This question is incomplete. Luckily, I found the same problem which is shown in the attached picture. To answer the question, we must know how the size and charge affect the lattice energy. The answer is: lattice energy increases with the increasing charge of the ions, and decreasing radius of the atoms.
<em>Therefore, the ranking would be: A < B < C</em>.
The required volume of water is 0.18 liters.
<h3>What is molarity?</h3>
Molarity of any solution is define as the number of moles of solute present in per liter of solution as;
M = n/V
Moles of solute will be calculated as:
n = W/M, where
W = given mass of HCl = 32g
M = molar mass of HCl = 36.4g/mol
n = 32 / 36.4 = 0.88 mole
Given molarity of solution = 4.80M
On putting all values in the above equation, we get
V = (0.88) / (36.4) = 0.18 L
Hence required volume of water is 0.18L.
To know more about volume & concentration, visit the below link:
brainly.com/question/26762947
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Answer:
I am going to chose your apple on your breakfast plate. Somebody plantes the apple seed. The apple tree takes time to grow. Once the apples are ready to be picked they are picked. Then the apples are sorted and go off to different schools like your school. Which is where you pick the apple up and put it on your plate.
Explanation:
Ionic Crystal. Ionic hydrogen.postive electron transfer to atoms
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)
