Answer:
[H₃O⁺] = 1.4 × 10⁻⁹ M.
Explanation:
NH₄Cl is a salt that dissolves well in water. The 2.5 M NH₄Cl will give an initial NH₄⁺ concentration of 2.5 M.
NH₃ is a weak base. It combines with water to produce NH₄⁺ and OH⁻. The opposite process can also take place. NH₄⁺ combines with OH⁻ to produce NH₃ and H₂O. The final H₃O⁺ concentration can be found from the OH⁻ concentration. What will be the final OH⁻ concentration?
Let the increase in OH⁻ concentration be x. The initial OH⁻ concentration at room temperature is 10⁻⁷ M.
Construct a RICE table for the equilibrium between NH₃ and NH₄⁺:
.
The 
value for ammonia is small. The value of x will be so small that at equilibrium, 
and 
.
![\displaystyle \text{K}_b = \frac{[{\text{NH}_4}^{+}]\cdot [{\text{OH}}^{-}]}{[\text{NH}_3]} \approx \frac{2.5\;(x + 1.0\times 10^{-7})}{1.0}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ctext%7BK%7D_b%20%3D%20%5Cfrac%7B%5B%7B%5Ctext%7BNH%7D_4%7D%5E%7B%2B%7D%5D%5Ccdot%20%5B%7B%5Ctext%7BOH%7D%7D%5E%7B-%7D%5D%7D%7B%5B%5Ctext%7BNH%7D_3%5D%7D%20%5Capprox%20%5Cfrac%7B2.5%5C%3B%28x%20%2B%201.0%5Ctimes%2010%5E%7B-7%7D%29%7D%7B1.0%7D)
.
.
![\displaystyle [\text{OH}^{-}] = x+1.0\times 10^{-7} = 1.8\times 10^{-5} /\left(\frac{2.5}{1.0}\right) = 7.2\times 10^{-6}\;\text{mol}\cdot\text{L}^{-}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%20x%2B1.0%5Ctimes%2010%5E%7B-7%7D%20%3D%201.8%5Ctimes%2010%5E%7B-5%7D%20%2F%5Cleft%28%5Cfrac%7B2.5%7D%7B1.0%7D%5Cright%29%20%3D%207.2%5Ctimes%2010%5E%7B-6%7D%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7BL%7D%5E%7B-%7D)
.
Again, 
at room temperature.
![\displaystyle [\text{H}_3\text{O}^{+}] = \frac{\text{K}_w}{[\text{OH}^{-}]}=\frac{1.0\times 10^{-14}}{7.2\times 10^{-6}} = 1.4\times 10^{-9} \;\text{mol}\cdot\text{L}^{-1}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5B%5Ctext%7BH%7D_3%5Ctext%7BO%7D%5E%7B%2B%7D%5D%20%3D%20%5Cfrac%7B%5Ctext%7BK%7D_w%7D%7B%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%7D%3D%5Cfrac%7B1.0%5Ctimes%2010%5E%7B-14%7D%7D%7B7.2%5Ctimes%2010%5E%7B-6%7D%7D%20%3D%201.4%5Ctimes%2010%5E%7B-9%7D%20%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7BL%7D%5E%7B-1%7D)
Answer:
132.17 g
Explanation:
The reaction given , in the question is -
CH₄ (g ) + 4 S ( g ) ---> CS₂ ( g ) + 2H₂S ( g )
From the reaction , 4 mole of S is required for the production of 1 mole of CS₂ .
since ,
Moles of CS₂ = given mass of CS₂ / Molecular weight of CS₂
Since ,
the Molecular weight of CS₂ = 76
Given , mass of CS₂ = 72.57 g
Moles of CS₂ = 72.57 / 76 = 0.95 mol
Since ,
The yield is 92.0 % .
Moles of S required = 4 * 0.95 mol / 0.92 = 4.13 moles
Mass of S required = 4.13 * 32 = 132.17 g .
Answer: there is nothing there to answer
Explanation:
The empirical formula for the compound is - P₂O₅
the empirical formula is the simplest ratio of whole numbers of components in a compound.
molecular formula is the actual ratio of components in a compound.
we have to first find the number of empirical units in the molecular formula
molecular mass - 283.89 g/mol
mass of empirical formula - 283.8 g
number of empirical units - 283.89 g/mol / 283.8 g
number of empirical units - 1.000
therefore empirical formula = molecular formula
molecular formula - P₂O₅
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