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:
the type of drink they consumed each day
<span>divide the 201g by the mol mass of the compound. Just add up the masses of the various element</span>
Answer:
c.
Explanation:
it moves in slow convection currents, hope this helps!
Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
M(Ca(NO3)2)= M(Ca) + M(N) + 6M(O)= 40.0 +14.0 +6*16.0 = 150 g/mol
15.0 g Ca(NO3)2 * 1mol/150 g = 0. 100 mol Ca(NO3)2
Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
1 mol 2 mol
0.100 mol 0.200 mol
We have 0.2 mol NO3⁻ in 300. mL=0.300 L of solution,
so
0.200 mol NO3⁻ / 0.300 L solution ≈ 0.667 mol NO3⁻ /L solution = 0.667 M
Concentration of NO3⁻ is 0.667 M.
