Answer : 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.
Solution : Given,
Mass of 
= 100 g
Molar mass of = 27 g/mole
Molar mass of 
= 28 g/mole
First we have to calculate moles of .
The given balanced chemical reaction is,
From the given reaction, we conclude that
2 moles of produced from 1 mole of
3.7 moles of produced from 
of
Now we have to calculate the mass of .
Mass of = Moles of × Molar mass of
Mass of = 1.85 mole × 28 g/mole = 51.8 g
Therefore, 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.
<span>They both have charged particles
They have the same attractive forces between particles
They have the same space between particles
They create magnetic and electric fields</span>
Answer:
T° freezing solution → -11.3°C
T° boiling solution → 103.1 °C
Explanation:
Assuming 100 % dissociation, we must find the i, Van't Hoff factor which means "the ions that are dissolved in solution"
This salt dissociates as this:
SnCl₄ (aq) → 1Sn⁴⁺ (aq) + 4Cl⁻ (aq) (so i =5)
The formula for the colligative property of freezing point depression and boiling point elevation are:
ΔT = Kf . m . i
where ΔT = T° freezing pure solvent - T° freezing solution
ΔT = Kb . m . i
where ΔT = T° boiling solution - T° boiling pure solvent
Freezing point depression:
0° - T° freezing solution = 1.86°C/m . 1.22 m . 5
T° freezing solution = - (1.86°C/m . 1.22 m . 5) → -11.3°C
Boiling point elevation:
T° boiling solution - 100°C = 0.512 °C/m . 1.22 m . 5
T° boiling solution = (0.512 °C/m . 1.22 m . 5) + 100°C → 103.1 °C
<h2>Freezing:</h2>
The transformation of liquid into solid when it reaches when it's temperature is below freezing point.
