Looks correct but the second to last I would of put abiotic and biotic factors but I don’t know what’s right for you
Using a more concentrated HCl solution and Crushing the CaCO₃ into a fine powder makes the reaction to occur at a faster rate.
<u>Explanation:</u>
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(aq) + CO₂(g)
When calcium carbonate reacts with hydrochloric acid, it gives out carbon-dioxide in the form of bubbles and there is a formation of calcium chloride in aqueous medium.
The rate of the reaction can be increased by
- Using a more concentrated HCl solution
- Crushing the CaCO₃ into a fine powder
When concentrated acid is used instead of dilute acid then the reaction will occur at a faster rate.
When CaCO₃ is crushed into a fine powder then the surface area will increases thereby increasing the rate of the reaction.
Answer:
9.93
Explanation:
Your value for Kw is incorrect. The correct value is 5.48 × 10^-14.
pH + pOH = pKw
3.30 + pOH = -log(5.84 × 10^-14) = 13.23
pOH = 13.23 - 3.30 = 9.93
The pOH of the solution is 9.93.
Answer:
Explanation:
1. Mass of acetylsalicylic acid (ASA)
2. Moles of ASA
HC₉H₇O₄ =180.16 g/mol
3. Concentration of ASA
4. Set up an ICE table
5. Solve for x
![K_{\text{a}} = \dfrac{\text{[H}_{3}\text{O}^{+}]\text{A}^{-}]} {\text{[HA]}} = 3.33 \times 10^{-4}\\\\\dfrac{x^{2}}{0.01757 - x} = 3.33 \times 10^{-4}\\\\\textbf{Check that }\mathbf{x \ll 0.01757}\\\\\dfrac{ 0.01757 }{3.33 \times 10^{-4}} = 53 < 400\\\\\text{The ratio is less than 400. We must solve a quadratic equation.}\\\\x^{2} = 3.33 \times 10^{-4}(0.01757 - x) \\\\x^{2} = 5.851 \times 10^{-6} - 3.33 \times 10^{-4}x\\\\x^{2} + 3.33 \times 10^{-4}x - 5.851 \times 10^{-6} = 0](https://tex.z-dn.net/?f=K_%7B%5Ctext%7Ba%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BH%7D_%7B3%7D%5Ctext%7BO%7D%5E%7B%2B%7D%5D%5Ctext%7BA%7D%5E%7B-%7D%5D%7D%20%7B%5Ctext%7B%5BHA%5D%7D%7D%20%3D%203.33%20%5Ctimes%2010%5E%7B-4%7D%5C%5C%5C%5C%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.01757%20-%20x%7D%20%3D%203.33%20%5Ctimes%2010%5E%7B-4%7D%5C%5C%5C%5C%5Ctextbf%7BCheck%20that%20%7D%5Cmathbf%7Bx%20%5Cll%200.01757%7D%5C%5C%5C%5C%5Cdfrac%7B%200.01757%20%7D%7B3.33%20%5Ctimes%2010%5E%7B-4%7D%7D%20%3D%2053%20%3C%20400%5C%5C%5C%5C%5Ctext%7BThe%20ratio%20is%20less%20than%20400.%20We%20must%20solve%20a%20quadratic%20equation.%7D%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%203.33%20%5Ctimes%2010%5E%7B-4%7D%280.01757%20-%20x%29%20%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%205.851%20%5Ctimes%2010%5E%7B-6%7D%20-%203.33%20%5Ctimes%2010%5E%7B-4%7Dx%5C%5C%5C%5Cx%5E%7B2%7D%20%2B%203.33%20%5Ctimes%2010%5E%7B-4%7Dx%20-%205.851%20%5Ctimes%2010%5E%7B-6%7D%20%3D%200)
6. Solve the quadratic equation.
7. Calculate the pH
![\rm [H_{3}O^{+}]= x \, mol \cdot L^{-1} = 0.002258 \, mol \cdot L^{-1}\\\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{0.002258} = \mathbf{2.65}\\\text{The pH of the solution is } \boxed{\textbf{2.65}}](https://tex.z-dn.net/?f=%5Crm%20%5BH_%7B3%7DO%5E%7B%2B%7D%5D%3D%20x%20%5C%2C%20mol%20%5Ccdot%20L%5E%7B-1%7D%20%3D%200.002258%20%5C%2C%20mol%20%5Ccdot%20L%5E%7B-1%7D%5C%5C%5Ctext%7BpH%7D%20%3D%20-%5Clog%7B%5Crm%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%20%3D%20-%5Clog%7B0.002258%7D%20%3D%20%5Cmathbf%7B2.65%7D%5C%5C%5Ctext%7BThe%20pH%20of%20the%20solution%20is%20%7D%20%5Cboxed%7B%5Ctextbf%7B2.65%7D%7D)
Lithium-Chloride, Carbon-oxide, Barium Bromide, Ferrous Iodide, and Ammonium Chloride. Did this help?
