C₀=2 mol/l
c₁=0.400 mol/l
v₁=100.0 ml = 0.1 l
c₁v₁=c₀v₀
v₀=c₁v₁/c₀
v(H₂O)=v₁-v₀
v₀=0.1*0.400/2=0.02 l = 20 ml
v(H₂O)=100 - 20 = 80 ml
It is necessary to mix 20 ml of the feed solution and 80 ml of water.
Answer:
- <u>Yes, it is 14. g of compound X in 100 ml of solution.</u>
Explanation:
The relevant fact here is:
- the whole amount of solute disolved at 21°C is the same amount of precipitate after washing and drying the remaining liquid solution: the amount of solute before cooling the solution to 21°C is not needed, since it is soluble at 37°C but not soluble at 21°C.
That means that the precipitate that was thrown away, before evaporating the remaining liquid solution under vacuum, does not count; you must only use the amount of solute that was dissolved after cooling the solution to 21°C.
Then, the amount of solute dissolved in the 600 ml solution at 21°C is the weighed precipitate: 0.084 kg = 84 g.
With that, the solubility can be calculated from the followiing proportion:
- 84. g solute / 600 ml solution = y / 100 ml solution
⇒ y = 84. g solute × 100 ml solution / 600 ml solution = 14. g.
The correct number of significant figures is 2, since the mass 0.084 kg contains two significant figures.
<u>The answer is 14. g of solute per 100 ml of solution.</u>
Boron: isotope data. Both isotopes ofBoron, B-10 and B-11, are used extensively in the nuclear industry. B-10 is used in the form of boric acid as a chemical shim in pressurized water reactors while in the form of sodium pentaborate it is used for standby liquid control systems in boiling water reactors
Answer:
When the yeast get warm water and some food to eat (in the form of sugar), they will become active. And as they eat the sugar and break it down for food, they release carbon dioxide, which fills up the balloon. Yeast is actually a type of fungus related to mushrooms.
Hope it helps!!!
