Answer:
neq N2O4 = 0.9795 mol.....P = 0.5 atm; T = 25°C
Explanation:
ni change eq.
N2O4 1 1 - x 0.8154.....P = 1 atm; T = 25°C
NO2 0 0 + x x
∴ x = neq = Peq.V / R.T.....ideal gas mix
if P = 0.5 atm, T = 25°C; assuming: V = 1 L
⇒ x = neq = ((0.5 atm)(1 L))/((0.082 atm.L/K.mol)(298 K))
⇒ x = neq = 0.0205 mol
⇒ neq N2O4 = 1 - x = 1 - 0.0205 = 0.9795 mol
Answer:
C
Explanation:
the enthalpy of reaction is independent of the reaction path
1) Calcium carbonate contains 40.0% calcium by weight.
M(CaCO₃)=100.1 g/mol
M(Ca)=40.1 g/mol
w(Ca)=40.1/100.1=0.400 (40.0%)!
2) Mass fraction of this is excessive data.
3) The solution is:
m(Ca)=1.2 g
m(CaCO₃)=M(CaCO₃)*m(Ca)/M(Ca)
m(CaCO₃)=100.1g/mol*1.2g/40.1g/mol=3.0 g
Answer:
To release 7563 kJ of heat, we need to burn 163.17 grams of propane
Explanation:
<u>Step 1</u>: Data given
C3H8 + 5O2 -----------> 3CO2 + 4H2O ΔH° = –2044 kJ
This means every mole C3H8
Every mole of C3H8 produces 2044 kJ of heat when it burns (ΔH° is negative because it's an exothermic reaction)
<u>Step 2: </u>Calculate the number of moles to produce 7563 kJ of heat
1 mol = 2044 kJ
x mol = 7563 kJ
x = 7563/2044 = 3.70 moles
To produce 7563 kJ of heat we have to burn 3.70 moles of C3H8
<u>Step 3: </u>Calculate mass of propane
Mass propane = moles * Molar mass
Mass propane = 3.70 moles * 44.1 g/mol
Mass propane = 163.17 grams
To release 7563 kJ of heat, we need to burn 163.17 grams of propane