Answer:
The molality of isoborneol in camphor is 0.53 mol/kg.
Explanation:
Melting point of pure camphor= T =179°C
Melting point of sample = 
= 165°C
Depression in freezing point = 
Depression in freezing point is also given by formula:
= The freezing point depression constant
m = molality of the sample
i = van't Hoff factor
We have: = 40°C kg/mol
i = 1 ( organic compounds)
The molality of isoborneol in camphor is 0.53 mol/kg.
False
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<u><em>on the basis of the reaction of heat in plastic , their are two types of plastics : </em></u>
<h2><u>
<em>Thermoplastic & Thermosetting plastics </em></u></h2>
<u><em>Thermoplastics : Plastics which easily get deformed (become soft )on heating and also get bend easily are known as thermoplastics . </em></u>
<u><em>examples : polythene , polyvinyl chloride and polystyrene .</em></u>
<u><em>thermosetting plastics : plastics which once moulded into a shape do not become soft on heating and cannot be moulded again are called thermosetting plastics .</em></u>
<u><em>examples : bakelite , melamine and formica</em></u>
Answer:
2.15 L
Explanation:
M(NaOCl) = 23.0 + 16.0 + 35.5 = 74.5 g/mol
120.3 g *1 mol/74.5 g = 1.615 mol NaOCl
0.750 M = 0.750 mol/L
1.615 mol * 1L/0.750 mol = 2.15 L
Answer:
explaination:
final answer thoughts:
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