Using value of 1.1×10−10 for the Ksp, calculate how many milligrams of Ag2CrO4 will dissolve in 13.5 mL of H2O?
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Using value of 1.1×10−10 for the Ksp, calculate how many milligrams of Ag2CrO4 will dissolve in...
Calculate the number of milligrams of Ag2CrO4 that would dissolve in 400.0mL of distilled water. Use your average Ksp from the experiment. (Average Ksp is 1.58x10^-11)
QUESTION 2 Kep(Ag2CrO4) = 9,0 x 10-12, Ksp(BaSO4) = 1.1 x 10-10, Ksp (AgSCN) - 1.2 x 10-12 What is the correct order of the solubility of the above ionic compounds? (from smallest to greatest) a. BaSO4 → AgSCN → Ag2C+04 b. AgSCN → Ag2CrO4 → BaSO4 c. AgSCN → BaSO4 → Ag2C104 d. No correct order is found. e. BaSO4 → Ag2Cr04 → AgSCN
what volume of water is needed to completely dissolve 4.0 g of Ag2CrO4 (Ksp = 8.0 x 10^-12, MM = 331.8 g/mol)
The Ksp for BaCO3 is 5.1 x 10-9. How many grams of BaCO3 will dissolve in 1000.mL of water? THÅR O ? Value | Units Submit Previous Answers Request Answer
If the Ksp of silver chromate (Ag2CrO4) is 1.1 x 10-12 and the silver ion concentration in the solution is 0.0005 M, what is the chromate ion concentration? 1.0 x 10-4 M 4.4 x 10-6 M 5.0 x 10-7 M 1.0 x 10-6 M 6.5 x 10-5 M
1a) Silver chromate, Ag2CrO4, has a Ksp of 8.99 × 10–12. Calculate the solubility in mol/L of silver chromate. 1b) Calculate the molar solubility of AgCl (Ksp = 1.6 × 10–10) in 0.0034 M sodium chloride at 25°C. 1c) If 30 mL of 5.0 × 10–4M Ca(NO3)2 are added to 70 mL of 2.0 × 10–4M NaF, will a precipitate occur? (Ksp of CaF2 = 4.0 × 10–11)
Silver phosphate, Ag3PO4, has a Ksp of 8.89x10–17 . How many milligrams of silver phosphate can be dissolved in 250 mL of pure water?
How many milligrams of silver bromide would dissolve in 1.0 liter of water?
If the Ksp for Ag2CrO4 is 7.1 x 10-12, will precipitation occur when 10 mL of 1.0 x 10-4 M AgNO3 is mixed with 10 mL of 1.0 x 10-3 M K2CrO4?
6.) (8) The Ksp for Calcium Fluoride is 1.46 x 10-10. How many milligrams of Ca2+(aq) can be present in 3.75 L of 0.25M NaF solution? (Molar mass of Ca = 40.08 g/mol) Final answer: