A, nonpoint souce pollution is "a main problem with water quality."
Answer:
The final volume of the sample of gas
= 0.000151 
Explanation:
Initial volume
= 200 ml = 0.0002
Initial temperature
= 296 K
Initial pressure
= 101.3 K pa
Final temperature
= 336 K
Final pressure
= K pa
Relation between P , V & T is given by

Put all the values in the above equation we get

= 0.000151 
This is the final volume of the sample of gas.
Answer:
D
Explanation:
The sample must contain impurity that is lower in atomic mass to sodium and since potassium has higher atomic mass to sodium, the answer is the sample contains NaCl and LiCl. We are sure already that the sample is not pure which rules out option a and option b contains sodium iodide which cannot contribute to the increase in chlorine
Answer:
D
Explanation:
We can use the mole ratio to calculate the partial pressure. The total number of moles is 0.2 + 0.2 + 0.1 = 0.5 moles
Now, we know that the mole fraction of the argon gas would be 0.2/0.5
The partial pressure is as follows. To calculate this, we simple multiply the number of moles by the total pressure.
0.2/0.5 * 5 = 1.0/0.5 = 2.00atm
D
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