mass . moles = mass
342.30 g/m . 0.381 m = 130.4 g
