Answer:
CH4 < CH3CH3 < CH3CH2CH3
Explanation:
Alkanes are saturated aliphatic hydrocarbons that undergoes intermolecular Van der waals forces. Van der waals forces are the attractive forces which make it possible for non-polar molecules to form liquids and solids.
Van der waals force are described as intermolecular forces arising from induced fluctuating dipoles in atoms and molecules brought about by movement of electrons around the atomic nucleus.
An example of the Van der waals force is the london dispersion force that occurs in the alkane family. It is the weakest of all electrical forces that act between atoms and molecules, These forces are responsible for liquefaction or solidification of non-polar substances at low temperature.
The strength of the intermolecular forces is based on the number of electrons surrounding the molecule and the surface area of the molecule. SO, in alkanes, the longer the carbon chain, the more stronger the intermolecular forces.
Answer:
1). 1 mole of Carbon burnt in air
C + O2 →CO2
1 mole of carbon produces 1 mole of CO2 which is 44g of CO2
2). 1 mole of carbon is burnt in 16g of dioxygen
32g of O2 = 44g of CO2
1g of O2 = 44/32
CO2 (Dioxygen is limiting reagent)
16g of O2 = 4/32 × 16 = 22g of CO2 in one mole
3) 2 moles of Carbon burnt in 16g of dioxygen
16g of dioxygen is available, and thus it can combine with 0.5 mol of carbon to give 22g of CO2
Answer:
8.8g of Al are necessaries
Explanation:
Based on the reaction, 2 moles of Al are required to produce 3 moles of hydrogen gas.
To solve this question we must find the moles of H2 in 11L at STP using PV = nRT. With these moles we can find the moles of Al required and its mass as follows:
<em>Moles H2:</em>
PV = nRT; PV/RT = n
<em>Where P is pressure = 1atm at STP; V is volume = 11L; R is gas constant = 0.082atmL/molK and T is absolute temperature = 273.15K at STP</em>
Replacing:
1atm*11L/0.082atmL/molK*273.15K = n
n = 0.491 moles of H2 must be produced
<em />
<em>Moles Al:</em>
0.491 moles of H2 * (2mol Al / 3mol H2) = 0.327moles of Al are required
<em />
<em>Mass Al -Molar mass: 26.98g/mol-:</em>
0.327moles of Al * (26.98g / mol) = 8.8g of Al are necessaries
The time taken for the isotope to decay is 46 million years.
We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:
- Original amount (N₀) = 50.25 g
- Amount remaining (N) = 16.75
- Number of half-lives (n) =?
2ⁿ = N₀ / N
2ⁿ = N₀ / N
2ⁿ = 50.25 / 16.75
2ⁿ = 3
Take the log of both side
Log 2ⁿ = 3
nLog 2 = Log 3
Divide both side by log 2
n = Log 3 / Log 2
n = 2
Finally, we shall determine the time.
- Half-life (t½) = 23 million years
- Number of half-lives (n) = 2
t = n × t½
t = 2 × 23
t = 46 million years
Learn more about half-life: brainly.com/question/25927447
