Answer:
Explanation:
Hello,
In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):
Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:
Best regards.
Answer:
0.56 liters
Explanation:
First we <u>convert 0.80 grams of O₂ into moles</u>, using its molar mass:
- 0.80 g ÷ 32 g/mol = 0.025 mol
At STP, 1 mol of any given mass occupies 22.4 L. With that information in mind we <u>calculate the volume that 0.025 moles of O₂ gas would occupy</u>:
- 0.025 mol * 22.4 L/mol = 0.56 L
Thus the answer is 0.56 liters.
Alright sorry you're getting the answer hours later, but i can help with this.
so you're looking for specific heat, the equation for it is <span>macaΔTa = - mbcbΔTb with object a and object b. that's mass of a times specific heat of a times final minus initial temperature of a equals -(mass of b times specific heat of b times final minus initial temperature of b)
</span>so putting in your values is, 755g * ca * (75 celsius - 84.5 celsius) = -(50g * cb * (75 celsius - 5 celsius))
well we know the specific heat of water is always 4180J/kg celsius, so put that in for cb
with a bit of simplification to the equation by doing everything on each side first you have, -7172.5 * ca = -14630000
divide both sides by -7172.5 so you can single out ca and you get, ca= 2039.74
add units for specific heat which are J/kg celsius and the specific heat of the material is 2039.74 J/kg celsius
Cause their are both have difference s
Hey there!:
Volume in mL :
1.71 L * 1000 => 1710 mL
Density = 0.921 g/mL
Therefore:
Mass = Density * Volume
Mass = 0.921 * 1710
Mass = 1574.91 g
