Answer:
The heat at constant pressure is -3,275.7413 kJ
Explanation:
The combustion equation is 2C₆H₆ (l) + 15O₂ (g) → 12CO₂ (g) + 6H₂O (l)
= (12 - 15)/2 = -3/2
We have;
![\Delta H = \Delta U + \Delta n_g\cdot R\cdot T](https://tex.z-dn.net/?f=%5CDelta%20H%20%3D%20%5CDelta%20U%20%2B%20%5CDelta%20n_g%5Ccdot%20R%5Ccdot%20T)
Where R and T are constant, and ΔU is given we can write the relationship as follows;
![H = U + \Delta n_g\cdot R\cdot T](https://tex.z-dn.net/?f=H%20%3D%20U%20%2B%20%5CDelta%20n_g%5Ccdot%20R%5Ccdot%20T)
Where;
H = The heat at constant pressure
U = The heat at constant volume = -3,272 kJ
= The change in the number of gas molecules per mole
R = The universal gas constant = 8.314 J/(mol·K)
T = The temperature = 300 K
Therefore, we get;
H = -3,272 kJ + (-3/2) mol ×8.314 J/(mol·K) ×300 K) × 1 kJ/(1000 J) = -3,275.7413 kJ
The heat at constant pressure, H = -3,275.7413 kJ.
Answer:
The expression to calculate the mass of the reactant is ![m = \frac{1.080kJ}{31.2kJ/g}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B1.080kJ%7D%7B31.2kJ%2Fg%7D)
Explanation:
<em>The amount of heat released is equal to the amount of heat released per gram of reactant times the mass of the reactant.</em> To keep to coherence between units we need to transform 1,080 J to kJ. We do so with proportions:
![1,080J.\frac{1kJ}{10^{3}J } =1.080kJ](https://tex.z-dn.net/?f=1%2C080J.%5Cfrac%7B1kJ%7D%7B10%5E%7B3%7DJ%20%7D%20%3D1.080kJ)
Then,
![1.080kJ=31.2\frac{kJ}{g} .m\\m = \frac{1.080kJ}{31.2kJ/g}](https://tex.z-dn.net/?f=1.080kJ%3D31.2%5Cfrac%7BkJ%7D%7Bg%7D%20.m%5C%5Cm%20%3D%20%5Cfrac%7B1.080kJ%7D%7B31.2kJ%2Fg%7D)
Answer:
viscosity
Explanation:
one one left is mercury and the other one is honey right?
Explanation:
I hope you interested about the chemical what they add in tea
<u>density</u> is a measure of mass per volume. <u>mass</u> is both a property of a physical body and a measure of its resistance to acceleration.
<em>hope this helps! ❤ from peachimin</em>