Answer:
The full amount (5.00 g) will be dissolved in 1 L of water at 25°C.
Explanation:
The molecular weight (MW) of Vanillin (C₈H₈O₃) is calculated from the chemical formula as follows:
MW(C₈H₈O₃) = (12 g/mol x 8) + (1 g/mol x 8) + (16 g/mol x 3) = 152 g/mol
If 0.070 mol of C₈H₈O₃ are soluble per liter of water at 25°C, the maximum mass that can be dissolved in 1 L is:
0.070 mol x 152 g/mol = 10.64 g
Since 5.00 g is lesser than the maximum amount that can be dissolved (10.64 g), the added amount will be completely dissolved in 1 L of water at 25°C.
vaporisation, Freezing, Condensation
Answer:
106 mL
Explanation:
In order to be able to answer this question, you must understand what the density of a substance tells you.
The density of a substance is nothing more than the mass of that substance that occupies one unit of volume.
In your case, the density of ethanol is given in Grams per milliliter, which means that one unit of volume will be
1 mL
.
So, ethanol has a density of
0.785 g mL
−
1
, which is equivalent to saying that if you take exactly
1 mL
of ethanol and weigh it, you will end up with a mass of
0.785 g
.
Now, you know that the volume you're using has a mass of
83.3 g
. Well, if you get
0.785 g
for every
1 mL
of ethanol, it follows that this much mass will correspond to a volume of
83.3
g ethanol
⋅
ethanol's density
1 mL
0.785
g ethanol
=
106.11 mL
Rounded to three sig figs, the answer will be
V
ethanol
=
106 mL
Hope this helps
Amines are organic compound obtained from ammonia by replacement of one or more hydrogen atoms by organic groups. If one of the 'H' atom of NH3 is replace by R = alkyl group, they are referred as primary amine. If two 'H' atoms of NH3 are replace by same or different alkyl group, they are referred as secondary amine.
