At STP conditions the volume of 1 mol of any ideal gas will be 22.4L
0.500 mol C3H3 x 22.4L / 1 mol = 11.2 L
Answer : The vapor pressure of propane at 
is 17.73 atm.
Explanation :
The Clausius- Clapeyron equation is :
where,
= vapor pressure of propane at = ?
= vapor pressure of propane at normal boiling point = 1 atm
= temperature of propane = 
= normal boiling point of propane = 
= heat of vaporization = 24.54 kJ/mole = 24540 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:
Hence, the vapor pressure of propane at is 17.73 atm.
Answer: potential energy is stored
Explanation:
Answer:
48.32 g of anhydrous MnSO4.
Explanation:
Equation of dehydration reaction:
MnSO4 •4H2O --> MnSO4 + 4H2O
Molar mass = 55 + 32 + (4*16) + 4((1*2) + 16)
= 223 g/mol
Mass of MnSO4 • 4H2O = 71.6 g
Number of moles = mass/molar mass
= 71.6/223
= 0.32 mol.
By stoichiometry, since 1 mole of MnSO4 •4H2O is dehydrated to give 1 mole of anhydrous MnSO4
Number of moles of MnSO4 = 0.32 mol.
Molar mass = 55 + 32 + (4*16)
= 151 g/mol.
Mass = 151 * 0.32
= 48.32 g of anhydrous MnSO4.
Answer:
+125.4 KJmol-1
Explanation:
∆H C4H10(g) = -2877.6kJ/mol
∆H C(s)=-393.5kJ/mol
∆H H2(g) = -285.8
∆H reaction= ∆Hproducts - ∆H reactants
∆H reaction= (-2877.6kJ/mol) - [4(-393.5kJ/mol) +5(-285.8)]
∆H reaction= +125.4 KJmol-1
