Answer:
335.25g
Explanation:
using equation of molarity
Answer:
Following are the responses to the given question:
Explanation:
calculating the HOI:
So, in the above compund there are three bounds or we can say that one is double and two bound is single.
Please find the attachment file of the Function group:
In the above given conditon compound doesn't include the alkene functional group that is absense of
, and compound include 
So, please find the attached file of the structure of the compound:
The answer is B because <span>It would be useful to memorize that sentence. Once you know that, you can figure out whatever else happens at the anode, the cathode, in the solution, and in the external circuit.</span>
Answer:
C6H14O3F
Explanation:
The first step is to divide each compound by its molecular weight
Carbon
= 39.10/12
= 3.258
Hydrogen
= 7.67/1
= 7.67
Oxygen
= 26.11/16
= 1.63
Phosphorous
= 16.82/31
= 0.542
Flourine
= 10.30/19
= 0.542
The next step is to divide by the lowes value
3.258/0.542
= 6 mol of C
7.67/0.542
= 14 mol of H
1.63/0.542
= 3 mol of O
0.542/0.542
= 1 mol of P
0.542/0.542
= 1 mol of F
Hence the molecular formula is C6H14O3F
