Explanation:
From the knowledge of law of multiple proportions,
mass ratio of S to O in SO:
mass of S : mass of O
= 32 : 16
= 32/16
= 2/1
mass ratio of S to O in SO2:
= mass of S : 2 × mass of O
= 32 : 2 × 16
= 32/32
= 1/1
ratio of mass ratio of S to O in SO to mass ratio of S to O in SO2:
= 2/1 ÷ 1/1
= 2
Thus, the S to O mass ratio in SO is twice the S to O mass ratio in SO2.
Explanation:
the answer and the working out is shown above, hope it helps.
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.
Assuming the question is asking for percent composition by mass,
The molar mass of sodium bicarbonate is:
22.990 + 1.008 + 12.011 + 3*15.999 = 84.006 g/mol
The molar mass of 3 oxygen atoms is:
3*15.999 = 47.997 g/mol
The percentage composition by mass is:
47.997/84.006 * 100% = 57.1% (3 s.f.)
