Yeah of-course!! It's valency by group most of the chemical property like electronegativity, ionization energy etc. by the combination of groups and periods...
Answer:
Theoretical yield of the reaction = 34 g
Excess reactant is hydrogen
Limiting reactant is nitrogen
Explanation:
Given there is 100 g of nitrogen and 100 g of hydrogen
Number of moles of nitrogen = 100 ÷ 28 = 3·57
Number of moles of hydrogen = 100 ÷ 2 = 50
Reaction between nitrogen and hydrogen yields ammonia according to the following chemical equation
N2 + 3H2 → 2NH3
From the above chemical equation for every mole of nitrogen that reacts, 3 moles of hydrogen will be required and 2 moles of ammonia will be formed
Now we have 3·57 moles of nitrogen and therefore we require 3 × 3·57 moles of hydrogen
⇒ We require 10·71 moles of hydrogen
But we have 50 moles of hydrogen
∴ Limiting reactant is nitrogen and excess reactant is hydrogen
From the balanced chemical equation the yield will be 2 × 3·57 moles of ammonia
Molecular weight of ammonia = 17 g
∴ Theoretical yield of the reaction = 2 × 3·57 × 17 = 121·38 g
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Answer:</h3>
The gas obeys the Boyle's law
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Explanation:</h3>
- According to Boyle's law, the volume of a fixed mass of a gas and the pressure are inversely proportional at constant absolute temperature.
- That is;

- Therefore,
, where k is a constant - At varying volume and pressures while keeping absolute temperature constant; k = P1V1 =P2V2
In this case, we are given;
Initial Volume of 13.0 L at initial pressure of 1.0 atm
New volume of 6.5 L at new pressure of 2.0 atm
But, K = PV
Therefore,
k1 = P1V1
= 1.0 atm × 13.0 L
= 13 atm.L
k2 = P2V2
= 2.0 atm × 6.5 L
= 13 atm.L
Thus, k1=k2
Thus, the gas obeys the Boyle's law
Answer:
Explanation:
From the given information:
The density of O₂ gas = 
here:
P = pressure of the O₂ gas = 310 bar
= 
= 305.97 atm
The temperature T = 415 K
The rate R = 0.0821 L.atm/mol.K
molar mass of O₂ gas = 32 g/mol
∴

= 287.37 g/L
To find the density using the Van der Waal equation
Recall that:
the Van der Waal constant for O₂ is:
a = 1.382 bar. L²/mol² &
b = 0.0319 L/mol
The initial step is to determine the volume = Vm
The Van der Waal equation can be represented as:

where;
R = gas constant (in bar) = 8.314 × 10⁻² L.bar/ K.mol
Replacing our values into the above equation, we have:



After solving;
V = 0.1152 L
∴

= 277.77 g/L
We say that the repulsive part of the interaction potential dominates because the results showcase that the density of the Van der Waals is lesser than the density of ideal gas.