AgNO3 ( aq ) + NaI ( aq ) = NaNO3 ( aq ) + AgI ( s ) .
Answer:
3.24 mol/L
Explanation:
Given that:
mass of Boron triiodide = 6664 grams
molar mass of BI_3 = 391.52 g/mol
Recall that:
number of moles = mass/molar mass
∴
number of moles = 6664 g /391.52 g/mol
number of moles = 17.02 mol
Also;
Molarity = moles for solute/liter for solution
= 17.02 mol/5.25 L
= 3.24 mol/L
Answer:
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Hope this helped.
Answer:c
Explanation: becasue if u add to it u get 5 or c
