Question

gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.(a) BaSiF6, 0.026 g/100 mL (contains SiF6 2− ions)(b) Ce(IO3)4, 1.5 × 10–2 g/100 mL(c) Gd2(SO4)3, 3.98 g/100 mL(d) (NH4)2PtBr6, 0.59 g/100 mL (contains PtBr6 2− ions)

Name: gives solubilities of the following compounds in grams per 100 ml of water because these compounds are only slightly soluble assume that the volume does not change on dissolution and calcula 17198
Uploaded: 2023-09-10T21:04:59-08:00
Duration: 2 min 13 s
Channel: Adi S
Description: gives solubilities of the following compounds in grams per 100 ml of water because these compounds are only slightly soluble assume that the volume does not change on dissolution and calcula 17198

          gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.(a) BaSiF6, 0.026 g/100 mL (contains SiF6 2− ions)(b) Ce(IO3)4, 1.5 × 10–2 g/100 mL(c) Gd2(SO4)3, 3.98 g/100 mL(d) (NH4)2PtBr6, 0.59 g/100 mL (contains PtBr6 2− ions)

Added by Karen F.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Adi S

09/10/2023

Step 1

026 g/100 mL, we need to convert it to moles per liter (M). Molar mass of BaSiF6 = 137.33 (Ba) + 28.09 (Si) + 6 * 19.00 (F) = 379.45 g/mol Solubility in moles per liter = (0.026 g/100 mL) * (1 mol/379.45 g) * (1000 mL/1 L) = 6.85 * 10^-4 M Now, we can write the Show more…

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gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.(a) BaSiF6, 0.026 g/100 mL (contains SiF6 2− ions)(b) Ce(IO3)4, 1.5 × 10–2 g/100 mL(c) Gd2(SO4)3, 3.98 g/100 mL(d) (NH4)2PtBr6, 0.59 g/100 mL (contains PtBr6 2− ions)