Answer: the name of the reaction is Combustion
Explanation: Combustion is a reaction in which a substance burns in air (oxygen) to produce CO2 and H20
Answer:
Model A
Explanation:
Model A represents an atom that is more reactive than the others represented.
Valence electrons actually determine the reactivity of elements. They also determine the properties of elements.
Elements with one valence electron are highly reactive because they need low energy to remove them. They can either gain more electrons to become stable or they share/give out their electrons.
Therefore, Model A is the correct answer because it has one valence electron and its valence electron is farther from the nucleus thereby this makes it more reactive.
According to Avagadros law 1 mole of a substance contains 6.02 × 10^23 particles.
Thus, 1 mole of C6H14 contains 6.02 × 10^ 23 molecules,
Therefore, 9.25 × 10^24 molecules will have;
(9.25 × 10^24)/ (6.02 ×10^23)
= 15.365 moles.
Answer:
Vapor pressure = 0.7566atm
Explanation:
The approach applied is raoult's law of ideal solution which says that the vapor pressure of a solvent in a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction in the solution.
Mathematically, pA = NA x P'A
pB = NB x p'B
And, pT = pA + pB, not all liquids obey raoult's law, the mixtures which obey roault's law are called IDEAL SOLUTIONS.
The step by step calculations is shown in the attachment
Answer:
-88.66 kJ/mol
Explanation:
The expressions of heat capacity (Cp,m) for C(s) and for H₂(g) are:
C(s): Cp,m/(J K-1 mol-1) = 16.86 + (4.77T/10³) - (8.54x10⁵/T²)
H₂(g): Cp,m/(J K-1 mol-1) = 27.28 + (3.26T/10³) + (0.50x10⁵/T²)
Cp = A + BT + CT⁻²
For the Kirchoff's Law:
ΔHf = ΔH°f +
Where ΔH°f is the enthalpy at 298 K, T1 is 298 K, T2 is the temperature given (373 K), and DCp is the variation of Cp (products less reactants). ΔH°f for ethene is -84.68 kJ/mol and the reaction is:
2C(s) + 3H₂(g) → C₂H₆
So, DCp:
dA = A(C₂H₆) - [2xA(C) + 3xA(H₂)] = 14.73 - [2x16.86 + 3x27.28] = -100.83
dB = B(C₂H₆) - [2xB(C) + 3xB(H₂)] = 0.1272 - [2x4.77x10⁻³ + 3x3.26x10⁻³] = 0.10788
dC = C(C₂H₆) - [2xC(C) + 3xC(H₂)] = 0 - (2x(-8.54x10⁵) + 3x0.50x10⁵) = 15.58x10⁵
dCp = -100.83 + 0.10788T + 15.58x10⁵T⁻²
= -3796.48 J/mol = -3.80 kJ/mol (solved by a graphic calculator)
ΔHf = -84.68 - 3.80
ΔHf = -88.66 kJ/mol