I: Current
V: Voltage
R: resistance
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Al(NO3)3(aq) + 3NaOH(s) --> Al(OH)3 (s) + 3NaNO3 (aq)
The precipitate here is Al(OH)3 (s), since the solid reactant is the precipitate in the aqueous solution. Usually, it is okay to assume in basic chemistry that the transition metal is going to be part of the compound that is the precipitate, especially in an acidic salt and a strong base reaction that we have here.
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<span>The density of the solution =1.05 g/ml.
</span><span>The total mass of the resulting solution is = 398.7 g (CaCl2 + water)
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Find moles of CaCl2 and water.
Molar mass of CaCl2 = 110 (approx.)
Moles of CaCl2 = 23.7 / 110 = 0.22
so, moles of Cl- ion = 2 x 0.22 = 0.44 (because each molecule of CaCl2 will give two Cl- ions)
Moles of water = 375 / 18 = 20.83
Now, Mole fraction of CaCl2 = (moles of CaCl2) / (total moles)
total moles = moles of Cl- ions + moles of Ca2+ ions + moles of water
= 0.44 + 0.22 + 20.83
=21.49
So, mole fraction = 0.44 / (21.49) = 0.02
Guess what !!! density is not used. No need
