Answer : The enthalpy change of dissolution of
is -54.0 kJ/mole
Explanation :
![q=m\times c\times \Delta T](https://tex.z-dn.net/?f=q%3Dm%5Ctimes%20c%5Ctimes%20%5CDelta%20T)
where,
q = heat released by the solution
c = specific heat of water = ![4.184J/g^oC](https://tex.z-dn.net/?f=4.184J%2Fg%5EoC)
m = mass of solution = mass of water + mass of
= 102.00 + 1.69 = 103.69 g
= change in temperature = ![T_2-T_1=(25.00-23.48)=1.52^oC](https://tex.z-dn.net/?f=T_2-T_1%3D%2825.00-23.48%29%3D1.52%5EoC)
Now put all the given values in the above formula, we get:
![q=103.69g\times 4.184J/g^oC\times 1.52^oC](https://tex.z-dn.net/?f=q%3D103.69g%5Ctimes%204.184J%2Fg%5EoC%5Ctimes%201.52%5EoC)
![q=659.4J=0.659kJ](https://tex.z-dn.net/?f=q%3D659.4J%3D0.659kJ)
Now we have to calculate the enthalpy change of dissolution of ![KClO_4](https://tex.z-dn.net/?f=KClO_4)
![\Delta H=-\frac{q}{n}](https://tex.z-dn.net/?f=%5CDelta%20H%3D-%5Cfrac%7Bq%7D%7Bn%7D)
where,
= enthalpy change of dissolution = ?
q = heat released = 0.659 kJ
m = mass of
= 1.69 g
Molar mass of
= 138.55 g/mol
![\text{Moles of }KClO_4=\frac{\text{Mass of }KClO_4}{\text{Molar mass of }KClO_4}=\frac{1.69g}{138.55g/mole}=0.0122mole](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DKClO_4%3D%5Cfrac%7B%5Ctext%7BMass%20of%20%7DKClO_4%7D%7B%5Ctext%7BMolar%20mass%20of%20%7DKClO_4%7D%3D%5Cfrac%7B1.69g%7D%7B138.55g%2Fmole%7D%3D0.0122mole)
![\Delta H=-\frac{0.659kJ}{0.0122mole}=-54.0kJ/mole](https://tex.z-dn.net/?f=%5CDelta%20H%3D-%5Cfrac%7B0.659kJ%7D%7B0.0122mole%7D%3D-54.0kJ%2Fmole)
Therefore, the enthalpy change of dissolution of
is -54.0 kJ/mole
We know that the equation for density is:
![D=\frac{m}{V}](https://tex.z-dn.net/?f=%20D%3D%5Cfrac%7Bm%7D%7BV%7D%20%20%20)
where D is the density, m is the mass in grams, and V is the volume.
Given two of the variables, we can then solve for density:
![D=\frac{1250g}{1500mL}=\frac{0.833g}{mL}](https://tex.z-dn.net/?f=%20D%3D%5Cfrac%7B1250g%7D%7B1500mL%7D%3D%5Cfrac%7B0.833g%7D%7BmL%7D%20%20)
Therefore, we now know that the density of the gasoline is 0.833g/mL.
He answers is B It equals the rate of the formation of SO3
At a temperature of 30 deg C, the vapour pressure of water
H2O is about 32 mm Hg. Therefore at a total pressure f 734 mm Hg, the partial
pressure of the Hydrogen gas collected is:
<span>P Hydrogen = 734 mm Hg – 32 mm Hg = 702 mm Hg</span>