Suppose 0.775 g of sodium iodide is dissolved in 100. mL of a 57.0 m M aqueous solution of silver nitrate. Calculate the final molarity of iodide anion in the solution. You can assume the volume of the solution doesn't change when the sodium iodide is dissolved in it. Be sure your answer has the correct number of significant digits.
Suppose 0.775 g of sodium iodide is dissolved in 100. mL of a 57.0 m M aqueous solution of silver nitrate.
Calculate the final molarity of iodide anion in the solution. You can assume the volume of the solution doesn't change when the sodium iodide is dissolved in it.
Be sure your answer has the correct number of significant digits.
The molarity of a solution is calculated by dividing the moles of solute by the volume of solution in liters.
A limiting reactant is the reactant that is available in less amount than the required amount and gets entirely consumed in a chemical reaction.
One mole of any substance is equivalent to its molar mass.
Write the balanced chemical equation for the reaction between sodium iodide (NaI) and silver nitrate (AgNO3).
One mole of sodium iodide and one mole of silver nitrate reacts to produce one mole of silver iodide (AgI) and sodium nitrate (NaNO3).
The sodium iodide mass is 0.775 g, and the volume and molarity of silver nitrate are 100.0 mL and 57.0 mM respectively.
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