Given that there is 48 liters of gasoline to be burned and that 45 kJ of energy is released per gram of gasoline burned, the amount of energy that the gasoline fuel produces can then be calculated, First, we convert 48 liters of gasoline to units of mass (grams) in order to use the given conversion of 45 kJ per gram of gasoline. To do this, we use the density of gasoline which is 0.77 g/mL. The following expression is then used:
48 L gasoline x 1000 mL/L x 0.77 g/mL x 45 kJ/g gasoline = 1663200 kJ
<span>The amount of energy produced by burning 48 L of gasoline was then determined to be 1663200 kJ. </span>
Well if anything was mixed in it would make his result different. if not, it could have changed the temperature when he mixed it. which would give him either a lower or higher answer
No, more than 1 paratical
Answer:
1.D) CH4 + 2O2 → CO2 + 2H2O
2C)2,1,2
3.D)SIO2 + 4HF → SIF4 + 2H2O
4.D)2As + 6NaOH → 2Na3AsO3 + 3H2
5.C)SiCI4 + 2H2O → SiO2 + 4HCI
Explanation:
equal number of atoms of each elements in both the sides of the chemical equation is required to have an equation in balanced state.
Answer: The heat required is 6.88 kJ.
Explanation:
The conversions involved in this process are :

Now we have to calculate the enthalpy change.
![\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bm%5Ctimes%20c_%7Bp%2Cs%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bn%5Ctimes%20%5CDelta%20H_%7Bfusion%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bn%5Ctimes%20%5CDelta%20H_%7Bvap%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cg%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= enthalpy change = ?
m = mass of ethanol = 25.0 g
= specific heat of solid ethanol= 0.97 J/gK
= specific heat of liquid ethanol = 2.31 J/gK
n = number of moles of ethanol = 
= enthalpy change for fusion = 5.02 KJ/mole = 5020 J/mole
= change in temperature
The value of change in temperature always same in Kelvin and degree Celsius.
Now put all the given values in the above expression, we get
![\Delta H=[25.0 g\times 0.97J/gK\times (-114-(-135)K]+0.534mole\times 5020J/mole+[25.0g\times 2.31J/gK\times (-50-(-114))K]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B25.0%20g%5Ctimes%200.97J%2FgK%5Ctimes%20%28-114-%28-135%29K%5D%2B0.534mole%5Ctimes%205020J%2Fmole%2B%5B25.0g%5Ctimes%202.31J%2FgK%5Ctimes%20%28-50-%28-114%29%29K%5D)
(1 KJ = 1000 J)
Therefore, the heat required is 6.88 kJ