1) number of moles of N2 = n/2
2) Number of moles of CH4 = n/2
3) Total number of moles of the mixture = n/2 + n/2 = n
4) Kg of N2
mass in grams = number of moles * molar mass
molar mass of N2 = 2 * 14.0 g/mol = 28 g/mol
=> mass of N2 in grams = (n/2) * 28 = 14n
mass of N2 in Kg = mass of N2 in grams * [1 kg / 1000g] = 14n/1000 kg = 0.014n kg
Answer: mass of N2 in kg = 0.014n kg
The problem applies Charles' law since constant pressure with varying volume and temperature are given. Assuming ideal gas law, the equation to be used is
=
. We make sure the temperatures are expressed in Kelvin, hence the given added with 273. The volume 2 is equal to 25.2881 liters.
Answer:
75 kJ/mol
Explanation:
The reactions occur at a rate, which means that the concentration of the reagents decays at a time. The rate law is a function of the concentrations and of the rate constant (k) which depends on the temperature of the reaction.
The activation energy (Ea) is the minimum energy that the reagents must have so the reaction will happen. The rate constant is related to the activation energy by the Arrhenius equation:
ln(k) = ln(A) -Ea/RT
Where A is a constant of the reaction, which doesn't depend on the temperature, R is the gas constant (8.314 J/mol.K), and T is the temperature. So, for two different temperatures, if we make the difference between the two equations:
ln(k1) - ln(k2) = ln(A) - Ea/RT1 - ln(A) + Ea/RT2
ln (k1/k2) = (Ea/R)*(1/T2 - 1/T1)
k1 = 8.3x10⁸, T1 = 142.0°C = 415 K
k2 = 6.9x10⁶, T2 = 67.0°C = 340 K
ln(8.3x10⁸/6.9x10⁶) = (Ea/8.314)*(1/340 - 1/415)
4.8 = 6.39x10⁻⁵Ea
Ea = 75078 J/mol
Ea = 75 kJ/mol
Answer:
The volume of reactant measured at STP left over is 409.9 L
Explanation:
Answer:
See attached picture.
Explanation:
Hello,
In this case, on the attached picture you will find the required line structures for the cis and trans configurations of the given compound (2-pentene). Take into account for the cis that the adjacent carbons to those having the double bond remain in the same plane, whereas for the trans one, the adjacent carbons remain in a different plane.
Regards-
