The balanced equation for the formation of ammonia is as follows
N₂ + 3H₂ ---> 2NH₃
stoichiometry of N₂ to H₂ is 1:3
we need to find the moles of N₂, volume of N₂ has been given
molar volume is where 1 mol of any gas occupies a volume of 22.4 L at STP.
if 22.4 L is occupied by 1 mol
then 3.5 L of gas is occupied by - 3.5 L / 22.4 L/mol = 0.16 mol
number of moles of N₂ present - 0.16 mol
1 mol of N₂ requires 3 mol of H₂
therefore 0.16 mol of N₂ requires - 3 x 0.16 = 0.48 mol of H₂
mass of H₂ required - 0.48 mol x 2 g/mol = 0.96 g
0.96 g of H₂ is required
Answer:
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Explanation:
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Answer:
Atomic radius of sodium = 227 pm
Atomic radius of potassium = 280 pm
Explanation:
Atomic radii trend along group:
As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.
Consider the example of sodium and potassium.
Sodium is present above the potassium with in same group i.e, group one.
The atomic number of sodium is 11 and potassium 19.
So potassium will have larger atomic radius as compared to sodium.
Atomic radius of sodium = 227 pm
Atomic radius of potassium = 280 pm
Answer : The pressure of the helium gas is, 1269.2 mmHg
Explanation :
To calculate the pressure of the gas we are using ideal gas equation:
where,
P = Pressure of 
gas = ?
V = Volume of gas = 210. mL = 0.210 L (1 L = 1000 mL)
n = number of moles = 0.0130 mole
R = Gas constant = 
T = Temperature of gas = 
Putting values in above equation, we get:
Conversion used : (1 atm = 760 mmHg)
Thus, the pressure of the helium gas is, 1269.2 mmHg
