Answer:
ΔH°c = -2219.9 kJ
Explanation:
Let's consider the combustion of propane.
C₃H₈(g) + 5 O₂(g) ⟶ 3 CO₂(g) + 4 H₂O(l)
We can find the standard enthalpy of the combustion (ΔH°c) using the following expression.
ΔH°c = [3 mol × ΔH°f(CO₂(g)) + 4 mol × ΔH°f(H₂O(l))] - [1 mol × ΔH°f(C₃H₈(g)) + 5 mol × ΔH°f(O₂(g))]
ΔH°c = [3 mol × (-393.5 kJ/mol) + 4 mol × (-285.8 kJ/mol)] - [1 mol × (-103.8 kJ/mol) + 5 mol × (0 kJ/mol)]
ΔH°c = -2219.9 kJ
The answer is (4) Add enough solvent to 30.0 g of solute to make 1.0 L solution. The molarity is calculated using volume of the solution. When solute dissolving, the total volume will change. So the final volume of solution need to be 1.0 L.
Answer: CuI₂ + Br₂
Explanation:
1) The activity series F > Cl > Br > I means that F is the most active and I is the least active of those four elements (the halogens, group 17 in the periodic table).
The activity is a measure of how eager is an element to react compared to other elements in the series in a single replacement reaction.
2) Choice 1: CuI₂ + Br₂
Since the activity of Br is higher than that of I, Br will react with CuI₂, displacing I, which will be left alone, as per this chemical equation:
CuI₂ + Br₂ → CuBr₂ + I₂
Being I less active than Br, it cannot displace Br in CuBr₂.
3) Choice 2: Cl₂ + AlF₃
Being Cl less active than F, the former will not displace the latter, and the reaction will not proceed.
4) Choice 3: Br₂ + NaCl
Again, being Br less active than Cl, the former will not displace the latter, and the reaction will not proceed.
5) Choice 4: CuF₂ + I₂
Once more, being I less active than F, the former will not displace the latter, and the reaction will not proceed.
Answer:
Explanation:
Ketcher 01232019462D 1 1.00000 0.00000 0 5 4 0 0 0 999 V2000 -0.0330 2.2250 0.0000 C 0 0 0 0 0 0 0 0 0 0 0.8330 2.7250 0.0000 C 0 0 0 0 0 0 0 0 0 0 1.6990 2.2250 0.0000 C 0 0 0 0 0 0 0 0 0 0 0.8330 3.7250 0.0000 C 0 0 0 0 0 0 0 0 0 0 1.6990 1.2250 0.0000 C 0 0 0 0 0 0 0 0 0 0 1 2 1 0 0 0 2 3 1 0 0 0 2 4 1 0 0 0 3 5 1 0 0 0 M END
Answer:
3.24 × 10^5 J/mol
Explanation:
The activation energy of this reaction can be calculated using the equation:
ln(k2/k1) = Ea/R x (1/T1 - 1/T2)
Where; Ea = the activation energy (J/mol)
R = the ideal gas constant = 8.3145 J/Kmol
T1 and T2 = absolute temperatures (K)
k1 and k2 = the reaction rate constants at respective temperature
First, we need to convert the temperatures in °C to K
T(K) = T(°C) + 273.15
T1 = 325°C + 273.15
T1 = 598.15K
T2 = 407°C + 273.15
T2 = 680.15K
Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea
ln(k2/k1) = Ea/R x (1/T1 - 1/T2)
ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)
ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4
7.831 = Ea(2.417 × 10^-5)
Ea = 3.24 × 10^5 J/mol