Answer:
Molar solubility of AgBr = 51.33 × 10⁻¹³
Explanation:
Given:
Amount of NaBr = 0.150 M
Ksp (AgBr) = 7.7 × 10⁻¹³
Find:
Molar solubility of AgBr
Computation:
Molar solubility of AgBr = Ksp (AgBr) / Amount of NaBr
Molar solubility of AgBr = 7.7 × 10⁻¹³ / 0.150
Molar solubility of AgBr = 51.33 × 10⁻¹³
Basically it means carbon dioxide is removed from the atmosphere and is trapped in e.g. calcium carbonate (like shells), rocks, oceans etc.
Answer: I remember looking out of the car as we pulled up to the Frantz school.”
“I thought maybe it was Mardi Gras.”
“As we walked through the crowd, I didn’t see any faces.”
“[The school] looked bigger and nicer than my old school.”
“The crowd behind us made me think this was an important place.”
“It must be college, I thought to myself.
Explanation: