Answer:
The empirical formula is .
Explanation:
Given that, the combustion of 29.5 mg produced 80.2 mg CO₂ and 16.4 mg of H₂O.
Mass of C
g
Mass of H
g
Mass of O =
g
Number of mole of C
mol
Number of mole of H
mol
Number of mole O
mol
Dividing by lowest integer
The empirical formula is .
Answer:
The volume of ammonia produced by 4.2 L of hydrogen and excess nitrogen is 2.8 L.
Explanation:
Given data:
Volume of H₂ = 4.2 L
Volume of NH₃ produced = ?
Solution:
First of all we will write the balance chemical equation:
N₂ + 3H₂ → 2NH₃
Hydrogen is limiting reactant and volume ratio is,
H₂ : NH₃
3 : 2
4.2 : 2/3 × 4.2 = 2.8 L
The volume of ammonia produced by 4.2 L of hydrogen and excess nitrogen is 2.8 L.
Answer : The fuel value and the fuel density of pentane is, 49.09 kJ/g and respectively.
Explanation :
Fuel value : It is defined as the amount of energy released from the combustion of hydrocarbon fuels. The fuel value always in positive and in kilojoule per gram (kJ/g).
As we are given that:
Fuel value =
Molar mass of pentane = 72 g/mol
Fuel value =
Fuel value = 49.09 kJ/g
Now we have to calculate the fuel density of pentane.
Fuel density = Fuel value × Density
Fuel density = (49.09 kJ/g) × (0.626g/mL)
Fuel density = 30.73 kJ/mL =
Thus, the fuel density of pentane is
Yes they do theyre always in a single phase
The molecular weight of an unknown gas is 613.8 kg.
<u>Explanation:</u>
RMS or root mean square velocity is given by the formula,
RMS =
where R is the gas constant = 8.314 kgm²s⁻²K⁻¹mol⁻¹
T is the temperature in Kelvin = 50 + 273 = 323 K
Molar mass of the Nitrogen gas in kg is 0.028 kg/mol
So we can find the RMs of the nitrogen by plugin the above values in the above equation, we will get,
RMS =
= 536.4 m/s
Now we have to take half of RMS as 268.2 m/s
Now we have to rearrange the equation to get M as,
M =
Now the temperature is T = 130 + 273 = 403 K
Plugin the values as,
M =
= 613.8 kg/mol
So the molecular weight of an unknown gas is 613.8 kg.