what is the concentration of Al in 1.24 M solution of AlBr3
Since AlBr3 is a salt which gets completely dissociated to give Al^+3 and 3 Br^- ions
So, AlBr3 ---- Al^+3 + 3Br^-
So, as 1 mole of AlBr3 produce 1 mole Al^+3 ions
So, concentration of Al^+3 ion = 1.24 M
while that for Br^- is = 3 x 1.24 = 3.72 M
Question: Calculate the mass,in grams, Of AlBr3 required to prepare exactly 250ml of a .281-M solution of AlBr3.
What is the total molar concentration of ions in a 0.90 M AL (SO4)3 solution? View Available Hint(s) 0 4.5 M O 1.8 M 428 g 16.0 g
A stock solution of 1.24 x 10-4 M dye is available in the labo- ratory. What volume of the stock solution must be transferred to the volumetric flask to prepare a 25.00 mL dilution with a concentration of 2.00 x 10^M? Perhaps the volume in question 2 is an inconvenient frac- |tional amount. What volume should the analyst dispense to simplify the preparation of the dilution?
2. A stock solution of 1.24 x 10^-4 M dye is available in the laboratory. What volume of the stock solution must be transferred to the volumetric flask to prepare a 25.00 mL dilution with a concentration of 2.00 x 10^-5 M? 3. Perhaps the volume in question 2 is an inconvenient fractional amount. What volume should the analyst dispense to simplify the preparation of the dilution?
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added to 2.50 L of solution that originally has [OH-] = 1*10^-3
Ksp(Al(OH)3 )=1.3*10^-33
Watch Video Ch15.LO3 and answer the following question about the common ion effect. Common Ion What is the concentration of Al3+ when 25 grams of Al(OH)3 is added to 2.50 L of solution that originally has [OH-] = 1 x 10-3 Ksp(Al(OH)3) = 1.3 x 10-33 2.63 x 10-M - • 1.3 x 10-30...
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What is oxidized in the following reaction? 2 Al (s) + 3 Br2 (l) → 2 AlBr3 (s) A- nothing is oxidized B- Br2 C- AlBr3 D- Al
General Chemistry II: Why is the solution formed from AlBr3 salt acidic?