Answer:
There are 0.93 g of glucose in 100 mL of the final solution
Explanation:
In the first solution, the concentration of glucose (in g/L) is:
15.5 g / 0.100 L = 155 g/L
Then a 30.0 mL sample of this solution was taken and diluted to 0.500 L.
- 30.0 mL equals 0.030 L (Because 30.0 mL ÷ 1000 = 0.030 L)
The concentration of the second solution is:
So in 1 L of the second solution there are 9.3 g of glucose, in 100 mL (or 0.1 L) there would be:
1 L --------- 9.3 g
0.1 L--------- Xg
Xg = 9.3 g * 0.1 L / 1 L = 0.93 g
there are 4 atoms in aluminum chloride
Answer:
89.4%
Explanation:
Initially, there is 5.0 of the acetanilide in 100 mL of water, then the solution is chilled at 0ºC. The solubility represents the amount that the solvent (water) can dissolve of the solute (acetanilide). So, at 0ºC, 100 mL of water can dissolve till 0.53 g of the compound, the rest will precipitate and will be recovered.
So, the mass that is recovered is 5.0 - 0.53 = 4.47 g
The percent recovery is:
(4.47/5)x100% = 89.4%
Answer is because
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Answer:
0.045 L or 45 mL
Explanation:
Moles = Mass/M.Mass
Moles = 10 g / 109.94 g/mol
Moles = 0.09 moles
Also,
Molarity = Moles / Vol in L
Or,
Vol in L = Moles / Molarity
Vol in L = 0.09 mol / 2 mol/L
Vol in L = 0.045 L
