Answer: The standard enthalpy of formation of liquid octane is -250.2 kJ/mol
Explanation:
The given balanced chemical reaction is,
First we have to calculate the enthalpy of reaction 
.
![\Delta H^o=[n_{O_2}\times \Delta H_f^0_{(O_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})+n_{O_2}\times \Delta H_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%2Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2O%29%7D%5D-%5Bn_%7BC_8H_%7B18%7D%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_8H_%7B18%7D%29%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%5D)
where,
We are given:
Putting values in above equation, we get:
![-1.0940\times 10^4=[(16\times -393.5)+(18\times -285.8)]-[(25\times 0)+(2\times \Delat H_f{C_8H_{18}(l)}]](https://tex.z-dn.net/?f=-1.0940%5Ctimes%2010%5E4%3D%5B%2816%5Ctimes%20-393.5%29%2B%2818%5Ctimes%20-285.8%29%5D-%5B%2825%5Ctimes%200%29%2B%282%5Ctimes%20%5CDelat%20H_f%7BC_8H_%7B18%7D%28l%29%7D%5D)
Thus the standard enthalpy of formation of liquid octane is -250.2 kJ/mol
We assume that this gas is an ideal gas. We use the ideal gas equation to calculate the amount of the gas in moles. It is expressed as:
PV = nRT
(672) (1/760) (36.52) = n (0.08206) ( 68 +273.15)
n = 1.15 mol of gas
Hope this answers the question. Have a nice day.
[OH⁻] = 1.6 × 10⁻⁸ mol / dm³
<h3>Explanation</h3>
By definition, ![[\text{H}^{+}] = 10^{-\text{pH}}](https://tex.z-dn.net/?f=%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%20%3D%2010%5E%7B-%5Ctext%7BpH%7D%7D)
, where ![[\text{H}^{+}]](https://tex.z-dn.net/?f=%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D)
is the concentration of proton in the solution.
pH = 6.2 for this solution. As a result, 
.
![[\text{H}^{+}] \cdot [\text{OH}^{-}] = \text{K}_w](https://tex.z-dn.net/?f=%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%20%5Ccdot%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%20%5Ctext%7BK%7D_w)
, where ![[\text{OH}^{-}]](https://tex.z-dn.net/?f=%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D)
the concentration of hydroxide ions and 
is the dissociation constant of water.
at 0.10 MPa and 25 °C. As a result, ![[\text{OH}^{-}] = \text{K}_w / [\text{H}^{+}] = 10^{14} / (6.31 \times 10^{-7}) = 1.6 \times 10^{-8} \; \text{mol}\cdot \text{dm}^{-3}](https://tex.z-dn.net/?f=%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%20%5Ctext%7BK%7D_w%20%2F%20%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%20%3D%2010%5E%7B14%7D%20%2F%20%286.31%20%5Ctimes%2010%5E%7B-7%7D%29%20%3D%201.6%20%5Ctimes%2010%5E%7B-8%7D%20%5C%3B%20%5Ctext%7Bmol%7D%5Ccdot%20%5Ctext%7Bdm%7D%5E%7B-3%7D)
.
Answer:
calcium phosphate has the formula Ca3(PO4)2, which has the mass of 310 grams/mol.
1 mol contains 310 grams
2.3*10^-4 moles contain 2.3*10^-4 * 310, which means 713*10^-4 grams, or 71.3 milligrams.
If you wrote the formula right and named the compound wrong, all you have to do is replace 310 with 278 and the answer will be 639.4*10^-4 grams, or 63.94 milligrams.
Answer:
the one with less thermal energy
Explanation:
thermal energy is heat
