It is an Alkene because it has a double bond, so it’ll have “ene” at the end. The simplest Alkene has 2 carbons.
2 carbons = “eth”
Look at that! Two carbons! It must be “ethene”
Answer:
0.924 g
Explanation:
The following data were obtained from the question:
Volume of CO2 at RTP = 0.50 dm³
Mass of CO2 =?
Next, we shall determine the number of mole of CO2 that occupied 0.50 dm³ at RTP (room temperature and pressure). This can be obtained as follow:
1 mole of gas = 24 dm³ at RTP
Thus,
1 mole of CO2 occupies 24 dm³ at RTP.
Therefore, Xmol of CO2 will occupy 0.50 dm³ at RTP i.e
Xmol of CO2 = 0.5 /24
Xmol of CO2 = 0.021 mole
Thus, 0.021 mole of CO2 occupied 0.5 dm³ at RTP.
Finally, we shall determine the mass of CO2 as follow:
Mole of CO2 = 0.021 mole
Molar mass of CO2 = 12 + (2×16) = 13 + 32 = 44 g/mol
Mass of CO2 =?
Mole = mass /Molar mass
0.021 = mass of CO2 /44
Cross multiply
Mass of CO2 = 0.021 × 44
Mass of CO2 = 0.924 g.
3/5 times 5/3x = 8*3/5. X=24/5 simplified would be x= 4.8 L.
Charles law gives the relationship between volume and temperature of gas.
It states that at constant pressure volume is directly proportional to temperature
Therefore
V/ T = k
Where V - volume T - temperature in kelvin and k - constant
V1/T1 = V2/T2
Parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
Substituting the values in the equation
267 L/ 480 K = V / 750 K
V = 417 L
Final volume is 417 L
I believe you just look at your periodic table for this value. I don't think there is any math involved.
Therefore one mole of Mg = 24.305g.
