A, nonpoint souce pollution is "a main problem with water quality."
Answer:
1.8 × 10⁻¹⁶ mol
Explanation:
(a) Calculate the solubility of the Sr₃(PO₄)₂
Let s = the solubility of Sr₃(PO₄)₂.
The equation for the equilibrium is
Sr₃(PO₄)₂(s) ⇌ 3Sr²⁺(aq) + 2PO₄³⁻(aq); Ksp = 1.0 × 10⁻³¹
1.2 + 3s 2s
![K_{sp} =\text{[Sr$^{2+}$]$^{3}$[PO$_{4}^{3-}$]$^{2}$} = (1.2 + 3s)^{3}\times (2s)^{2} = 1.0 \times 10^{-31}\\\text{Assume } 3s \ll 1.2\\1.2^{3} \times 4s^{2} = 1.0 \times 10^{-31}\\6.91s^{2} = 1.0 \times 10^{-31}\\s^{2} = \dfrac{1.0 \times 10^{-31}}{6.91} = 1.45 \times 10^{-32}\\\\s = \sqrt{ 1.45 \times 10^{-32}} = 1.20 \times 10^{-16} \text{ mol/L}\\](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5Ctext%7B%5BSr%24%5E%7B2%2B%7D%24%5D%24%5E%7B3%7D%24%5BPO%24_%7B4%7D%5E%7B3-%7D%24%5D%24%5E%7B2%7D%24%7D%20%3D%20%281.2%20%2B%203s%29%5E%7B3%7D%5Ctimes%20%282s%29%5E%7B2%7D%20%3D%20%201.0%20%5Ctimes%2010%5E%7B-31%7D%5C%5C%5Ctext%7BAssume%20%7D%203s%20%5Cll%201.2%5C%5C1.2%5E%7B3%7D%20%5Ctimes%204s%5E%7B2%7D%20%3D%201.0%20%5Ctimes%2010%5E%7B-31%7D%5C%5C6.91s%5E%7B2%7D%20%3D%201.0%20%5Ctimes%2010%5E%7B-31%7D%5C%5Cs%5E%7B2%7D%20%3D%20%5Cdfrac%7B1.0%20%5Ctimes%2010%5E%7B-31%7D%7D%7B6.91%7D%20%3D%201.45%20%5Ctimes%2010%5E%7B-32%7D%5C%5C%5C%5Cs%20%3D%20%5Csqrt%7B%201.45%20%5Ctimes%2010%5E%7B-32%7D%7D%20%3D%201.20%20%5Ctimes%2010%5E%7B-16%7D%20%5Ctext%7B%20mol%2FL%7D%5C%5C)
(b) Concentration of PO₄³⁻
[PO₄³⁻] = 2s = 2 × 1.20× 10⁻¹⁶ mol·L⁻¹ = 2.41× 10⁻¹⁶ mol·L⁻¹
(c) Moles of PO₄³⁻
Moles = 0.750 L × 2.41 × 10⁻¹⁶ mol·L⁻¹ = 1.8 × 10⁻¹⁶ mol
Answer:
110ml
Explanation:
<em>Using the dilution equation, C1V1 = C2V2</em>
<em>Where C1 is the initial concentration of solution</em>
<em>C2 is final concentration of solution</em>
<em>V1 is intital volume of solution</em>
<em>V2 is final volume of solution.</em>
From the question , C1=6M, C2=0.5M, V1=10ml, V2=?



volume of water added = final volume -initial volume
= 120-10
=110ml
Answer:
It is required answer.
Explanation:
Given that :
1. using balanced chemical equation:
ammonium acetate:
The balanced equation is:
NH₃ + H₂O ===> NH₄OH
when ammonia gas dissolves in water then we get the base in form of ammonium hydroxide.
When NH₄OH reacts with CH₃COOH then we get ammonium acetate and water
NH₄OH + CH₃COOH ===> [CH₃COO]- & NH₄+ & H₂O
So, we can say that,
when we are adding an acid and a base together then we get the product of H₂O and given elements.
2. addition of barium hydroxide to sulfuric acid:
the balanced equation is
H₂SO4+ Ba(OH)₂--> BaSO₄+ 2H₂O
when acid and base reacts together than we get barium sulphate and water
when sulfuric acid and barium hydroxide.
Hence, it is required answer.