Answer:
fH = - 3,255.7 kJ/mol
Explanation:
Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.
Qsystem + Qcombustion = 0
Qsystem = heat capacity*ΔT
10000*(25.000 - 20.826) + Qc = 0
Qcombustion = - 41,740 J = - 41.74 kJ
So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:
n = mass/molar mass = 1/ 78
n = 0.01282 mol
fH = -41.74/0.01282
fH = - 3,255.7 kJ/mol
Answer:
I think the answer is increases
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!
Answer:
Substance B, boiling point of 105 °C
Explanation:
Non volatile substances have high boiling points
Answer:
The percentage by mass of element X is 25 %.
The percentage by mass of element Y is 75 %
Explanation:
X + Y ⇒ XY
3.5 g of element X + 10.5 g of element Y = 14g in total
⇒Element X : 3.5g / 14g = 0.25 ⇒ x 100 % = 25 %
⇒Element Y : 10.5 / 14 = 0.75 ⇒ x 100 % = 75 %
The percentage by mass of element X is 25 %.
The percentage by mass of element Y is 75 %