Combustion equation of n-hexane:
2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O
a)
Assuming we have 100 moles of air,
Oxygen = 20.9 moles
n-hexane required = 20.9/19 x 2
= 2.2 moles
LFL = Half of stoichometric amount = 2.2 / 2 = 1.1
LFL n-hexane = 1.1%
b)
1.1 volume percent required for LFL
1.1% x 1
= 0.0011 m³ of n-hexane required
Answer:
410 nm
Explanation:
The highest frequency is associated with the shortest wavelength. Of those listed, 410 nm is the shortest.
If I have 0.725 moles of gas at a temperature of 105 K and a pressure of 3.75 atmospheres the volume of the gas 1.66 litres.
Explanation:
Data given:
number of moles of the gas = 0.725
temperature = 105 K
pressure = 3.75 atm
volume of the gas =?
R = 0.08206 Latm/mole Kelvin
Applying the ideal gas law to calculate the volume of the given gas:
PV = nRT
rearranging the equation to calculate volume:
V = 
putting the values in the equation:
V = 
V = 1.66 Litres.
At a temperature of 105 K and pressure of 3.75 atm, 0.725 moles of gas occupy 1.66 litres of volume.
Answer : The correct option is, (B) 2, 3 and 6.
Explanation :
Combustion reaction : It is a type of reaction where a hydrocarbon react with an oxygen molecule to give carbon dioxide, water as a product.
In general,

For example : Ethane react with oxygen to give carbon dioxide and water.

In the given list,
is a reactant and
are the products.
Therefore, the correct answer is, (B) 2, 3, and 6