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
Learn more about empirical formula:
brainly.com/question/24297883
#SPJ1
Answer:
The electrons in an atom move around the nucleus in regions known as electron shells.
By international agreement, absolute zero is defined as precisely; 0 K on the Kelvin scale, which is a thermodynamic (absolute) temperature scale; and –273.15 degrees Celsius on the Celsius scale.
Answer:
Explanation:
E = hc/λ
h = planck's constant = 6.66 x 10 ^ -34
c = speed of light = 2.98 *10^8 m/s
E = (6.66 x 10 ^ -34 )(2.98 *10^8)/32
= 19.85 * 10 ^-26
=0.62 x 10^-26
= 6.2 x 10^-27 J
Answer:
The final concentration is 0.226 M.
Explanation:
In this problem the dilution has occurred two times. So the symbol initial concentration, concentration after first dilution and the concentration after second dilution is given as C1, C2, and C3 respectively. Same can be done for the volume i.e, V1, V2, and V3 will be the initial volume, the volume taken after first dilution, and the volume after second dilution.
So, let's use the dilution formula for the first dilution


Now, for the second dilution
