Option E, Real gas particles have more complex interactions than ideal gas particles.
In ideal gases, there is absolutely no interaction between any atoms. At all. Atoms simply don't bump into each other in ideal gases.
Obviously, you know that's unrealistic. In real gases, atoms collide into each other all the time.
-T.B.
Excess reactant : Na
NaCl produced : = 16.497 g
<h3>Further explanation</h3>
Given
Reaction(balanced)
2Na + Cl₂⇒ 2NaCl
20 g Na
10 g Cl₂
Required
Excess reactant
NaCl produced
Solution
mol Na(Ar = 23 g/mol) :
= 20 : 23 = 0.87
mol Cl₂(MW=71 g/mol):
= 10 : 71 g/mol = 0.141
mol : coefficient :
Na = 0.87 : 2 = 0.435
Cl₂ = 0.141 : 1 = 0.141
Limiting reactant : Cl₂(smaller ratio)
Excess reactant : Na
Mol NaCl based on mol Cl₂, so mol NaCl :
= 2/1 x mol Cl₂
= 2/1 x 0.141
= 0.282
Mass NaCl :
= 0.282 x 58.5 g/mol
= 16.497 g
Answer:
V = 22.42 L/mol
N₂ and H₂ Same molar Volume at STP
Explanation:
Data Given:
molar volume of N₂ at STP = 22.42 L/mol
Calculation of molar volume of N₂ at STP = ?
Comparison of molar volume of H₂ and N₂ = ?
Solution:
Molar Volume of Gas:
The volume occupied by 1 mole of any gas at standard temperature and pressure and it is always equal to 22.42 L/ mol
Molar volume can be calculated by using ideal gas formula
PV = nRT
Rearrange the equation for Volume
V = nRT / P . . . . . . . . . (1)
where
P = pressure
V = Volume
T= Temperature
n = Number of moles
R = ideal gas constant
Standard values
P = 1 atm
T = 273 K
n = 1 mole
R = 0.08206 L.atm / mol. K
Now put the value in formula (1) to calculate volume for 1 mole of N₂
V = 1 x 273 K x 0.08206 L.atm / mol. K / 1 atm
V = 22.42 L/mol
Now if we look for the above calculation it will be the same for H₂ or any gas. so if we compare the molar volume of 1 mole N₂ and H₂ it will be the same at STP.
Answer: 0.05
Explanation:Divide the length (1.5 cm) by the number of turns (30)
Number of moles : n₂ = 1.775 moles
<h3>Further explanation</h3>
Given
Moles = n₁ = 1.4
Volume = V₁=22.4 L
V₂=28.4 L
Required
Moles-n₂
Solution
Avogadro's hypothesis, at the same temperature and pressure, the ratio of gas volume will be equal to the ratio of gas moles
The ratio of gas volume will be equal to the ratio of gas moles
Input the values :
n₂ = (V₂ x n₁)/V₁
n₂ = (28.4 x 1.4)/22.4
n₂ = 1.775 moles
