For this problem, we use Graham's Effusion Law to find out the rate of effusion of chlorine gas. The formula is as follows:
R₁/R₂ = √(M₂/M₁)
Let 1 be N₂ while 2 be Cl₂
255/R₂ = √(28/70.8)
Solving for R₂,
R₂ = 405.5 s
<em>Thus, it would take 405.5 s to effuse chlorine gas.</em>
Answer:
k = [F2]² [PO]² / [P2] [F2O]²
Explanation:
In a chemical equilibrium, the equilibrium constant expression is written as the ratio between the molar concentration of the products over the molar concentration of the reactants. Each species powered to its reaction coefficient. For the equilibrium:
P2(g) + 2F2O(g) ⇄ 2PO(g) + 2F2(g)
The equilibrium constant, k, is:
k = [F2]² [PO]² / [P2] [F2O]²
Answer:
105.8 g of Na would be required
Explanation:
Let's think the reaction:
2Na(s) + Cl₂(g) → 2NaCl (s)
1 mol of chlorine reacts with 2 moles of sodium
Then, 2.3 moles of Cl₂ would react with (2.3 .2) / 1 = 4.6 moles
Let's determine the mass of them.
4.6 mol . 23 g/mol = 105.8 g
382.85 Celsius is the temperature does 0.750 moles of an ideal gas occupy a volume of 35.9 L at 114 kPa.
Explanation:
Given data:
number of moles of the gas = 0.75 moles
volume of the gas = 35.9 liters
pressure of the gas = 114 KPa or 1.125 atm
R = 0.0821 latm/moleK
temperature of the gas T = ?
The equation used to calculate temperature from above data is ideal gas law equation.
the equation is :
PV = nRT
T = 
Putting the values in the above rewritten equation:
T = 
T = 655.9 K
To convert kelvin into celsius, formula used is
K = 273.15+ C
putting the values in the equation
C = 656 - 273.15
= 382.85 Celsius
Answer:
pH = 9.48
Explanation:
We have first to realize that NH₃ is a weak base:
NH₃ + H₂O ⇔ NH₄⁺ + OH⁻ Kb = 1.8 x 10⁻⁵
and we are adding this weak base to a solution of NH₄NO₃ which being a salt dissociates 100 % in water.
Effectively what we have here is a buffer of a weak base and its conjugate acid. Therefore, we need the Henderson-Hasselbach formula for weak bases given by:
pOH = pKb + log ( [ conjugate acid ] / [ weak base ]
mol NH₃ = 0.139 L x 0.39 M = 0.054 mol
mol NH₄⁺ = 0.169 L x 0.19 M = 0.032 mol
Now we have all the information required to calculate the pOH ( Note that we dont have to calculate the concentrations since in the formula they are a ratio and the volume will cancel out)
pOH = -log(1.8 x 10⁻⁵) + log ( 0.032/0.054) = 4.52
pOH + pH = 14 ⇒ pH = 14 - 4.52 = 9.48
The solution is basic which agrees with NH₃ being a weak base.
