Answer:
Ethane would have a higher boiling point.
Explanation:
In this case, for the lewis structures, we have to keep in mind that all atoms must have <u>8 electrons</u> (except hydrogen). Additionally, each carbon would have <u>4 valence electrons</u>, with this in mind, for methane we have to put the hydrogens around the carbon, and with this structure, we will have 8 electrons for the carbon. In ethane, we will have a bond between the carbons, therefore we have to put three hydrogens around each carbon to obtain 8 electrons for each carbon.
Now, the main difference between methane and ethane is an <u>additional carbon</u>. In ethane, we have an additional carbon, therefore due to this additional carbon, we will have <u>more area of interaction</u> for ethane. If we have more area of interaction we have to give <u>more energy</u> to the molecule to convert from liquid to gas, so, the ethane will have a higher boiling point.
I hope it helps!
Answer:
5.6 seconds
Explanation:
The reaction follows a zero-order in dinitrogen monoxide
Rate = k[N20]^0 = change in concentration/time
[N20]^0 = 1
Time = change in concentration of N2O/k
Initial number of moles of N2O = 300 mmol = 300/1000 = 0.3 mol
Initial concentration = moles/volume = 0.3/4 = 0.075
Number of moles after t seconds = 150 mmol = 150/1000 = 0.15 mol
Concentration after t seconds = 0.15/4 = 0.0375 M
Change in concentration of N2O = 0.075 - 0.0375 = 0.0375 M
k = 0.0067 M/s
Time = 0.0375/0.0067 = 5.6 s
1. The empirical formula of the hydrocarbon is CH₃
2. The molecular formula of the hydrocarbon is C₂H₆
<h3>How to determine the mass of Carbon </h3>
- Mass of CO₂ = 1.47 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 1.47
Mass of C = 0.4 g
<h3>How to determine the mass of H</h3>
- Mass of compound = 0.5 g
- Mass of C = 0.4 g
- Mass of H = ?
Mass of H = (mass of compound) – (mass of C)
Mass of H = 0.5 – 0.4
Mass of H =0.1 g
<h3>1. How to determine the empirical formula </h3>
- C = 0.4 g
- H = 0.1 g
- Empirical formula =?
Divide by their molar mass
C = 0.4 / 12 = 0.03
H = 0.1 / 1 = 0.1
Divide by the smallest
C = 0.03 / 0.03 = 1
H = 0.1 / 0.03 = 3
Thus, the empirical formula of the compound is CH₃
<h3>2. How to determine the molecular formula</h3>
- Empirical formula = CH₃
- Molar mass = 30 g/mol
- Molecular formula =?
Molecular formula = empirical × n = mass number
[CH₃]n = 30
[12 + (3×1)]n = 30
15n = 30
Divide both side by 15
n = 30 / 15
n = 2
Molecular formula = [CH₃]n
Molecular formula = [CH₃]₂
Molecular formula = C₂H₆
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Answer: option B. The kinetic energy of gas molecules is directly proportional to the Kelvin temperature of the gas.
Explanation:
The kinetic theory of gases explains the behavior and properties of gases from a molecular perspective.
Specifically and explicity, the kinetic theory of gases states that gases are constituted by particles (molecules) and that the average kinetic energy of the particles is proportional to the absolute temperature (Kelvin scale) of the gas. Furthermore, the temperature of all the (ideal) gases is the same at a given temperature.
Hence, you know that the higher the temperature of the gas, the higher the kinetic energy and the average speed of the molecules.
Other postulates of the kinetic theory of gases are that: i) the volume of the particles is neglectible; ii) the particles do not exhibit intermolecular attraction or repulsion; iii) the particles are in continuous random motion in straight paths, until they collide with other particles or the walls of the vessel, and iv) the collisions are elastic (the energy is conserved).
Answer: The forward and reverse reactions eventually reach the same rate.
Explanation:
The reactions which do not go on completion and in which the reactant forms product and the products goes back to the reactants simultaneously are known as equilibrium reactions.
Equilibrium state is the state when reactants and products are present but the concentrations does not change with time.
The equilibrium is dynamic in nature and the reactions are continuous in nature. Rate of forward reaction is equal to the rate of backward reaction and it appears as it has stopped.
