you can google it and it pops up right away
Answer:
C
Explanation:
I just aswered this question and i got it right
Answer:
Explanation:
In this case, we have to start with the <u>chemical reaction</u>:
So, if we start with <u>10 mol of cyclohexanol</u> (
) we will obtain 10 mol of cyclohexanol (
). So, we can calculate the grams of cyclohexanol if we<u> calculate the molar mass:</u>
With this value we can calculate the grams:
Now, we have as a product 500 mL of . If we use the <u>density value</u> (0.811 g/mL). We can calculate the grams of product:
Finally, with these values we can calculate the <u>yield</u>:
%= (405.5/820)*100 = 49.45 %
See figure 1
I hope it helps!
The empirical formula for the caproic acid, given the combustion analysis data is C₃H₆O
We'll begin bey obtaining the mass of carbon, hydrogen and oxygen in the compound. This is illustrated below:
How to determine the mass of C
- Mass of CO₂ = 9.78 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 9.78
Mass of C = 2.67 g
How to determine the mass of H
- Mass of H₂O = 20.99 g
- Molar mass of H₂O = 18 g/mol
- Molar of H = 2 × 1 = 2 g/mol
- Mass of H =?
Mass of H = (2 / 18) × 4
Mass of H = 0.44 g
How to determine the mass of O
- Mass of compound = 4.30 g
- Mass of C = 2.67 g
- Mass of H = 0.44 g
- Mass of O =?
Mass of O = (mass of compound) – (mass of C + mass of H)
Mass of O = 4.30 – (2.67 + 0.44)
Mass of O = 1.19 g
<h3>How to determine the empirical formula </h3>
The empirical formula of the compound can be obtained as follow:
- C = 2.67 g
- H = 0.44 g
- O = 1.19 g
- Empirical formula =?
Divide by their molar mass
C = 2.67 / 12 = 0.2225
H = 0.44 / 1 = 0.44
O = 1.19 / 16 = 0.074
Divide by the smallest
C = 0.2225 / 0.074 = 3
H = 0.44 / 0.0744 = 6
O = 0.074 / 0.074 = 1
Thus, the empirical formula of the compound is C₃H₆O
Learn more about empirical formula:
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