Explanation:
It is given that volume of is 300 mL and molarity is 0.200 M.
Volume of NaCl is 200 mL and molarity is 0.050 M.
The chemical reaction will be as follows.
for is given as .
As, molarity is number of moles present in liter of solution.
Hence, moles of (aq) will be calculated as follows.
moles of (aq) =
= 0.06 mol
=
= 0.120 M
Mole of =
= 0.010 M
Now, Q =
=
=
As, Q < hence, there will be no formation of precipitate.
I'm taking the test right now; I'll edit this when I find out if I was right! (I chose "solvent" but I wanna be sure.)
Edit: Solvent is the correct answer!
Salinity has units of grams NaCl or salt per kilogram solution. We can use the density given and the molar mass of the salt to convert from salinity to molarity. We do as follows:
( 5.6 g / kg ) ( 1.03 kg / L ) ( 1 mol / 58.44 g ) = 0.0987 mol NaCl / L
The answer is either A or C. Sorry I'm not much help :)
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