The pressure increases as temperature increases, as a gas is heated it expands, so in a confined container (constant volume) pressure must increase, this can be observed through the universal gas law (pv=nRT)
Answer : The energy removed must be, -67.7 kJ
Solution :
The process involved in this problem are :
The expression used will be:
![\Delta H=[m\times c_{p,g}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,l}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bm%5Ctimes%20c_%7Bp%2Cg%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bm%5Ctimes%20%5CDelta%20H_%7Bvap%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= heat released by the reaction = ?
m = mass of benzene = 125 g
= specific heat of gaseous benzene = 
= specific heat of liquid benzene = 
= enthalpy change for vaporization = 
Molar mass of benzene = 78.11 g/mole
Now put all the given values in the above expression, we get:
![\Delta H=[125g\times 1.06J/g.K\times (353.0-(425.0))K]+125g\times -434.0J/g+[125g\times 1.73J/g.K\times (335.0-353.0)K]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B125g%5Ctimes%201.06J%2Fg.K%5Ctimes%20%28353.0-%28425.0%29%29K%5D%2B125g%5Ctimes%20-434.0J%2Fg%2B%5B125g%5Ctimes%201.73J%2Fg.K%5Ctimes%20%28335.0-353.0%29K%5D)
Therefore, the energy removed must be, -67.7 kJ
Answer:
0.1593 L.
Explanation:
- We can use the general law of ideal gas: PV = nRT.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and P are constant, and have two different values of V and T:
<em>P₁V₁T₂ = P₂V₂T₁</em>
<em></em>
P₁ = 600 torr/760 = 0.789 atm, V₁ = 185.0 mL = 0.185 L, T₁ = 25.0°C + 273 = 298.0 K.
P₂ (at STP) = 1.0 atm, V₂ = ??? L, T₂ (at STP = 0.0°C) = 0.0°C + 273 = 273.0 K.
<em>∴ V₂ = P₁V₁T₂/P₂T₁</em> = (0.789 atm)(0.185 mL)(298.0 K)/(1.0 atm)(273.0 K) = <em>0.1593 L.</em>
Answer:
Moles of H₂S needed = 6.2 mol
Moles of SO₂ produced = 6.2 mol
Explanation:
Given data:
Number of moles of O₂ = 9.3 mol
Moles of H₂S needed = ?
Moles of SO₂ produced = ?
Solution:
Chemical equation:
2H₂S + 3O₂ → 2SO₂ + 2H₂O
Now we will compare the moles of oxygen with H₂S.
O₂ : H₂S
3 : 2
9.3 : 2/3×9.3 = 6.2 mol
Now we will compare the moles of SO₂ with both reactant.
O₂ : SO₂
3 : 2
9.3 : 2/3×9.3 = 6.2 mol
H₂S : SO₂
2 : 2
6.2 : 6.2 mol
So 6.2 moles of SO₂ are produced.
