Answer:
Explanation:
For the reaction
C2H5OH (l) + 3 O2(g) = 2CO2(g) + 3 H2O
We can calculate the standard molar enthalpy of combustion using the standard enthalpies of formation of the species involved in the reaction according to Hess law:
ΔHºc = 2ΔHºf CO2 (g) + 3ΔHºfH2O(l) - ( ΔHºf C2H5OH (l) - 3ΔHºfO2 (g) )
( we were not give the water state but we know we are at standard conditions so it is in its liquid state )
The ΔHºfs can be found in appropiate reference or texts.
ΔHºc = 2ΔHºf CO2 (g)+ 3ΔHºfH2O(l) - ( ΔHºf C2H5OH (l) -+3ΔHºfO2 (g) )
= [ 2 ( -393.52 ) + 3 ( -285.83 ) ] - [( -276.2 + 0 ) ] kJ
ΔHºc = -1368.33 kJ
2 is right am 70 percent sure
<span>Data:
pH = 5.2
[H+] = ?
Knowing that: (</span><span>Equation to find the pH of a solution)</span>
![pH = -log[H+]](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH%2B%5D)
<span>
Solving:
</span>
![5.2 = - log [H+]](https://tex.z-dn.net/?f=5.2%20%3D%20-%20log%20%5BH%2B%5D)
Knowing that the exponential is the opposite operation of the logarithm, then we have:
![[H+] = 10^{-5.2}](https://tex.z-dn.net/?f=%5BH%2B%5D%20%3D%2010%5E%7B-5.2%7D)
![\boxed{\boxed{[H+] = 6.30*10^{-6}}}\end{array}}\qquad\quad\checkmark](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cboxed%7B%5BH%2B%5D%20%3D%206.30%2A10%5E%7B-6%7D%7D%7D%5Cend%7Barray%7D%7D%5Cqquad%5Cquad%5Ccheckmark)
Answer:
A budget is an estimation of revenue and expenses over a specified future period of time and is usually compiled and re-evaluated on a periodic basis. Budgets can be made for a person, a group of people, a business, a government, or just about anything else that makes and spends money.
