100°C because all the molecules are moving the fastest past each other
Answer:
40% of the ammonia will take 4.97x10^-5 s to react.
Explanation:
The rate is equal to:
R = k*[NH3]*[HOCl] = 5.1x10^6 * [NH3] * 2x10^-3 = 10200 s^-1 * [NH3]
R = k´ * [NH3]
k´ = 10200 s^-1
Because k´ is the psuedo first-order rate constant, we have the following:
b/(b-x) = 100/(100-40) ; 40% ammonia reacts
b/(b-x) = 1.67
log(b/(b-x)) = log(1.67)
log(b/(b-x)) = 0.22
the time will equal to:
t = (2.303/k) * log(b/(b-x)) = (2.303/10200) * (0.22) = 4.97x10^-5 s
Convert moles to mass.
mass C = 0.2 mol * 12 g / mol = 2.4 g
mass H = 0.4 mol * 1 g / mol = 0.4 g
So mass left for O = 6 g – (2.4 g + 0.4 g) = 3.2 g
Calculating for moles O given mass:
moles O = 3.2 g / (16 g / mol) = 0.2 moles
Answer:
<span>0.2 moles O</span>
Answer:
molecular weight (Mb) = 0.42 g/mol
Explanation:
mass sample (solute) (wb) = 58.125 g
mass sln = 750.0 g = mass solute + mass solvent
∴ solute (b) unknown nonelectrolyte compound
∴ solvent (a): water
⇒ mb = mol solute/Kg solvent (nb/wa)
boiling point:
- ΔT = K*mb = 100.220°C ≅ 373.22 K
∴ K water = 1.86 K.Kg/mol
⇒ Mb = ? (molecular weight) (wb/nb)
⇒ mb = ΔT / K
⇒ mb = (373.22 K) / (1.86 K.Kg/mol)
⇒ mb = 200.656 mol/Kg
∴ mass solvent = 750.0 g - 58.125 g = 691.875 g = 0.692 Kg
moles solute:
⇒ nb = (200.656 mol/Kg)*(0.692 Kg) = 138.83 mol solute
molecular weight:
⇒ Mb = (58.125 g)/(138.83 mol) = 0.42 g/mol
<h3>
Answer:</h3>
The Equilibrium would shift to produce more NO
<h3>
Explanation:</h3>
The reaction is;
N₂(g) + O₂(g) ⇆ 2NO(g)
- When a reaction is at equilibrium then the forward reaction rate will be equivalent to the reverse reaction rate. Additionally, the concentration of the reactants and products are the same.
- From Le Chatelier's principle, additional reactants favor the formation of more products while additional products favor the formation of more reactants.
- For example, when more oxygen is added then more Nitrogen (II) oxide will be formed.
- Oxygen is a reactant and when increased it favors forward reaction which leads to the formation of more NO which is the product.
