Answer:
b) 0.47
Explanation:
MwC5H12 = 72.15g/mol
⇒mol C5H12 = (10.0)*(mol/72.15)=0.1386molC5H12
MwC6H14=86.18g/mol
⇒molC6H14=(20.0)*(mol/86.18)=0,232
MwC6H6=78.11g/mol
⇒molC6H6=(10.0)*(mol/78.11)=0.128molC6H6
<h3>XC6H14=(0.232)/(0.1386+0.232+0,128)=0.465≅0.47</h3>
Answer:
Close to the calculated endpoint of a titration - <u>Partially open</u>
At the beginning of a titration - <u>Completely open</u>
Filling the buret with titrant - <u>Completely closed</u>
Conditioning the buret with the titrant - <u>Completely closed</u>
Explanation:
'Titration' is depicted as the process under which the concentration of some substances in a solution is determined by adding measured amounts of some other substance until a rection is displayed to be complete.
As per the question, the stopcock would remain completely open when the process of titration starts. After the buret is successfully placed, the titrant is carefully put through the buret in the stopcock which is entirely closed. Thereafter, when the titrant and the buret are conditioned, the stopcock must remain closed for correct results. Then, when the process is near the estimated end-point and the solution begins to turn its color, the stopcock would be slightly open before the reading of the endpoint for adding the drops of titrant for final observation.
3.25 kg in g = 3.25 * 1000 = 3250 g
Molar mass C₂H₆O₂ = 62.0 g/mol
Mass solvent = 7.75 kg
Number of moles:
n = mass solute / molar mass
n = 3250 / 62.0
n = 52.419 moles
Molality = moles of solute / kilograms of solvent
M = 52.419 / 7.75
M = 6.7637 mol/kg
hope this helps!
Answer:
2.65 M
Explanation:
Convert grams of K₂CO₃ to moles. The molar mass is 138.205 g/mol.
(110 g)/(138.205 g/mol) = 0.796 mol
Convert milliliters of solution to liters.
300 mL = 0.300 L
Divide moles of K₂CO₃ by liters of solution.
0.796 mol/0.300 L = 2.65 mol/L = 0.265 M