Chapter 4 Homework 1. The compound aluminum acetate is a strong electrolyte. Wri

Question

Chapter 4 Homework 1. The compound aluminum acetate is a strong electrolyte. Write the reaction when solid aluminum acetate is put into water 2. The e compound chromium (IlI) ulfate is a strong electrolyte. Write the reaction when solid chromium (IIl) sulfate is put into water. 3 The compound calcium acetate is a strong electrolyte. Write the reaction when solid calcium acetate is put into water You need to make an aqueous solution of 0.121 M zinc chloride for an experiment in lab, using a 300 mL volumetric flask. How much solid zinc chloride should you add in 4. S. How many militers of an aqueous solution of 0.146 M copper () iodide is needed to 6. In the laboratory you dissolve 18.1 g of magnesium nitrate in a volumetric flask and add 7. In the laboratory you dissolve 12.1 g of zinc chloride in a volumetric flask and add water grams? obtain 6.83 grams of the salt? water to a total volume of 125 mL. What is the molarity of the solution? to a total volume of 375 mL. a. b. c. What is the molarity of the solution? What is the concentration of the zinc cation? What is the concentration of the chloride anion? 8. In the laboratory you dissolve 14.6 g of copper (l) iodide in a volumetric flask and add water to a total volume of 250 ml a. What is the molarity of the solution? b. What c. What is the concentration of the iodide anion? is the concentration of the copper (I) cation? 9. In the laboratory you dissolve 18.1 g of magnesium nitrate in a volumetric flask and add water to a total volume of 125 mL a. What is the molarity of the solution? b. What is the concentration of the magnesium cation? c. What is the concentration of the nitrate anion? to a total volume of 50.0 mL. What is the concentration of the dilute solution? acid. How much concentrated acid must you add to obtain a total volume of 100 mL of 10. In the laboratory you dilute 4.47 ml of a concentrated 6.00 M hydrobromic acid solution 11. You wish to make a 0.201 M nitric acid solution from a stock solution of 3.00 M nitric 12. In the laboratory you dilute 2.48 ml of a concentrated 12.0 M perchloric acid solution to 13. In the laboratory, a student dilutes 11.1 mL of a 7.20 M nitric acid solution to a total 14. How many milliters of 8.81 M perchloric acid solution should be used to prepare 3.50 L 15. In the laboratory, a student adds 41.5 mL of water to 25.6 mL of a 0.825 M hydroiodic the dilute solution? a total volume of 75.0 mL. What is the concentration of the dilute solution? volume of 100.0 mL. What is the concentration of the diluted solution? of 0.600 M HCIO4? acid solution. What is the concentration of the diluted solution? Assume the volumes are additive. ons

Explanation / Answer

1. The dissolution equation is basically the split of the compound into its ions with their respective charges. Aluminum acetate, with formula Al(CH3COO)3, will split into 1 Al ion and 3 acetate ions. We'll use the single arrow because it is a strong electrolyte:

Al(CH3COO)3 (s) Al3+ + 3CH3COO-

2. In this case, the formula of Chromium (III) sulfate is Cr2(SO4)3, so when disolved, it will split into 2 Cr(III) ions and 3 sulfate ions:

Cr2(SO4)3 (s) 2Cr3+ + 3SO42-

3. The formula of calcium acetate is Ca(CH3COO)2, so when dissolved in water, it will split into 1 Ca ion and 2 acetate ions:

Ca(CH3COO)2 (s) Ca2+ + 2CH3COO-

4. We need to make 300mL (0.3L) of 0.121M ZnCl2 solution (molar mass of ZnCl2 = 136.29 g/mol).

Use the formula:

M = n/L

Where M = molarity, n = moles and L = liters of solution.

We can expand the formula knowing that n = g/MM (g = grams, MM = molar mass).

M = (g/MM)/L

Rearrange:

M = g/MM*L

Solve for g:

g = M*MM*L

Substitute the values:

g = (0.121 mol/L)*(136.29 g/mol)*(0.3L)

g = 4.95 g ZnCl2

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